{"id":22651,"date":"2024-01-24T09:39:44","date_gmt":"2024-01-24T04:09:44","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22651"},"modified":"2024-01-25T14:31:16","modified_gmt":"2024-01-25T09:01:16","slug":"ap-inter-2nd-year-physics-study-material-chapter-4","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-inter-2nd-year-physics-study-material-chapter-4\/","title":{"rendered":"AP Inter 2nd Year Physics Study Material Chapter 4 Electric Charges and Fields"},"content":{"rendered":"

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material<\/a> 4th Lesson Electric Charges and Fields Textbook Questions and Answers.<\/p>\n

AP Inter 2nd Year Physics Study Material 4th Lesson Electric Charges and Fields<\/h2>\n

Very Short Answer Questions<\/span><\/p>\n

Question 1.
\nWhat is meant by the statement ‘charge is quantized’?
\nAnswer:
\nThe minimum charge that can be transferred from one body to the other is equal to the charge of the electron (e = 1.602 \u00d7 10-19<\/sup>C). A charge always exists as an integral multiple of the charge of the electron (q = ne). Therefore charge is said to be quantized.<\/p>\n

Question 2.
\nRepulsion is the sure test of charging than attraction. Why?
\nAnswer:
\nA charged body may attract a neutral body and also an oppositely charged body. But it always repels a like-charged body. Hence repulsion is the sure test of electrification.<\/p>\n

\"AP<\/p>\n

Question 3.
\nHow many electrons constitute 1 C of charge ?
\nAnswer:
\nn = \\(\\frac{q}{e}\\) = \\(\\frac{1}{1.6 \\times 10^{-19}}\\) = 6.25 \u00d7 1018<\/sup> electrons<\/p>\n

Question 4.
\nWhat happens to the weight of a body when it is charged positively ?
\nAnswer:
\nWhen a body positively charged it must loose some electrons. Hence weight of the body will decrease.<\/p>\n

Question 5.
\nWhat happens to the force between two charges if the distance between them is
\na) halved
\nb) doubled ?
\nAnswer:
\nFrom Coulombs law, F \u221d \\(\\frac{1}{\\mathrm{~d}^2}\\), so
\na) When distance is reduced to half, force increases by four times.
\n\"AP
\nb) When distance is doubled, then force is reduced by four times.
\n\"AP<\/p>\n

Question 6.
\nThe electric lines of force do not intersect. Why ?
\nAnswer:
\nThey do not intersect because if they intersect, at the point of intersection, intensity of electric field must act in two different directions, which is impossible.<\/p>\n

Question 7.
\nConsider two charges + q and -q placed at B and C of an equilateral triangle ABC. For this system, the total charge is zero. But the electric field (intensity) at A which is equidistant from B and C is not zero. Why ?
\nAnswer:
\nCharges are scalars, but the .electrical intensities are vectors and add vectorially.<\/p>\n

\"AP<\/p>\n

Question 8.
\nElectrostatic field lines of force do not form closed loops. If they form closed loops then the work done in moving a charge along a closed path will not be zero. From the above two statements can you guess the nature of electrostatic force ?
\nAnswer:
\nIt is conservative force.<\/p>\n

Question 9.
\nState Gauss’s law in electrostatics.
\nAnswer:
\nGauss’s law : It states that “the total electric flux through any closed surface is equal to – \\(\\frac{1}{\\varepsilon_0}\\) times net charge enclosed by the surface”.
\n\\(\\oint \\overrightarrow{\\mathrm{E}} \\cdot \\overrightarrow{\\mathrm{ds}}\\) = \\(\\frac{\\mathrm{q}}{\\varepsilon_0}\\)<\/p>\n

Question 10.
\nWhen is the electric flux negative and when is it positive ?
\nAnswer:
\nElectric flux \u03d5 = \\(\\vec{E} \\cdot \\vec{A}\\). If angle between \\(\\overrightarrow{\\mathrm{E}}\\) and \\(\\overrightarrow{\\mathrm{A}}\\) is 180\u00b0, then flux will have a ‘-ve’ sign. We consider the flux flowing out of the surface as positive and flux entering into the surface as negative.<\/p>\n

Question 11.
\nWrite the expression for electric intensity due to an infinite long charged wire at a distance radial distance r from the wire.
\nAnswer:
\nThe electric intensity due to an infinitely long charged wire E = \\(\\frac{\\lambda}{2 \\pi \\varepsilon_0 r}\\) the conductor.
\nWhere \u03bb = Uniform linear charge density
\nr = Distance of the point from the conductor.<\/p>\n

Question 12.
\nWrite the expression for electric intensity due to an infinite plane sheet of charge.
\nAnswer:
\nThe electric intensity due to an infinite plane sheet of charge is E = \\(\\frac{\\sigma}{2 \\varepsilon_0}\\).<\/p>\n

Question 13.
\nWrite the expression for electric intensity due to a charged conducting spherical shell at points outside and inside the shell.
\nAnswer:
\na) Intensity of electric field at any point inside a spherical shell is zero.
\nb) Intensity of electric field at any point outside a uniformly charged spherical shell is
\nE = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\cdot \\frac{q}{r^2}\\)<\/p>\n

Short Answer Questions<\/span><\/p>\n

Question 1.
\nState and explain Coulomb’s inverse square law in electricity.
\nAnswer:
\nCoulomb\u2019s law – Statement: Force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. The force acts along the straight line joining the two charges.
\nExplanation : Let us consider two charges q1<\/sub> and q2<\/sub> be separated by a distance r.
\nThen F \u221d q1<\/sub>q2<\/sub> and F \u221d \\(\\frac{1}{\\mathrm{r}^2}\\) or F \u221d \\(\\frac{q_1 q_2}{r^2}\\)
\n\"AP
\n\u2234 F = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\cdot \\frac{q_1 q_2}{r^2}\\) where \\(\\frac{1}{4 \\pi \\varepsilon_0}\\) = 9 \u00d7 109<\/sup> Nm2<\/sup>C-2<\/sup>
\nIn vector form, in free space \\(\\overrightarrow{\\mathrm{F}}\\) = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\frac{\\mathrm{q}_1 \\mathrm{q}_2}{\\mathrm{r}^2} \\hat{\\mathrm{r}}\\). Here \\(\\hat{\\mathrm{r}}\\) is a unit vector.
\n\u03b50<\/sub> is called permittivity of free space.
\n\u03b50<\/sub> = 8.85 \u00d7 10-12<\/sup> C2<\/sup>\/N-m2<\/sup> or Farad\/meter.
\nWhere \u03b5 is called permittivity of the medium.<\/p>\n

\"AP<\/p>\n

Question 2.
\nDefine intensity of electric field at a point. Derive an expression for the intensity due to a point charge.
\nAnswer:
\nIntensity of electric field (E) : Intensity of electric field at any point in an electric field is defined as the force experienced by a unit positive charge placed at that point.
\nExpression :
\n1) Intensity of electric field is a vector. It’s direction is along the direction’ of motion of positive charge.
\n2) Consider point charge q. Electric field will exist around that charge. Consider any point P in that electric field at a distance r from the given charge. A test charge q0<\/sub> is placed at R
\n3) Force acting on q0<\/sub> due to q is F = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\cdot \\frac{q q_0}{r^2}\\)
\n4) Intensity of electric field at that point is equal to the force experienced by a test charge q0<\/sub>.
\nIntensity of electric field, E = \\(\\frac{\\mathrm{F}}{\\mathrm{q}_0}\\)
\n\"AP
\nE = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\cdot \\frac{q}{r^2}\\) N\/C or V\/m<\/p>\n

Question 3.
\nDerive the equation for the couple acting on a electric dipole in a uniform electric field.
\nAnswer:
\n1) A pair of opposite charges separated by a small distance is called dipole.
\n2) Consider the charge of dipole are -q and +q coulomb and the distance between them is 2a.
\n3) Then the electric dipole moment P is given by P = q \u00d7 2a = 2aq. It is a vector. It’s direction is from -q to +q along the axis of dipole.
\n4) It is placed in a uniform electric field E, making an angle 0 with field direction as shown in fig.
\n5) Due to electric field force on +q is F = +qE and force on -q is F = -qE.
\n\"AP
\n6) These two equal and opposite charges constitute torque or moment of couple.
\ni. e., torque, \\(\\tau\\) = \u22a5r<\/sup> distance \u00d7 magnitude of one of force
\n\u2234 \\(\\tau\\) = (2a sin \u03b8)qE = 2aqE sin \u03b8 = PE sin \u03b8
\nIn vector form, \\(\\vec{\\tau}\\) = \\(\\overrightarrow{\\mathrm{P}}\\) \u00d7 \\(\\overrightarrow{\\mathrm{E}}\\)<\/p>\n

Question 4.
\nDerive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.
\nAnswer:
\nElectric field at a point on the axis of a dipole :
\n1) Consider an electric dipole consisting of two charges -q and +q separated by a distance ‘2a’ with centre ‘O’.
\n\"AP
\n2) We shall calculate electric field E at point P on the axial line of dipole, and at a distance OP = r.
\n3) Let E1<\/sub> and E2<\/sub> be the intensities of electric field at P due to charges +q and -q respectively.
\n4)
\n\"AP
\nThe resultant intensity at P is E = E1<\/sub> – E2<\/sub> [\u2235 They are opposite and E1<\/sub> > E2<\/sub>]
\n\"AP
\nIf r > > a then a2<\/sup> can be neglected in comparision to r2<\/sup>.
\n\"AP
\nIn vector form, \\(\\overrightarrow{\\mathrm{E}}\\) = \\(\\frac{2 \\overrightarrow{\\mathrm{P}}}{4 \\pi \\varepsilon_0 \\mathrm{r}^3}\\)<\/p>\n

\"AP<\/p>\n

Question 5.
\nDerive an expression for the intensity of the electric field at a point on the equatorial plane of an electric dipole. (A.P. Mar. ’19, ’15)
\nAnswer:
\nElectric field intensity on equitorial line of electric dipole:
\n1) Consider an electric dipole consisting of two charges -q and +q separated by a distance ‘2a’ with centre at ‘O’.
\n2) We shall calculate electric field E at P on equitorial line of dipole and at a distance OP = r.
\n\"AP
\n3) Let E1<\/sub> and E2<\/sub> be the electric fields at P due to charges +q and -q respectively.
\n4) The \u22a5r<\/sup> components (E1<\/sub> sin \u03b8 and E2<\/sub> sin \u03b8) cancel each other because they are equal and opposite. The ||el<\/sup> components (E1<\/sub> cos \u03b8 and E2<\/sub> cos \u03b8) are in the same direction and hence add up.
\n5) The resultant field intensity at point P is given by E = E1<\/sub> cos \u03b8 + E2<\/sub> cos \u03b8
\n\"AP
\n6) From figure, cos \u03b8 = \\(\\frac{a}{\\left(r^2+a^2\\right)^{1 \/ 2}}\\)
\n\u2234 E = \\(\\frac{\\mathrm{P}}{4 \\pi \\varepsilon_0} \\times \\frac{1}{\\mathrm{r}^3}\\)
\n7) If r >> a, then a2 <\/sup>can be neglected in comparison to r2<\/sup>. Then
\nE = \\(\\frac{\\mathrm{P}}{4 \\pi \\varepsilon_0} \\times \\frac{1}{\\mathrm{r}^3}\\)
\nIn vector form \\(\\overrightarrow{\\mathrm{E}}\\) = \\(\\frac{\\overrightarrow{\\mathrm{P}}}{4 \\pi \\varepsilon_0 \\mathrm{r}^3}\\)<\/p>\n

Question 6.
\nState Gauss’s law in electrostatics and explain its importance.
\nAnswer:
\nGauss’s law : The total-electric flux through any closed surface is equal to \\(\\frac{1}{\\varepsilon_0}\\) times the net charge enclosed by the surface.
\nTotal electric flux,
\n\"AP
\nHere q is the total charge enclosed by the surface ‘S’, \\(\\oint\\) represents surface integral of the closed surface.<\/p>\n

Importance :<\/p>\n

    \n
  1. Gauss’s law is very useful in. calculating the electric field in case of problems where it is possible to construct a closed surface. Such surface is called Gaussian surface.<\/li>\n
  2. Gauss’s law is true for any closed surface, no matter what its shape or size.<\/li>\n
  3. Symmetric considerations in many problems make the application of Gauss’s law much easier.<\/li>\n<\/ol>\n

    Long Answer Questions<\/span><\/p>\n

    Question 1.
    \nDefine electric flux. Applying Gauss’s law and derive the expression for electric intensity due to an infinite long straight charged wise. (Assume that the electric field is everywhere radial and depends only on the radial distance r of the point from the wire.)
    \nAnswer:
    \nElectric flux : The number of electric lines of force passing perpendicular to the area is known as electric flux (\u03d5). Electric flux \u03d5 = \\(\\overrightarrow{\\mathrm{E}} \\cdot \\overrightarrow{\\mathrm{A}}\\). So flux is a scalar.<\/p>\n

    Expression for E due to an infinite long straight charged wire :<\/p>\n

    1) Consider an infinitely long thin straight wire with uniform linear charge density ‘\u03bb’.
    \n2) Linear charge density \u03bb = \\(\\frac{\\text { change q }}{\\text { length } l}\\) \u21d2 \u03bbl —– (1)
    \n3) Construct a coaxial cylindrical gaussion surface of length T and radius ‘r’. Due to symmetry we will assume that electric field is radial i.e., normal to the conducting wire.
    \n4) The flat surfaces AB and CD are \u22a5r<\/sup> to the wire. Select small area ds1<\/sub> and ds2<\/sub> on the surface as AB and CD.
    \nThey are \u22a5r<\/sup> to \\(\\overrightarrow{\\mathrm{E}}\\). So flux coming out through them is zero.
    \nSince flux \u03d5 = \\(\\oint \\vec{E} \\cdot d \\vec{s}\\) = Eds cos 90\u00b0 = 0
    \n\"AP
    \n5) So flux coming out through the cylindrical surface ABCD is taken into account.<\/p>\n

    6) From Gauss’s law
    \n\"AP<\/p>\n

    7) From (2) and (3), E(2\u03c0rl) = \\(\\frac{Q}{\\varepsilon_0}\\) = \\(\\frac{\\lambda \/}{\\varepsilon_0}\\) (\u2235 Q = \u03bbl)
    \n\u2234 E = \\(\\frac{\\lambda l}{2 \\pi \\varepsilon_0 \\mathrm{r} l}=\\frac{1}{2 \\pi \\varepsilon_0} \\frac{\\lambda}{\\mathrm{r}}\\)<\/p>\n

    8) Therefore electric intensity due to an infinitely long conducting wire E = \\(\\frac{\\lambda}{2 \\pi \\varepsilon_0 r}\\).<\/p>\n

    \"AP<\/p>\n

    Question 2.
    \nState Gauss’s law in electrostatics. Applying Gauss’s law derive the expression for electric intensity due to an infinite plane sheet of charge.
    \nAnswer:
    \nGauss’s law: The total electric flux through any closed surface is equal to \\(\\frac{1}{\\varepsilon_0}\\) times the net charge enclosed by the surface, i.e.,
    \n\"AP
    \n\"AP<\/p>\n

    Expression for E due to an infinite plane sheet of charge :<\/p>\n

      \n
    1. Consider an infinite plane sheet of charge. Let the charge distribution is uniform on this plane.<\/li>\n
    2. Uniform charge density on this surface \u03c3 = \\(\\frac{\\mathrm{dq}}{\\mathrm{dS}}\\) where dq is the charge over an infinite small area ds.<\/li>\n
    3. Construct a horizontal cylindrical Gaussian surface ABCD perpendicular to the plane with length 2r.<\/li>\n
    4. The flat surfaces BC and AD are parallel to the plane sheet and are at equal distance from the plane.<\/li>\n
    5. Let area of these surfaces are dS1<\/sub> and dS2<\/sub>. They are parallel to \\(\\overrightarrow{\\mathrm{E}}\\). So flux through these two surfaces is
      \n\"AP ——– (1)
      \nWhere S is area of plane surface AD or BC. Both are equal in area and intensity.<\/li>\n
    6. Consider cylindrical surface of AB and CD. Let their areas are say dS3<\/sub> and dS4<\/sub>. These surfaces are \u22a5lr<\/sup> to electric intensity \\(\\overrightarrow{\\mathrm{E}}\\).<\/li>\n
    7. So angle between \\(\\overrightarrow{\\mathrm{E}}\\) and d\\(\\overrightarrow{\\mathbf{s}_3}\\) or dS4<\/sub> is 90\u00b0. Total flux through these, surfaces is zero. Since
      \n\"AP<\/li>\n
    8. From Gauss’s law total flux, \u03d5 = \\(\\oint \\overrightarrow{\\mathrm{E}} \\cdot \\mathrm{d} \\overrightarrow{\\mathrm{S}}\\) = 2ES = \\(\\frac{\\mathrm{q}}{\\varepsilon_0}\\)
      \n\u2234 2ES = \\(\\frac{\\sigma S}{\\varepsilon_0}\\) [\u2235 \\(\\text { Q }\\) = \u03c3 \u00d7 S]<\/li>\n
    9. Therefore intensity of electric field due to an infinite plane sheet of charge E = \\(\\frac{\\sigma}{2 \\varepsilon_0}\\)<\/li>\n<\/ol>\n

      Question 3.
      \nApplying Gauss’s law derive the expression for electric intensity due to a charged conducting spherical shell at
      \n(i) a point outside the shell
      \n(ii) a point on the surface of the shell and
      \n(iii) a point inside the shell.
      \nAnswer:
      \nExpression for E due to a charged conducting spherical shell:<\/p>\n

        \n
      1. Consider a uniformly charged spherical shell. Let total charge on it is ‘q’ and its radius is R.
        \n\"AP<\/li>\n
      2. Since the shell is uniformly charged, the intensity of electric field at any point depends on radial distance ‘r’ from centre ‘O’. The direction of E is away from the centre along the radius.<\/li>\n<\/ol>\n

        i) E at a point outside the shell:<\/p>\n

        1) Consider a point at a distance ‘r’ outside the sphere. Construct a Gaussian surface with ‘r’ as radius (where r > R).<\/p>\n

        2) Total flux coming out of this sphere is
        \n\"AP
        \n\"AP
        \n\"AP
        \n3) Therefore at any point outside the sphere, E = \\(\\frac{\\sigma}{\\varepsilon_0} \\frac{\\mathrm{R}^2}{\\mathrm{r}^2}\\)<\/p>\n

        ii) E at a point on the surface of shell:<\/p>\n

        1) Construct a Gaussian surface with radius r = R.<\/p>\n

        2)
        \n\"AP
        \n\"AP
        \n3) Therefore intensity at any point on surface of the sphere E = \\(\\frac{\\sigma}{\\varepsilon_0}\\)<\/p>\n

        iii) E at a point inside the shell :<\/p>\n

        1) Consider a point P inside the shell. Construct a Gaussian surface with radius r (where r < R). There is no charge inside the shell. So from Gauss\u2019s law \\(\\oint_{\\mathrm{S}} \\overrightarrow{\\mathrm{E}} \\cdot \\mathrm{d} \\overrightarrow{\\mathrm{S}}\\) = \\(\\frac{\\mathrm{q}}{\\varepsilon_0}\\)
        \n\"AP
        \n2) Therefore, intensity of electric field at any point inside a charged shell is zero.<\/p>\n

        Textual Exercises<\/span><\/p>\n

        Question 1.
        \nTwo small identical balls, each of mass 0.20 g, carry identical charges and are suspended by two threads of equal lengths. The balls position themselves at equilibrium such that the angle between the threads is 60\u00b0. If the distance between the balls is 0.5 m, find the charge on each ball.
        \nSolution:
        \nGiven m = 0.20 g = 0.2 \u00d7 10-3<\/sup> kg; \u03b8 = 60\u00b0 \u21d2 \u03b1 = \\(\\frac{\\theta}{2}\\) = 30\u00b0
        \nr = 0.5 m, Let q1<\/sub> = q2<\/sub> = q
        \n\"AP<\/p>\n

        \"AP<\/p>\n

        Question 2.
        \nAn infinite number of charges-each of magnitude q are placed on x-axis at distance of 1, 2, 4, 8, …….. meter from the origin respectively. Find intensity of the electric field at origin.
        \nSolution:
        \nLet q1<\/sub> = q2<\/sub> = q3<\/sub> = q4<\/sub> = ……. = q
        \nr1<\/sub> = 1; r2<\/sub> = 2; r3<\/sub> = 4; r4<\/sub> = 8, …….
        \nThe resultant electric field at origin ‘O’ is given by
        \nE = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\frac{q_1}{r_1^2}\\) + \\(\\frac{1}{4 \\pi \\varepsilon_0} \\frac{\\mathrm{q}_2}{\\mathrm{r}_2^2}\\) + \\(\\frac{1}{4 \\pi \\varepsilon_0} \\frac{q_3}{r_3^2}\\) + \\(\\frac{1}{4 \\pi \\varepsilon_0} \\frac{\\mathrm{q}_4}{\\mathrm{r}_4^2}\\) + ……..
        \n\"AP<\/p>\n

        Question 3.
        \nA clock face has negative charges -q, -2q, -3q, ….. -12q fixed at the position of the corresponding numerals on the dial. The clock hands do not disturb the net field due to the point charges. At what time does the hour hand point in the direction of the electric field at the centre of the dial ?
        \nSolution:
        \nLet distance of each charge from unit charge at centre ‘O’ = r.
        \nResultant electric field of each charge, E = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\frac{6 q}{r^2}\\) [\u2235 -6q – (-12q)]
        \nLet OX be the reference axis. The angles of resultant fields with OX-axis are shown.
        \n\"AP
        \nResultant field along OX-axis = \\(\\left(0+\\frac{1}{2}+\\frac{\\sqrt{3}}{2}+1+\\frac{\\sqrt{3}}{2}+\\frac{1}{2}\\right)\\)i = (2 + \\(\\sqrt{3}\\))i
        \nResultant field along OY-axis = \\(\\left(1+\\frac{\\sqrt{3}}{2}+\\frac{1}{2}+0-\\frac{1}{2}-\\frac{\\sqrt{3}}{2}\\right) \\hat{\\mathrm{j}}\\)
        \n= 1\\(\\hat{\\mathrm{i}}\\)
        \n\u2234 Resultant electric field, ER(OH)<\/sub> = (2 + \\(\\sqrt{3}\\))\\(\\hat{i}\\) + 1\\(\\hat{j}\\)
        \nThe direction of resultant field (OH) is given by, tan \u03b8 = \\(\\frac{|\\mathrm{OY}|}{|\\mathrm{OX}|}\\)
        \n\u21d2 tan \u03b8 = \\(\\frac{1}{2+\\sqrt{3}}\\) = tan 15\u00b0
        \n\u21d2 \u03b8 = 15\u00b0, with OX-axis
        \n\u2234 The hour hand shows at the centre of the dial is at 9.30.<\/p>\n

        Question 4.
        \nConsider a uniform electric field E = 3 \u00d7 103<\/sup> N\/C.
        \n(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ?
        \n(b) What is the flux through the same square if the normal to its plane makes a 60\u00b0 angle with the x – axis ?
        \nSolution:
        \na) Given E = 3 \u00d7 103<\/sup> N\/C
        \nS = 102<\/sup> cm2<\/sup> = 102<\/sup> \u00d7 (10-2<\/sup>m)2<\/sup> = 10-2<\/sup>m2<\/sup>
        \n\u03b8 = 0\u00b0
        \n\u03d5 = ES cos \u03b8
        \n= 3 \u00d7 103<\/sup> \u00d7 10-2<\/sup> \u00d7 cos 0\u00b0
        \n\u2234 \u03d5 = 30 Nm2<\/sup>C-1<\/sup>
        \n\"AP
        \nb) If \u03b8 = 60\u00b0, \u03d5 = ES cos \u03b8
        \n= 3 \u00d7 103<\/sup> \u00d7 10-2<\/sup> \u00d7 cos 60\u00b0
        \n\u2234 \u03d5 = 15 Nm2<\/sup>C-1<\/sup><\/p>\n

        Question 5.
        \nThere are four charges, each with a magnitude Q. Two are positive and two are negative. The charges are fixed to the comers of a square of side ‘L’, one to each comer, in such a way that the force on any charge is directed toward the center of the square. Find the magnitude of the net electric force experienced by any charge ?
        \nSolution:
        \n\"AP<\/p>\n

        Question 6.
        \nThe electric field in a region is given by \\(\\overrightarrow{\\mathbf{E}}\\) = a\\(\\hat{\\mathbf{i}}\\) + b\\(\\hat{\\mathbf{j}}\\). Here a and b are constants. Find the net flux passing through a square area of side L parallel to y-z plane.
        \nSolution:
        \nGiven \\(\\overrightarrow{\\mathrm{E}}\\) = a\\(\\hat{\\mathrm{i}}\\) + b\\(\\hat{\\mathrm{j}}\\)
        \n\\(\\vec{S}\\) = L2<\/sup>\\(\\hat{\\mathrm{i}}\\)
        \n\"AP
        \n\u03d5 = \\(\\overrightarrow{\\mathrm{E}} \\cdot \\overrightarrow{\\mathrm{S}}\\) = (a\\(\\hat{i}\\) + b\\(\\hat{j}\\)) .L2<\/sup>\\(\\hat{i}\\)
        \n\u2234 \u03d5 = aL2<\/sup> [\u2234 \\(\\hat{i}\\). \\(\\hat{i}\\) = 1 and \\(\\hat{i}\\). \\(\\hat{j}\\) = 0]<\/p>\n

        \"AP<\/p>\n

        Question 7.
        \nA hollow spherical shell of radius r has a uniform charge density \u03c3. It is kept in a cube of edge 3r such that the centre of the cube coincides with the center of the shell. Calculate the electric flux that comes out of a face of the cube.
        \nSolution:
        \nFor spherical shell, charge = q (say)
        \nRadius = r
        \nCharge density = \u03c3 = \\(\\frac{q}{A}\\) = \\(\\frac{\\mathrm{q}}{4 \\pi \\mathrm{r}^2}\\)
        \n\u2234 Charge on spherical shell, q = 4\u03c0r2<\/sup>\u03c3
        \nFlux through one of the face of a cube,
        \n\u03d5E<\/sub> = \\(\\frac{1}{6} \\frac{\\mathrm{q}}{\\varepsilon_0}\\) = \\(\\frac{1}{6} \\times \\frac{4 \\pi r^2 \\sigma}{\\varepsilon_0}\\) = \\(\\frac{2 \\pi \\mathrm{r}^2 \\sigma}{3 \\varepsilon_0}\\)<\/p>\n

        Question 8.
        \nAn electric dipole consists of two equal and opposite point charge +Q and -Q, separated by a distance 2l. P is a point collinear with the charges such that its distance from the positive charge is half of its distance from the negative charge. Calculate electric intensity at P.
        \nSolution:
        \nDistance of P from -Q = d (say)
        \nDistance of P from +Q = d\/2
        \n\"AP<\/p>\n

        Question 9.
        \nTwo infinitely long thin straight wires having uniform linear charge densities \u03bb and 2\u03bb are arranged parallel to each other at a distance r apart. Calculate intensity of the electric field at a point midway between them.
        \nSolution:
        \nDistance between two parallel infinite long thin straight wires = r.
        \nElectric field due to infinite long thin straight wire, E = \\(\\frac{\\lambda}{2 \\pi \\varepsilon_0 r}\\)
        \n\"AP
        \n\u2234 Electric intensity at mid point, E = E2<\/sub> – E1<\/sub> = 2E1<\/sub> – E1<\/sub> = E
        \n\u2234 E = \\(\\frac{\\lambda}{\\pi \\varepsilon_0 \\mathrm{r}}\\)<\/p>\n

        Question 10.
        \nTwo infinitely long thin straight wires having uniform linear charge densities e and 2e are arranged parallel to each other at a distance r apart. Find the intensity of the electric field at a point midway between them.
        \nSolution:
        \nFor first infinitely long straight wire, linear charge density \u03bb = e.
        \nFor second infinitely long straight wire, linear charge density \u03bb’ = 2e
        \nDistance between two infinite parallel straight wires = r.
        \nDistance of point P from 1st<\/sup> and 2nd<\/sup> wire = \\(\\frac{\\mathrm{r}}{2}\\)
        \nElectric field intensity at P due 1st<\/sup> wire, E1<\/sub> = \\(\\frac{\\lambda}{2 \\pi \\varepsilon_0\\left(\\frac{\\mathrm{r}}{2}\\right)}=\\frac{\\mathrm{e}}{\\pi \\varepsilon_0 \\mathrm{r}}\\) —— (1)<\/p>\n

        Electric field intensity at P due 2nd<\/sup> wire, E2<\/sub> = \\(\\frac{\\lambda^{\\prime}}{2 \\pi \\varepsilon_0\\left(\\frac{r}{2}\\right)}=\\frac{2 \\mathrm{e}}{\\pi \\varepsilon_0 \\mathrm{r}}\\)
        \n\u2234 E2<\/sub> = 2E1<\/sub> [\u2235 from(1)]
        \n\u2234 Electric field intensity at middle point due to second infinitely long wire
        \nE2<\/sub> = \\(\\frac{2 \\lambda}{\\pi \\varepsilon_0 \\mathrm{r}}\\)<\/p>\n

        \"AP<\/p>\n

        Question 11.
        \nAn electron of mass m and charge e is fired perpendicular to a uniform electric field of intensity E with an initial velocity u. If the electron tranverses a distance x in the field in the direction of firing, find the transverse displacement y it suffers.
        \nSolution:
        \nGiven me<\/sub> = m; q = e; d = x; ux<\/sub> = u; uy<\/sub> = 0
        \nElectric field between the plates = E
        \nTime taken travel in the field, t = \\(\\frac{d}{u_x}\\) = \\(\\frac{\\mathbf{X}}{\\mathbf{u}}\\)
        \n\"AP
        \nForce on electron F = qE = eE
        \nAcceleration of electron, a = \\(\\frac{F}{m}\\) = \\(\\frac{\\mathrm{eE}}{\\mathrm{m}}\\)
        \nTransverse displacement of electron y = uy<\/sub>t + \\(\\frac{1}{2} \\mathrm{at}^2\\)
        \n\u21d2 y = 0 + \\(\\frac{1}{2}\\left(\\frac{e E}{m}\\right)\\left(\\frac{x}{u}\\right)^2\\)
        \n\u2234 y = \\(\\frac{\\mathrm{eEx}^2}{2 \\mathrm{mu}^2}\\)<\/p>\n

        Additiona Exercises<\/span><\/p>\n

        Question 1.
        \nWhat is the force between two small charged spheres having charges of 2 \u00d7 10-7<\/sup> C and 3 \u00d7 10-7<\/sup> C placed 30 cm apart in air ?
        \nSolution:
        \nGiven, q1<\/sub> = 2 \u00d7 10-7<\/sup> C; q2<\/sub> = 3 \u00d7 107<\/sup> C; d = 30 cm = 30 \u00d7 10-2<\/sup> m = 3 \u00d7 10-1<\/sup>m
        \n\"AP
        \nAs q1<\/sub>, q2<\/sub> are positive charges, the force between them is repulsive.<\/p>\n

        Question 2.
        \nThe electrostatic force on a small sphere of charge 0.4 \u03bcC due to another small sphere of charge -0.8 \u03bcC in air is 0.2 N.
        \n(a) What is the distance between the two spheres ?
        \n(b) What is the force on the second sphere due to the first ?
        \nSolution:
        \na) Given q1<\/sub> = 0.4 \u03bcc ;
        \n= 0.8 \u00d7 10-6<\/sup>C
        \nq2<\/sub> = 0.8 \u03bcc; F = 0.2 N = 0.4
        \n= 0.4 \u00d7 10-6<\/sup>m
        \n0.2 = \\(\\frac{9 \\times 10^9 \\times 0.4 \\times 10^{-6} \\times 0.8 \\times 10^{-6}}{\\mathrm{r}^2}\\)
        \nr2<\/sup> = 16 \u00d7 9 \u00d7 10-4<\/sup>
        \nr = 4 \u00d7 3 \u00d7 10-2<\/sup> = 12 \u00d7 10-2<\/sup> m
        \n\u2234 Distance between two charges, r = 12 cm<\/p>\n

        b) Electrostatic force between two charges obeys the Newton’s third law. i.e., force on q1<\/sub> due to q2<\/sub> = force on q2<\/sub> due to q1<\/sub>
        \nf12<\/sub> = f21<\/sub> = 0.2N<\/p>\n

        \"AP<\/p>\n

        Question 3.
        \nCheck that the ratio ke2<\/sup>\/G me<\/sub>mp<\/sub> is dimensionless. Look up a table of Physical Constants and determine the value of this ratio. What does the ratio signify ?
        \nSolution:
        \ni) In electrostatics, Fe<\/sub> = \\(\\frac{\\mathrm{Kq}_1 \\mathrm{q}_2}{\\mathrm{r}^2}\\) = \\(\\frac{\\mathrm{Ke}^2}{\\mathrm{r}^2}\\) ……. (1)
        \nWhere q1<\/sub> = q2<\/sub> = e
        \nIn gravitation, Fg<\/sub> = \\(\\frac{\\mathrm{Gm}_1 \\mathrm{~m}_2}{\\mathrm{r}^2}\\) = \\(\\frac{\\mathrm{Gm}_{\\mathrm{e}} \\mathrm{m}_{\\mathrm{p}}}{\\mathrm{r}^2}\\) …. (2)
        \nWhere m1<\/sub> = me<\/sub> ; m2<\/sub> = mp<\/sub>
        \n\"AP
        \nThus the given ratio is dimensionless.<\/p>\n

        ii) We know that e = 1.6 \u00d7 10-19<\/sup> C ; G = 6.67 \u00d7 10-11<\/sup> N-m2<\/sup>C2<\/sup>
        \n\"AP<\/p>\n

        Question 4.
        \na) Explain the meaning of the statement ‘electric charge of a body is quantized’.
        \nb) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e, large scale charges ?
        \nAnswer:
        \na) The electric charge of a body is quantized means that the charge on a body can occur in some particular values only. Charge on any body is the integral multiple of charge on an electron because the charge of an electron is the elementary charge in nature. The charge on any body can be expressed by the formula q = \u00b1 ne. Where n = number of electrons transferred and e = charge on one electron. The cause of quantization is that only integral number of electrons can be transferred from one body to other<\/p>\n

        b) We can ignore the quantization of electric charge when dealing with macroscopic charges because the charge on one electron is 1.6 \u00d7 10-19<\/sup> C in magnitude, which is very small as compared to the large scale change.<\/p>\n

        Question 5.
        \nWhen a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
        \nAnswer:
        \nAccording to law of conservation of charge, “charge can neither be created nor be destroyed but it can be transferred from one body to another body”. Before rubbing the two bodies they both are neutral i.e., the total charge of the system is zero. When the glass rod is rubbed with a silk cloth, some electrons are transferred from glass rod to silk cloth. Hence glass rod attains positive charge and silk cloth attains same negative charge.<\/p>\n

        Again the total charge of the system is zero, i.e., the charge before rubbing is same as the charge after rubbing. This is consistent with the law of conservation of charge. Here we can also say that charges can be created only in equal and unlike pairs.<\/p>\n

        Question 6.
        \nFour point charges qA<\/sub> = 2 \u00b5C, qB<\/sub> = -5 \u00b5C, qC<\/sub> = 2 \u00b5C and qD<\/sub> = -5 \u00b5C are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 \u00b5C placed at the centre of the square?
        \nSolution:
        \nLet the centre of the square is at O.
        \nThe charge placed on the centre is \u00b5C
        \nAB = BC = CD = DA = 10 cm; AC = \\(\\sqrt{2}\\) \u00d7 10 = 10\\(\\sqrt{2}\\)cm
        \nAO = BO = CO = DO = \\(\\frac{10 \\sqrt{2}}{2}\\) = 5\\(\\sqrt{2}\\) cm
        \n\"AP
        \n\"AP
        \nHere we observe that, FA<\/sub> = -FC<\/sub> and FD<\/sub> = -FB<\/sub>
        \n\u2234 The net resultant force on 1 \u00b5C is
        \nF = FA<\/sub> + FB<\/sub> + FC<\/sub> + FD<\/sub>
        \n= -FC<\/sub> + FB<\/sub> + FC<\/sub> – FB<\/sub>
        \n= 0.<\/p>\n

        Question 7.
        \na) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not ?
        \nb) Explain why two field lines never cross each other at any point ?
        \nAnswer:
        \na) An electrostatic field line represents the actual path travelled by a unit positive charge in an electric field. If the line have sudden breaks it means the unit positive test charge Jumps from one place to another which is not possible. It also means that electric field becomes zero suddenly at the breaks which is not possible. So, the field line cannot have any sudden breaks.<\/p>\n

        b) If two field lines cross each other, then we can draw two tangents at the point of intersection which indicates that (as tangent drawn at any point on electric line of force gives the direction of electric field at that point) there are two directions of electric field at a particular point, which is not possible at the same instant. Thus, two field lines never cross each other at any point.<\/p>\n

        \"AP<\/p>\n

        Question 8.
        \nTwo point charges qA<\/sub> = 3 \u03bcC and qB<\/sub> = -3 \u03bcC are located 20 cm apart in vaccum.
        \na) What is the electric field at the midpoint O of the line AB joining the two charges ?
        \nb) If a negative test charge of magnitude 1.5 \u00d7 10-9<\/sup> C is placed at this point, what is the force experienced by the test charge ?
        \nSolution:
        \na) Given qA<\/sub> = 3 \u03bcC = 3 \u00d7 10-6<\/sup> C; qB<\/sub> = -3 \u03bcC = -3 \u00d7 10-6<\/sup>C
        \n\"AP
        \nFrom fig. AO = OB = 10 cm = 0.1 m
        \nElectric field at midpoint ‘O’ due to qA<\/sub>
        \n\"AP
        \nThe direction of EA<\/sub> is A to O.
        \nElectric field at midpoint ‘O’ due to qB<\/sub> at B is
        \n\"AP
        \nThe direction of EB<\/sub> is O to B.
        \nNow we see that EA<\/sub> and EB<\/sub> are in same direction. So, the resultant electric field at O is E. Hence,
        \nE = EA<\/sub> + EB<\/sub> = 2.7 \u00d7 106<\/sup> + 2.7 \u00d7 106<\/sup> = 5.4 \u00d7 106<\/sup> N\/C :
        \nThe direction of E will be from O to B or toward B.<\/p>\n

        b) Let test charge q0<\/sub> = -1.5 \u00d7 10-9<\/sup> C is placed at midpoint O’.
        \nElectric field intensity at ‘O’ is E = 5.4 \u00d7 106<\/sup>
        \n\"AP
        \nForce F = Eq<\/sub> = 5.4 \u00d7 106<\/sup> \u00d7 -1.5 \u00d7 10-9<\/sup> N
        \n= -8.1 \u00d7 103<\/sup>N
        \nThe direction of force is from O to A.<\/p>\n

        Question 9.
        \nA system has two charges qA<\/sub> = 2.5 \u00d7 10-7<\/sup> C, and qB<\/sub> = -2.5 \u00d7 10-7<\/sup> C located at points A(0, 0, -15 cm) and B(0, 0, +15 cm). What are the total charge and electric dipole moment of the system ?
        \nSolution:
        \nGiven A(0, 0, -15 cm) and B(0, 0, 15 cm)
        \nqA<\/sub> = 2.5 \u00d7 10-7<\/sup>C
        \nqB<\/sub> = -2.5 \u00d7 10-7<\/sup> C
        \nAB = 2a = length of the dipole
        \n= 30 cm = 30 \u00d7 10-2<\/sup> m
        \n\"AP
        \nThe total charge q on the dipole is
        \nq = qA<\/sub> + qB<\/sub> = 2.5 \u00d7 10-7<\/sup>C – 2.5 \u00d7 10-7<\/sup>C = 0
        \nThe electric dipolemoment
        \nP = Any charge (qA<\/sub>) \u00d7 length of dipole (2a)
        \n= 2.5 \u00d7 10-7<\/sup> \u00d7 10 \u00d7 10-2<\/sup>
        \n\u2234 P = 7.5 \u00d7 10-8<\/sup> C-m
        \nThe direction of P is from negative charge to positive charge that is along B to A.<\/p>\n

        Question 10.
        \nAn electric dipole with dipole moment 4 \u00d7 10-9<\/sup> Cm is aligned at 30\u00b0 with the direction of a uniform electric field of magnitude 5 \u00d7 104<\/sup> NC-1<\/sup>. Calculate the magnitude of the torque acting on the dipole.
        \nSolution:
        \nGiven, P = 4 \u00d7 10-9<\/sup> C-m; E = 5 \u00d7 104<\/sup> N\/C; \u03b8 = 30\u00b0,
        \nTorque, \\(\\tau\\) = PE sin \u03b8
        \n\"AP
        \n= 4 \u00d7 10-9<\/sup> \u00d7 5 \u00d7 104<\/sup> sin 30\u00b0 = \\(\\frac{20 \\times 10^{-5}}{2}\\) = 10-4<\/sup>N-m
        \nThe direction of torque is \u22a5r<\/sup> to both electric field and dipole moment.<\/p>\n

        Question 11.
        \nA polythene piece rubbed with wool is found to have a negative charge 3 \u00d7 10-7<\/sup> C.
        \na) Estimate the number of electrons transferred (from which to which ?)
        \nb) Is there a transfer of mass from wool to polythene ?
        \nSolution:
        \na) Given, charge on Polythene, q = -3 \u00d7 10-7<\/sup> C
        \ne = -1.6 \u00d7 10-19<\/sup> C
        \nNo. of electrons transferred, n = \\(\\frac{\\mathrm{q}}{\\mathrm{e}}\\) = \\(\\frac{-3 \\times 10^{-7}}{-1.6 \\times 10^{-19}}\\)
        \n\u2234 n = 1.875 \u00d7 1012<\/sup> [\u2235 q = \u00b1 ne]
        \nElectrons are transferred from wool to polythene.
        \nSo wool gets positive charge and polythene gets negative charge.<\/p>\n

        b) The number of electrons transferred = 1.875 \u00d7 1012<\/sup>
        \nThe mass of one electron, me<\/sub> = 9.1 \u00d7 10-3<\/sup> kg
        \nMass transferred from wool to polythene M = n \u00d7 me<\/sub>
        \nM = 1.875 \u00d7 1012<\/sup> \u00d7 9.1 \u00d7 10-31<\/sup> = 1.8 \u00d7 10-18<\/sup> kg<\/p>\n

        \"AP<\/p>\n

        Question 12.
        \na) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each 6.5 \u00d7 10-7<\/sup> C ? The radii of A and B are negligible compared to the distance of
        \nseparation.
        \nb) What is the force of repulsion if each sphere is charged double the above amount and the distance between them is halved?
        \nSolution:
        \n\"AP
        \na) Given, qA<\/sub> = 6.5 \u00d7 10-7<\/sup>C ; qB<\/sub> = 6.5 \u00d7 10-7<\/sup>C
        \nr = AB = 50 cm = 50 \u00d7 10-2<\/sup>m
        \n\"AP
        \nb)
        \n\"AP
        \nThis force is also repulsive in nature because both the charges are similar (positive) in nature.<\/p>\n

        Question 13.
        \nSuppose the spheres A and B in Exercise – 12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with second and finally removed from both. What is the new force of repulsion between A and B?
        \nSolution:
        \nGiven qA<\/sub> = 6.5 \u00d7 10-7<\/sup>C;
        \nqB<\/sub> = 6.5 \u00d7 10-7<\/sup> C; qC<\/sub> = 0
        \n\"AP
        \nAfter contact of A and C, the charges will be divided equally on both of them. Then final charge on A, then
        \n\\(\\mathrm{q}_{\\mathrm{A}}^{\\prime}\\) = \\(\\frac{\\mathrm{q}_{\\mathrm{A}}+\\mathrm{q}_{\\mathrm{C}}}{2}\\) = \\(\\frac{6.5 \\times 10^{-7}+0}{2}\\)
        \n= 3.25 \u00d7 10-7<\/sup>C
        \nSimilarly charge on C, \\(\\mathrm{q}_{\\mathrm{c}}^{\\prime}\\) = 3.25 \u00d7 10-7<\/sup> C
        \nAfter contact of B and C, the charges will be divided equally on both of them.<\/p>\n

        Then final charge on B, \\(q_B^{\\prime}\\) = \\(\\frac{\\mathrm{q}_{\\mathrm{B}}+\\mathrm{q}_{\\mathrm{C}}^{\\prime}}{2}\\) = \\(\\frac{6.5 \\times 10^{-7}+3.25 \\times 10^{-7}}{2}\\) = 4.875 \u00d7 10-7<\/sup> C
        \nSimilarly final charge one, \\(q_C^{\\prime \\prime}\\) = 4.875 \u00d7 10-7<\/sup> C
        \n\"AP<\/p>\n

        Question 14.
        \nFigure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio ?
        \n\"AP
        \nAnswer:
        \nWe know that a positively charged particle is attracted towards the negatively charged plate and a negatively charged particle is attracted towards the positively charged plate.<\/p>\n

        Here, particle 1 and particle 2 are attracted towards positive plate that means particle 1 and particle 2 are negatively charged. Particle 3 is attracted towards negatively charged plate so it is positively charged. As the deflection in the path of a charged particle is directly proportional to the charge\/mass ratio.
        \ny \u221d \\(\\frac{\\mathrm{q}}{\\mathrm{m}}\\)
        \nHere, the deflection in particle 3 is maximum, so the charge to mass ratio of particle 3 is maximum.<\/p>\n

        \"AP<\/p>\n

        Question 15.
        \nConsider a uniform electric field E = 3 \u00d7 103<\/sup> \\(\\hat{\\mathbf{i}}\\) N\/C.
        \n(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ?
        \n(b) What is the flux through the same square if the normal to its plane makes a 60\u00b0 angle with the x-axis ?
        \nSolution:
        \nGiven \\(\\overrightarrow{\\mathrm{E}}\\) = 3 \u00d7 103<\/sup> \\(\\hat{\\mathbf{i}}\\) N\/C<\/p>\n

        a) As the surface is in Y – Z plane, so the area vector (normal to the square) is along X – axis
        \nArea S = 10 \u00d7 10 = 100 cm2<\/sup> = 10-2<\/sup> m2<\/sup>
        \n\"AP
        \nArea vector \\(\\vec{S}\\) = 10-2<\/sup> \\(\\hat{\\mathbf{i}}\\) m2<\/sup>
        \n\u03d5 = \\(\\overrightarrow{\\mathrm{E}} \\cdot \\overrightarrow{\\mathrm{S}}\\) = (3 \u00d7 103<\/sup> \\(\\hat{\\mathbf{i}}\\)). (10-2<\/sup>i)
        \n\u2234 \u03d5 = 3 \u00d7 103<\/sup> \u00d7 10-2<\/sup> = 30N-m2<\/sup>\/c
        \n\"AP<\/p>\n

        b) \\(\\overrightarrow{\\mathrm{E}}\\) = 3 \u00d7 103<\/sup> \\(\\hat{\\mathbf{i}}\\) N\/C ; \\(\\vec{S}\\) = \\(\\hat{\\mathbf{i}}\\) m2<\/sup> ; \u03b8 = 60\u00b0
        \n\u03d5 = \\(\\overrightarrow{\\mathrm{E}}\\) . \\(\\overrightarrow{\\mathrm{S}}\\) = ES cos 60\u00b0 = 3 \u00d7 103<\/sup> \u00d7 10-2<\/sup> \u00d7 cos 60\u00b0
        \n\u2234 \u03d5 = 15 N – m2<\/sup>\/C<\/p>\n

        Question 16.
        \nWhat is the net flux of the uniform electric field of Exercise -15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes ?
        \nAnswer:
        \nAs we know that the number of lines entering in the cube is the same as that the number of lines leaving the cube. So, no flux is remained on the cube and hence, the net flux over the cube is zero.<\/p>\n

        Question 17.
        \nCareful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 \u00d7 103<\/sup> Nm2<\/sup>\/C.
        \n(a) What is the net charge inside the box ?
        \n(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box ? Why or Why not ?
        \nSolution:
        \na) Given, \u03d5 = 8.0 \u00d7 103<\/sup> N – m2<\/sup>\/C
        \n\u03b50<\/sub> = 8 \u00d7 103<\/sup> \u00d7 8.854 \u00d7 10-12<\/sup>
        \n\u2234 q = 0.07 \u03bcc
        \nThe flux is outward hence the charge is positive in nature<\/p>\n

        b) Net outward flux = 0
        \nThen, we can conclude that the net charge inside the box is zero. i.e., the box may have either zero charge or have equal amount of positive and negative charges. It means we cannot conclude that there is no charge inside the box.<\/p>\n

        Question 18.
        \nA point charge +10 \u03bcC is a ‘distance 5 cm directly above the centre of a square of side 10 cm, as shown in fig. What is the magnitude of the electric flux through the square ? (Hint: Think of the square as one face of a cube with edge 10 cm).
        \n\"AP
        \nSolution:
        \nLet the charge q is placed at the centre of cube as shown in fig.
        \nThe total flux enclosed through the cube is \u03d5 = \\(\\frac{q}{\\varepsilon_0}\\)
        \nThe flux enclosed by one face \u03d5 = \\(\\frac{1}{6}\\) of total flux.
        \n[\u2235 Cube has 6 faces]
        \n\"AP
        \n\u03d5 = \\(\\frac{\\phi}{6}\\) = \\(\\frac{1}{6} \\frac{\\mathrm{q}}{\\varepsilon_0}\\)
        \nHere q = 10 \u03bcC = 10 \u00d7 10-6<\/sup>C ; \u03b50<\/sub> = 8.854 \u00d7 10-12<\/sup>C2<\/sup> – N-1<\/sup>-m-2<\/sup>
        \n\u2234 \u03d5 = \\(\\frac{1}{6} \\times \\frac{10 \\times 10^{-6}}{8.854 \\times 10^{-12}}\\)
        \n= 1.88 \u00d7 105<\/sup> N-m2<\/sup>\/C<\/p>\n

        Question 19.
        \nA point charge of 2.0 \u03bcC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface ?
        \nSolution:
        \nGiven, q = 2.0 \u03bcC = 2.0 \u00d7 10-6<\/sup>C
        \n\u03b50<\/sub> = 8.854 \u00d7 10-12<\/sup> C2<\/sup>-N-1<\/sup> – m-2<\/sup>
        \nThe net flux through the surface,
        \n\u03d5 = \\(\\frac{\\mathrm{q}}{\\varepsilon_0}\\) = \\(\\frac{2 \\times 10^{-6}}{8.854 \\times 10^{-12}}\\) = 2.26 \u00d7 105<\/sup>N-m2<\/sup>\/C<\/p>\n

        \"AP<\/p>\n

        Question 20.
        \nA point charge causes an electric flux of -1.0 \u00d7 103<\/sup> Nm2<\/sup>\/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge,
        \n(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface ?
        \n(b) What is the value of the point charge ?
        \nSolution:
        \na) From Gauss’s law, \u03d5 = \\(\\frac{\\mathrm{q}}{\\varepsilon_0}\\)
        \nElectric flux \u03d5 depends on charge q.
        \nIt is independent of radius of Gaussian surface. Hence the radius of Gaussian surface were doubled, flux does not change.<\/p>\n

        b) \u03d5 = – 1.0 \u00d7 103<\/sup> N-m2<\/sup>\/c ; \u03b50<\/sub> = 8.854 \u00d7 10-12<\/sup> e2<\/sup>-N-1<\/sup>-m-2<\/sup>
        \nq = \u03d5\u03b50<\/sub> = -1.0 \u00d7 103<\/sup> \u00d7 8.854 \u00d7 10-12<\/sup> = -8.85 \u00d7 10-9<\/sup>C.
        \n\u2234 The value of point charge, q = -8.85 \u00d7 10-9<\/sup>C<\/p>\n

        Question 21.
        \nA conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 \u00d7 103<\/sup> N\/C and points radially inward, what is the net charge on the sphere?
        \nSolution:
        \nE = 1.5 \u00d7 103<\/sup> N\/C; r = 20 cm = 20 \u00d7 10-2<\/sup>m.
        \nE = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\cdot \\frac{q}{r^2}\\)
        \n1.5 \u00d7 103<\/sup> = \\(\\frac{9 \\times 10^9 \\times \\mathrm{q}}{\\left(20 \\times 10^{-2}\\right)^2}\\)
        \nq = \\(\\frac{1.5 \\times 10^3 \\times 20 \\times 20 \\times 10^{-4}}{9 \\times 10^9}\\) = 6.67 \u00d7 10-9<\/sup>C.<\/p>\n

        Question 22.
        \nA uniformly charged conducting sphere of 2.4 m diameter has a surface charge density, of 80.0 \u03bcC\/m2<\/sup>.
        \n(a) Find the charge on the sphere,
        \n(b) What is the total electric flux leaving the surface of the sphere ?
        \nSolution:
        \na) Given D = 2.4 m; r = \\(\\frac{\\mathrm{D}}{2}\\) = 1.2 m
        \n\u03c3 = 80 \u00b5c\/m2<\/sup> = 80 \u00d7 10-6<\/sup> C\/m2<\/sup>
        \n\u03c3 = \\(\\frac{\\mathrm{q}}{4 \\pi r^2}\\) \u21d2 q = \u03c3 4\u03c0r2<\/sup>
        \n\u21d2 q = 80 \u00d7 10-6<\/sup> \u00d7 4 \u00d7 3.14 \u00d7 1.2 \u00d7 1.2
        \n\u2234 q = 1.45 \u00d7 10-3<\/sup>C<\/p>\n

        b) \u03d5 = \\(\\frac{Q}{\\varepsilon_0}\\) = \\(\\frac{1.4 \\times 10^{-3}}{8.854 \\times 10^{-12}}\\) = 1.6 \u00d7 108<\/sup>N-m2<\/sup>\/C
        \nThus, the flux leaving the surface of sphere is 1.6 \u00d7 108<\/sup> N – m2<\/sup>\/c<\/p>\n

        Question 23.
        \nAn infinite line charge produces a field of 9 \u00d7 104<\/sup> N\/C at a distance of 2 cm. Calculate the linear charge density.
        \nSolution:
        \nGiven r = 2 cm = 2 \u00d7 10-2<\/sup>m ; E = 9 \u00d7 104<\/sup> N\/C
        \n\"AP
        \n\"AP
        \nThus, the linear charge density is 10-7<\/sup> C\/m.<\/p>\n

        Question 24.
        \nTwo large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 \u00d7 10-22<\/sup> C\/m2<\/sup>. What is E :
        \n(a) in the outer region of the first plate,
        \n(b) in the outer region of the second plate and
        \n(c) between the plates ?
        \nSolution:
        \nGiven \u03c3A<\/sub> = 127.0 \u00d7 10-22<\/sup> C\/m2<\/sup>
        \n\u03c3B<\/sub> = 17.0 \u00d7 10-22<\/sup> C\/m2<\/sup>
        \n\"AP
        \n\"AP<\/p>\n

        Question 25.
        \nAn oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 \u00d7 104<\/sup> NC-1<\/sup> in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm-3<\/sup>. Estimate the radius of the drop, (g = 9.81 ms-2<\/sup>; e = 1.60 \u00d7 10-19<\/sup>C).
        \nSolution:
        \nGiven n = 12; E = 2.55 \u00d7 104<\/sup> N\/C
        \np = 1.26 g\/cm3<\/sup> = 1.26 \u00d7 103<\/sup> kg\/m3<\/sup>
        \ne = 1.6 \u00d7 10-19<\/sup>C ; g = 9.81 ms-2<\/sup>
        \n\"AP
        \nAs the oil drop is stationary,
        \nElectrostatic force = Gravitational force
        \n\u21d2 qE = mg
        \nneE = \\(\\frac{4}{3} \\pi r^3 \\mathrm{\\rho g}\\)
        \nr3<\/sup> = \\(\\frac{3 \\mathrm{neE}}{4 \\pi \\rho \\mathrm{g}}\\) = \\(\\frac{3 \\times 12 \\times 1.6 \\times 10^{-19} \\times 2.55 \\times 10^4}{4 \\times 3.14 \\times 1.26 \\times 10^3 \\times 9.8}\\)
        \nr = 0.94 \u00d7 10-18<\/sup>
        \nr = [0.94 \u00d7 10-18<\/sup>]\\(\\frac{1}{3}\\)<\/sup> = 9.81 \u00d7 10-7<\/sup>m
        \n\u2234 Radius of the drop = 9.81 \u00d7 10-7<\/sup> m.<\/p>\n

        \"AP<\/p>\n

        Question 26.
        \nWhich among the curves shown in Fig. cannot possibly represent electrostatic field lines ?
        \n\"AP
        \nSolution:
        \na) According to the properties of electric lines of force, the lines should be always \u22a5r to the surface of a conductor as they starts or they ends. Here, some of the lines are not \u22a5r to the surface, thus it cannot represent the electrostatic field lines.<\/p>\n

        b) According to the property of electrostatic field lines, they never start from negative charge, here some of the lines start from negative charge. So, it cannot represent the electrostatic field lines.<\/p>\n

        c) As the property of electric field lines that they start outwards from positive charge. Hence, it represents the electrostatic field lines.<\/p>\n

        d) By the property of electric field lines, two electric field lines never intersect each other. Here, two lines intersect. So it does not represent the electric field lines.<\/p>\n

        e) By the property of electric field lines that they are not in the form of closed loops. Here, the lines form closed loop. So, it does not represent the electric field lines.<\/p>\n

        Question 27.
        \nIn a certain region of space, electric field is along the Z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive Z-direction, at the rate of 105<\/sup> NC-1<\/sup> per metre. What are the force and torque experienced by a system having a total dipolemoment equal to 10-7<\/sup> Cm in the negative Z-direction ?
        \nSolution:
        \nThe electric field increases in positive Z – direction. dE
        \n\\(\\frac{\\mathrm{dE}}{\\mathrm{dZ}}\\) = 105<\/sup> N\/C-m
        \nThe direction of dipolemoment is in the negative Z-direction
        \n\"AP
        \nSo the negative charge q is placed at A and positive charge q is placed at B as the direction of dipole moment is from negative charge to positive charge.
        \nPZ<\/sub> = -10-7<\/sup>C-m<\/p>\n

        The negative sign shows its direction in negative Z – axis. According to the basic definition of electric field, F = qdE Now, multiplying and dividing by dz,
        \nF = q\\(\\frac{\\mathrm{dE}}{\\mathrm{dz}} \\cdot \\mathrm{dz}\\) .dz = q.dz\\(\\frac{\\mathrm{dE}}{\\mathrm{dz}}\\)
        \nqdz = dipolement pz<\/sub>, as the length of the dipole is dz.<\/p>\n

        \u2234 F = Pz<\/sub>. \\(\\frac{\\mathrm{dE}}{\\mathrm{dz}}\\) = -10-7<\/sup> \u00d7 105<\/sup> = -10-2<\/sup>N
        \nTorque, \\(\\tau\\) = PE sin \u03b8 (\u2235 \u03b8 = 180\u00b0 angle between P and E)
        \n\\(\\tau\\) = PE sin 180\u00b0 = 0
        \nThus the force is -10-2<\/sup> N and the torque is 0.<\/p>\n

        Question 28.
        \na) A conductor A with a cavity as shown in Fig. (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor, (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q (Fig. (b)). (c) A sensitive instrument is o he shielded from the strong electrostatic fields in its environment. Suggest a possible way.
        \n\"AP
        \nSolution:
        \na) As we know the property of conductor that the net electric field inside a charged conductor is zero, i.e., E = 0.
        \nNow let us choose a Gaussian surface lying completely inside the conductor enclosing the cavity.
        \nSo, from Gauss’s theorem \\(\\oint \\text { E. dS }\\) = \\(\\frac{\\mathrm{q}}{\\varepsilon_0}\\)
        \nAs E = 0 \u21d2 \\(\\frac{q}{\\varepsilon_0}\\) = 0 \u21d2 q = 0
        \nThat means the charge inside the cavity is zero. Thus, the entire charge Q on the conductor must appear on the outer surface of the conductor.<\/p>\n

        b) As the conductor B carrying a charge +q inserted in the cavity, the charge -q is induced on the metal surface of the cavity and then charge +q induced on the outside surface of the conductor A. Initially the outer surface of A of A has a charge Q and now it has a charge +q induced. So the total charge on the outer surface of A is Q + q.<\/p>\n

        c) To protect any sensitive instrument from electrostatic field, the sensitive instrument must be put in the metallic cover. This is known as electrostatic shielding.<\/p>\n

        \"AP<\/p>\n

        Question 29.
        \nA hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (\u03c3\/\u03b5\u03b50<\/sub>)\\(\\hat{\\mathbf{n}}\\), where \\(\\hat{\\mathbf{n}}\\) is the unit vector in the outward normal direction and \u03c3 is the surface charge density near the hole.
        \nSolution:
        \nSurface charge density near the hole = \u03c3
        \nUnit vector = \\(\\hat{\\mathbf{n}}\\) (normal directed outwards)
        \nLet P be the point on the hole.
        \nThe electric field at point P closed to the surface of conductor, according to Gauss’s theorem,
        \n\"AP
        \n\\(\\oint \\mathrm{E} \\cdot \\mathrm{dS}\\) = \\(\\frac{q}{\\varepsilon_0}\\)
        \nWhere q is the charge near the hole.
        \nE ds cos \u03b8 = \\(\\frac{\\sigma \\mathrm{dS}}{\\varepsilon_0}\\) (\u2234 \u03c3 = \\(\\frac{\\mathrm{q}}{\\mathrm{dS}}\\) \u2234q = \u03c3 dS) where dS = area<\/p>\n

        \u2234 Angle between electric field and area vector is 0\u00b0.
        \nEdS = \\(\\frac{\\sigma \\mathrm{dS}}{\\varepsilon_0}\\)
        \nE = \\(\\frac{\\sigma}{\\varepsilon_0}\\)
        \nE = \\(\\frac{\\sigma}{\\varepsilon_0} \\hat{\\mathrm{n}}\\)<\/p>\n

        This electric field is due to the filled up hole and the field due to the rest of the charged conductor. The two fields inside the conductor are equal and opposite. So, there is no electric field inside the conductor. Outside the conductor, the electric fields are equal and are in the same direction.
        \nSo, the electric field at P due to each part = \\(\\frac{1}{2} \\mathrm{E}\\) = \\(\\frac{\\sigma}{2 \\varepsilon_0} \\hat{n}\\)<\/p>\n

        \"AP<\/p>\n

        Question 30.
        \nObtain the formula for the electric field due to a long thin wire of uniform linear charge density \u03bb without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
        \nSolution:
        \nLet us consider a long thin wire of linear charge density \u03bb. We have to find the resultant electric field due to this wire at point P.
        \nNow, consider a very small element of length dx at a distance x from C.
        \n\"AP
        \nThe charge on this elementary portion of length dx
        \nq = \u03bb dx ——- (1)<\/p>\n

        Electric field intensity at point P due to the elementary portion
        \ndE = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\cdot \\frac{\\mathrm{q}}{(\\mathrm{OP})^2}\\) = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\frac{\\lambda d x}{(\\mathrm{OP})^2}\\) [\u2235 from (1)]
        \nNow, in \u0394PCO (PO)2<\/sup> = (PC)2<\/sup> + (CO)2<\/sup>
        \n(OP)2<\/sup> = r2<\/sup> + x2<\/sup>
        \ndE = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\frac{\\lambda d \\mathbf{x}}{\\left(x^2+r^2\\right)}\\) ——- (2)<\/p>\n

        The components of dE are dE cos \u03b8 along PD and dE sin \u03b8 along PF.
        \nHere, there are so many elementary portion. So all the dE sin \u03b8 components balance each other. The resultant electric field at P is due to only dE cos \u03b8 components.
        \nThe resultant electric field due to elementary component, dE’ = dE cos \u03b8<\/p>\n

        dE’ = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\cdot \\frac{\\lambda d x}{\\left(x^2+r^2\\right)} \\cos \\theta\\) —— (3)
        \nIn \u0394OCP tan \u03b8 = \\(\\frac{x}{r}\\) \u21d2 x = r tan \u03b8
        \nDifferentiating with respect to \u03b8, we get dx = r sec2<\/sup> \u03b8 d\u03b8
        \nPutting in equation (3), we get
        \n\"AP
        \nAs the wire is of infinite length, so integrate within the limits –\\(\\frac{\\pi}{2}\\) to \\(\\frac{\\pi}{2}\\), we get
        \n\"AP<\/p>\n

        Question 31.
        \nIt is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge +(2\/3)e and the ‘down’ quark (denoted by d) of charge (-1\/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.
        \nSolution:
        \nFor the protons, the charge on it is +e let the number of up quarks are a, then the number of down quarks are (3 – a) as the total number of quarks are 3.
        \nSo, ax<\/sub> up quark charge + (3 – a) down quark charge = +e
        \na \u00d7 \\(\\frac{2}{3} \\mathrm{e}\\) + (3 – a)\\(\\left(\\frac{-\\mathrm{e}}{3}\\right)\\) = e
        \n\\(\\frac{2 \\mathrm{ae}}{3}\\) – \\(\\frac{(3-\\mathrm{a}) \\mathrm{e}}{3}\\) = e
        \n2a – 3 + a = 3
        \n3a = 6
        \na = 2
        \nThus, in the proton there are two up quarks and one down quark.
        \n\u2234 Possible quark composition for proton = uud<\/p>\n

        For the neutron, the charge on neutron is 0.
        \nLet the number of up quarks are b and the number of down quarks are (3 – b)
        \nSo, bx<\/sub> up quark charge + (3 – b) \u00d7 down quark charge = 0
        \nb\\(\\left(\\frac{2 \\mathrm{e}}{3}\\right)\\) + (3 – b)\\(\\left(\\frac{-\\mathrm{e}}{3}\\right)\\) = 0
        \n2b – 3 + b = 0
        \n3b = 3
        \n\u2234 b = 1
        \nThus, in neutron, there are one up quark and two down quarks.
        \n\u2234 Possible quark composition for neutrons = udd.<\/p>\n

        Question 32.
        \na) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e, where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
        \nb) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
        \nSolution:
        \na) Let us consider that initially the test charge is in the stable equilibrium. When the test charge is displaced from the null point (where, E = 0) in any direction, it must experience a restoring force towards the null point.
        \nThis means that there is a net inward flux through a closed surface around the null point According to the Gauss’s theorem, the net electric flux through a surface net enclosing any charge must be zero. Hence, the equilibrium is not stable.<\/p>\n

        b) The middle point of the line joining two like charges is a null point. If we displace a test Charge slightly along the
        \nline, the restoring force try to bring the test charge back to the centre. If we displace the test charge normal to the line, the net force on the test charge takes it further away from the null point. Hence the equilibrium is not stable.
        \n\"AP<\/p>\n

        Question 33.
        \nA particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed Vx<\/sub> (as in the fig.). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL2<\/sup>\/(2m \\(\\mathbf{V}_{\\mathbf{x}}^2\\)).
        \nCompare this motion with motion of a projectile in gravitational field discussed in section 4.10 of 1st<\/sup> Year Textbook of Physics.
        \nSolution:
        \nMass of particle = m
        \nCharge of particle = -q
        \nSpeed of particle = Vx<\/sub>
        \nLength of plates = L
        \n\"AP
        \nElectric field between the plates = E (from positive plate to negative plate).
        \nLet the deflection in the path of the charge – q is Y, because the force acting in +Y axis direction. The direction of force is from negative plate to positive plate because the charge is negative in nature.<\/p>\n

        Let us discuss the motion in Y axis direction. Initial velocity u = 0
        \nAcceleration a = \\(\\frac{F}{m}\\) = \\(\\frac{+\\mathrm{qE}}{\\mathrm{m}}\\)
        \nDeflection Y = ?
        \nTime = \\(\\frac{\\text { Distance }}{\\text { Velocity }}\\) = \\(\\frac{\\mathrm{L}}{\\mathrm{V}_{\\mathrm{x}}}\\)
        \nUsing second equation of motion,
        \nS = ut + \\(\\frac{1}{2} \\mathrm{at}^2\\)at
        \nPutting the values y = 0 + \\(\\frac{1}{2} \\times\\left(+\\frac{\\mathrm{qE}}{\\mathrm{m}}\\right) \\frac{\\mathrm{L}^2}{\\mathrm{~V}_{\\mathrm{x}}^2}\\)
        \nY = \\(\\frac{\\mathrm{qEL}^2}{2 \\mathrm{mV}_{\\mathrm{x}}^2}\\)<\/p>\n

        In the case of projectile motion y = \\(\\frac{1}{2} \\mathrm{gt}^2\\). Thus, it is exactly similar to the projectile motion in the gravitational field.<\/p>\n

        \"AP<\/p>\n

        Question 34.
        \nSuppose that the particle is an electron projected with velocity Vx<\/sub> = 2.0 \u00d7 106<\/sup> ms-1<\/sup>. If E between the plates separated by 0.5 cm is 9.1 \u00d7 102<\/sup> N\/C, where will the electron strike the upper plate ? (|e| = 1.6 \u00d7 10-19<\/sup> C, me<\/sub> = 9.1 \u00d7 10-31<\/sup> kg.)
        \nSolution:
        \nGiven Vx<\/sub> = 2 \u00d7 106<\/sup> m\/s; E = 9.1 \u00d7 102<\/sup> N\/C
        \nq = e = 1.6 \u00d7 10-19<\/sup> C; me<\/sub> = 9.1 \u00d7 10-31<\/sup> kg
        \nd = 0.5 cm = 0.5 \u00d7 10-2<\/sup> m = 5 \u00d7 10-3<\/sup> m
        \nThe electron will strike the upper plate at its other end at X = L as it get deflected.
        \n\"AP<\/p>\n","protected":false},"excerpt":{"rendered":"

        Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 4th Lesson Electric Charges and Fields Textbook Questions and Answers. AP Inter 2nd Year Physics Study Material 4th Lesson Electric Charges and Fields Very Short Answer Questions Question 1. What is meant by the statement ‘charge is quantized’? Answer: The minimum charge that can be … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22651"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22651"}],"version-history":[{"count":2,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22651\/revisions"}],"predecessor-version":[{"id":22653,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22651\/revisions\/22653"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22651"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22651"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22651"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}