{"id":34920,"date":"2022-11-22T13:20:31","date_gmt":"2022-11-22T07:50:31","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=34920"},"modified":"2022-11-22T13:20:31","modified_gmt":"2022-11-22T07:50:31","slug":"ts-inter-2nd-year-physics-study-material-chapter-4","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-4\/","title":{"rendered":"TS Inter 2nd Year Physics Study Material Chapter 4 Electric Charges and Fields"},"content":{"rendered":"

Telangana TSBIE\u00a0TS Inter 2nd Year Physics Study Material<\/a> 4th Lesson Electric Charges and Fields Textbook Questions and Answers.<\/p>\n

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

Very Short Answer Type 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 another is equal to the charge of an electron ‘e’.<\/p>\n

So charge always exists as an integral multiple of charge of electron i.e., Q = ne. (1 e = 1.6 \u00d7 10-19<\/sup> C). Therefore charge in quantized.<\/p>\n

Question 2.
\nRepulsion is the sure test of charging than attraction. Why?
\nAnswer:
\nA charged body can attract opposite charged body and also a neutral body. But repu\u00aclsion is only between two charged bodies of same polarity. Hence repulsion is sure test for charging than attraction.<\/p>\n

Question 3.
\nHow many electrons constitute 1 C of charge?
\nAnswer:
\nChargeq = ne; q = 1 C; e- 1.6 \u00d7 10-19<\/sup> C.
\n\"TS<\/p>\n

Question 4.
\nWhat happens to the weight of a body when it is charged positively?
\nAnswer:
\nWhen a body is positively charged it looses electrons, hence its weight decreases (or) it looses weight.<\/p>\n

\"TS<\/p>\n

Question 5.
\nWhat happens to the force between two charges if the distance between them is a) halved b) doubled?
\nAnswer:
\n\"TS
\n‘F’ increases four times its initial value<\/p>\n

b) Let d1<\/sub> = d and d2<\/sub> = 2d and F1<\/sub> = F, F2<\/sub> = ?
\n\"TS
\nForce between them reduces to 1\/4 of initial value.<\/p>\n

Question 6.
\nThe electric lines of force do not intersect. Why?
\nAnswer:
\nThe tangent drawn at a point to the line of force gives direction of electric field. If two lines intersect at point of intersection field will be in two different directions, which is not possible. Therefore electric lines of force do not intersect.<\/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:
\nCharge is a scalar so total charge Q = q + (- q) = 0. But electric field intensity is a vector and they must be add up vectorially at any given point. So at the point A of equilateral triangle<\/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:
\nThe electrostatic force is conservative force.<\/p>\n

Question 9.
\nState Gauss’s law in electrostatics. Explain its importance. [AP June 15; TS Mar. 15, May 15]
\nAnswer:
\nDef :
\nThe total electrical flux through any closed surface is equal to \\(\\frac{1}{\\varepsilon_0}\\) times the charge enclosed by the surface.
\n\"TS
\n\u03b50<\/sub> = permittivity of free space.<\/p>\n

Importance :
\nIt is used to find the electric intensity due to various bodies due to charge distributions.<\/p>\n

Question 10.
\nWhen is the electric flux negative and when is it positive?
\nAnswer:
\nElectric flux, \u03a6 = \\(\\overline{\\mathrm{E}}.\\overline{\\mathrm{A}}\\) = EA cos \u03b8.
\nwhere \u03b8 = angle between \\(\\overline{\\mathrm{E}}\\) and \\(\\overline{\\mathrm{A}}\\).
\nIf 0\u00b0 < \u03b8 \u2264 90\u00b0. \u21d2 \u03a6 is positive.
\nWhen 90\u00b0 < \u03b8 \u2264 180\u00b0. \u21d2 \u03a6 is 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 at a point due to an infinitely long charged wire (E) = \\(\\frac{\\lambda}{2 \\pi \\varepsilon_0 r}\\), \u03bb = the linear charge density of the wire.
\nr = the radial distance of the point from the axis of the wire.<\/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}\\), \u03c3 – is surface charge density of the sheet.<\/p>\n

\"TS<\/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:
\nElectric field intensity due to a charged conducting spherical shell
\na) At a point outside the shell intensity of electric field E = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\frac{\\mathrm{q}}{\\mathrm{r}^2}=\\frac{\\sigma}{\\varepsilon_0} \\frac{\\mathrm{R}^2}{\\mathrm{r}^2}\\)
\nq = \u03c3A = \u03c3 4\u03c0R\u00b2<\/p>\n

b) At a point inside the shell intensity of electric field E = 0.
\nBecause potential inside a conducting shell is zero.<\/p>\n

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

Question 1.
\nState and explain Coulomb’s inverse square law in electricity. [AP May 18, 17; TS Mar. 17, ’14]
\nAnswer:
\nCoulomb’s Law :
\nThe force of attraction or repulsion between two charges is directly proportional to the product of the two charges and is inversely proportional to square of the distance between them. This force acts along the line joining the two charges.
\nExplanation:
\n\"TS
\n\u03b50<\/sub> = permittivity of free space.
\n\u03b50<\/sub> = 8.85 \u00d7 10-12<\/sup>Nm\u00b2\/C\u00b2<\/p>\n

This electrostatic force between the charges depends on the nature of the medium between them.
\nIn any medium F = \\(\\frac{1}{4 \\pi \\varepsilon} \\frac{\\mathrm{q_1q_2}}{\\mathrm{r}^2}\\)
\n\u03b5 = permittivity of that medium.
\n\\(\\frac{\\mathrm{F_{vacuum}}}{\\mathrm{F_{med}}}= \\frac{\\varepsilon}{\\varepsilon_0} \\) = \u03b5r<\/sub> = k<\/p>\n

where \u03b5r<\/sub> (or) k is called relative permittivity or dielectric constant.<\/p>\n

Note :- This force is an action and reaction pair i.e., \\(\\overline{\\mathrm{F}}_{21}=-\\overline{\\mathrm{F}}_{21}\\)
\nIn vector form of Coulomb’s law is
\n\"TS<\/p>\n

Question 2.
\nDefine intensity of electric field at a point. Derive an expression for the intensity due to a point charge. [AP Mar. ’16]
\nAnswer:
\nIntensity of electric field: It is defined as the force on a unit positive charge when placed in the electric field.<\/p>\n

Proof :-
\nConsider a point charge ‘Q’ at O’, electric field will exist around that charge.
\nP = point at a distance r from the charge Q,
\nq0<\/sub> = Test charge placed at that point.
\n\"TS
\nDue to a negative charge field is towards it.
\nNote : The electric field due to a point charge is non-uniform.<\/p>\n

Question 3.
\nDerive the equation for the couple acting on an electric dipole in a uniform electric field. [TS Mar. ’19, May ’18; AP May ’16, ’14]
\nAnswer:
\nConsider electric dipole of moment ‘p’ in an uniform electric field ‘E’, situated at an angle \u03b8 with the field.<\/p>\n

The positive charge experiences force “qE” in the direction of field and negative charge experiences as a force – qE opposite to the direction of field. Net force on the dipole is zero. But these two forces will constitute a couple, they will produce torque on the dipole.
\n\"TS
\nMagnitude of torque (\u03c4) = force \u00d7 perpendicular distance
\n= qE (AC) = qE (2a sin \u03b8) = q(2a) E sin \u03b8 = pE sin \u03b8
\n\\(\\bar{\\tau}=\\overline{\\mathrm{p}} \\times \\overline{\\mathrm{E}}\\)
\nWhen \\(\\overline{\\mathrm{p}}\\) and \\(\\overline{\\mathrm{E}}\\) are in the plane of the paper then direction of torque is normal to the plane of paper.
\nIf \u03b8 = 90\u00b0 \u21d2 \u03c4max<\/sub> = pE.
\nThe electric dipole moment of a dipole is equal to the torque acting on it when placed in a uniform electric field.<\/p>\n

\"TS<\/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. [AP Mar. 19, 18, 17, 16, May 16; TS Mar. May 16]
\nAnswer:
\nConsider an electric dipole with charges q, – q with separation ‘2a’ between them.
\nLet p = a point on its axial line at a distance r from the mid point of the dipole
\nEaxial<\/sub> = intensity of electric field at p
\nEaxial<\/sub> = E+q<\/sub> + E-q<\/sub>.
\nThe electric field at p due to the charge + q
\n\"TS<\/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.
\nAnswer:
\nConsider an electric dipole with charges q, -q with a separation ‘2a’ between them. Consider a point p’ on the equatorial of the dipole at a distance r from the centre of the dipole. Electric field at p is the resultant of E+q<\/sub> and E-q<\/sub>.
\nThe electric field due to the charge +q at p
\n\"TS<\/p>\n

These are equal in magnitude. The components of E+q<\/sub> and E-q<\/sub> normal to the axis of the dipole cancel each other, the components along the axis will add up.
\n\"TS
\n\"TS<\/p>\n

Question 6.
\nState Gauss’s law in electrostatics and explain its importance. [TS Mar. \u2019 18,’ 15, May ‘ 17; AF June 15, Mar. 15]
\nAnswer:
\nGauss’s Law :
\nThe total electric flux through any closed surface is equal to \\(\\frac{1}{\\varepsilon_0}\\) times the net charge enclosed by the surface.
\n\"TS
\nwhere \u03b50<\/sub> = permittivity of free space,
\nq = total cherge enclosed by the surface.<\/p>\n

Importance:<\/p>\n

    \n
  1. Using Gauss law we can find field due to a distribution of charge.<\/li>\n
  2. Gauss’s law is often useful towards a much easier calculation of the electrostatic field when the system has symmetry.<\/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 wire. (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 :
    \nThe number of electric lines of force passing normally through a given surface is called “electric flux” (\u03a6).<\/p>\n

    Expression for electric intensity :
    \nLet us consider an infinitely long thin straight wire having linear charge density \u03bb. (\u2235 \u03bb= Q\/L)<\/p>\n

    Consider a cylindrical Gaussian surface ABCD of length ‘l’ and radial distance r. The electric field \\(\\overline{\\mathrm{E}}\\) is radial and which perpendicular to the length of the wire.<\/p>\n

    The flux through the flat surfaces AB and CD are zero. (\u2235 \\(\\overrightarrow{E}\\) \u22a5 \\(\\overrightarrow{A}\\))
    \n\"TS<\/p>\n

    The flux through the curved surface ABCD is given by
    \n\"TS<\/p>\n

    \u03bb is positive \u21d2 direction will be radially outwards
    \n\u03bb is negative \u21d2 direction will be radially inwards 2<\/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 :
    \nThe total electrical flux through any closed surface is equal to 1\/\u03b50<\/sub> times the net charge enclosed by that surface.<\/p>\n

    Expression for electric intensity:
    \n\"TS
    \nConsider an infinite plane sheet ABCD of uniform surface charge density ‘\u03c3’.
    \nSurface charge density
    \n\"TS
    \nTake a Gaussian surface in the form of a rectangular parallelopiped.
    \n\"TS
    \nAssume the area of cross section of the two surfaces (1) and (2) be ‘S’. These two surfaces only will contribute to electric flux, since \\(\\overrightarrow{E}\\) and area vector \\(\\overrightarrow{ds}\\) are parallel. The remaining surfaces will give rise to zero flux as \\(\\overrightarrow{E}\\) and \\(\\overrightarrow{ds}\\) are perpendicular.
    \nThe total flux through the surface
    \n\"TS
    \nThis field is independent of distance of the point, this field is a uniform field.<\/p>\n

    Question 3.
    \nApplying Gauss’s law derive the expression for electric intensity due to a charged conducting spherical shell at (i) a point outside the shell (ii) a point on the surface of the shell and (iu) a point inside the shell.
    \nAnswer:
    \nConsider a charged spherical shell of radius R and of uniform surface charge density \u03c3.
    \n\"TS
    \n1) Field out side the shell:
    \nConsider a point P outside the shell with a radius vector \\(\\overrightarrow{r}\\). (\\(\\overrightarrow{r}\\) > \\(\\overrightarrow{R}\\))
    \nNow consider a Gaussian surface which is spherical of radius r.
    \nAs all the points on this surface are at same distance.
    \nThe flux through the Gaussian surface
    \n\"TS
    \nThe charge enclosed by the
    \nGaussian surface is q = \u03c3.(4\u03c0r\u00b2)
    \n\"TS<\/p>\n

    2) Field at a point on the shell:
    \nIf the point lies on the surface of the shell
    \n\u21d2 r = R.
    \n\"TS<\/p>\n

    3) Field at a point inside the shell:
    \nConsider a spherical Gaussian surface passing through P inside the shell with centre as \u2019O’.
    \n\"TS
    \nThe flux through this surface is
    \n\"TS
    \nBut the there is no charge enclosed by the surface i.e., q = 0.
    \n\"TS
    \n\u2234 The field inside a uniformly charged thin shell at all points inside is zero.<\/p>\n

    Intext Question and Answer<\/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 0.5m, find the charge on each ball.
    \nAnswer:
    \nMass of each ball = 0.20g = 20 \u00d7 10-4<\/sup>kg
    \nAngle between them \u03b8 = 60\u00b0
    \nSeparation between balls = 0.5m
    \n\"TS
    \nAt equilibrium
    \nElectrostatic force F is
    \nbalanced by component of weight mg sin \u03b8.
    \n\"TS
    \n\"TS<\/p>\n

    Question 2.
    \nAn infinite number of charges each of magnitude q are placed on x – axis at distances of 1, 2, 4,8, …………. meter from the origin respectively. Find intensity of the electric field at origin.
    \nAnswer:
    \nThe charges are as shown.
    \n\"TS<\/p>\n

    Question 3.
    \nA clock face has negative charges-q, -2q, -3q, …….. – 12q fixed at the position of the corresponding numberals 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?
    \nAnswer:
    \nNegative charges are arranged on the clock as shown.
    \n\"TS
    \nCharge arrangement
    \nElectric field due to – q = \\(\\frac{1}{4 \\pi \\varepsilon_0} \\frac{\\mathrm{q}}{\\mathrm{r}^2}\\) = say k<\/p>\n

    Where r is distance from centre ‘O’ to the numbers on dial
    \nField due to- 2q = 2k; due to -3q = 3k ……….. Field due to -12q = 12k
    \nElectric field is directed as shown in figure.
    \n\"TS
    \nThey are as shown in figure.
    \nConsider E7<\/sub> and E9<\/sub> magnitude of each = 6k angle between them is 60\u00b0.
    \n\u2234 Resultant of E7<\/sub> & E9<\/sub> is along E8<\/sub>.
    \nMagnitude is
    \n\\(\\sqrt{6 \\mathrm{k}^2+6 \\mathrm{k}^2+2 \\times 6 \\mathrm{k} \\times 6 \\mathrm{k} \\times \\cos \\theta}\\) say x.
    \nField along E8<\/sub> = 6k + x …………. (1)
    \nConsider E10<\/sub> and E12<\/sub> their magnitudes are 6k and 6k angle between them is 60\u00b0.
    \nResultant of E10<\/sub>, E12<\/sub> is along E11<\/sub>.
    \nResultant field along
    \nE11<\/sub> = \\(\\sqrt{6 \\mathrm{k}^2+6 \\mathrm{k}^2+2 \\times 6 \\mathrm{k} \\times 6 \\mathrm{k} \\cdot \\cos 60^{\\circ}}\\) say y.
    \nNow x, y are equal.
    \nTotal field along E11<\/sub> = 6k + y ……….. (2)
    \nFrom eq 1, 2 magnitudes of E8<\/sub>, E11<\/sub> are equal and angle between them is 90\u00b0.
    \n\u2234 Angle of resultant \u03b8 = 45\u00b0 with E8<\/sub>.
    \nIn clock Anglb between each digit say 8 & 9 = 30\u00b0
    \n9 & 10 = 30\u00b0 i.e., 1 hour corresponds to 30\u00b0 angle so angle 45\u00b0 \u21d2 1\\(\\frac{1}{2}\\) hour
    \nDirection of resultant field = 8 + 1\\(\\frac{1}{2}\\) = 9\\(\\frac{1}{2}\\) hours = 9 hours 30 minutes.<\/p>\n

    \"TS<\/p>\n

    Question 4.
    \nConsider a uniform electric field E = 3 \u00d7 10\u00b3 N\/C. (a) What is the flux of this field through a square of 10 cm on aside whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60\u00b0 angle with the x – axis?
    \nAnswer:
    \nIntensity of electric field E = 3 \u00d7 10\u00b3 N\/C along x-axis.
    \nArea of Square = 1\u00b2 = 10 \u00d7 10 = 100 cm\u00b2
    \n= 100 \u00d7 10-4<\/sup> m\u00b2 = 10-2<\/sup>m\u00b2<\/p>\n

    (a) When plane of square is parallel to y – z plane it is perpendicular to x-axis
    \n\u21d2 0 = 90\u00b0
    \n\u2234 Flux through square \u03a6 = \\(\\overline{\\mathrm{E}}.\\overline{\\mathrm{A}}\\) (or)
    \n\u03a6 = EA cos \u03b8
    \n\u03a6 = 3 \u00d7 10\u00b3 \u00d7 10-2<\/sup> = 3 \u00d7 10 = 30 vm<\/p>\n

    (b) When square makes an angle \u03b8 = 60\u00b0
    \nwith x- axis, \u03b8 =60\u00b0.
    \n\"TS<\/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 corners of a square of side ‘L’, one to each corner, 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?
    \nAnswer:
    \nGiven two charges are +ve and two charges are -ve.
    \n\"TS
    \nForce on any charge is directed towards centre.
    \nSide of square = L at point 3
    \nConsider charge 3 total forces on it are
    \n\"TS<\/p>\n

    Question 6.
    \nThe electric field in a region is given by \\(\\overrightarrow{\\mathbf{E}}=\\mathbf{a} \\hat{\\mathbf{i}}+\\mathbf{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.
    \nAnswer:
    \nElectric field \\(\\overrightarrow{\\mathbf{E}}=\\mathbf{a} \\hat{\\mathbf{i}}+\\mathbf{b} \\hat{\\mathbf{j}}\\)
    \nSide of square = L
    \n\u2234 Area of square = L\u00b2
    \nGive square is parallel to y – z plane \u21d2 it is perpendicular x – axis \u21d2 Area vector
    \n\"TS<\/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 center of the cube coincides with the center of the shell. Calculate the electric flux that comes out of a face of the cube.
    \nAnswer:
    \n(Charge density on sphere = \u03c3. But \u03c3 = \\(\\frac{Q}{A}\\)
    \nArea of spere A = 4\u03c0r\u00b2
    \n\u21d2 Charge Q = 4\u03c0r\u00b2\u03c3
    \nFor a point out side the sphere it seems to be concentrated at centre.
    \n\u2234 Charge at centre of cube Q = 4\u03c0r\u00b2\u03c3)
    \nFrom gauss’s law total flux comming out of
    \n\"TS<\/p>\n

    Question 8.
    \nAn electric dipole consists of two equal and opposite point charges + Q and – Q, separated by a distance 2l. P is a point collinear with the charges such that is distance from the positive charge is half of its distance from the negative charge.
    \nAnswer:
    \nEach charge on dipole = q, -q
    \nSeparation between charges = 2l
    \n\"TS
    \nP is on the line joining the charges \u21d2 it is on axial line.
    \nGiven : distance from +ve’ charge = \\(\\frac{1}{2}\\) distance from -ve’ charge.
    \nFrom given data d – l = \\(\\frac{1}{2}\\) ( d + l) \u21d2 2d – 2l = d + l (or) d = 3l
    \nIntensity of electric field at any point on
    \n\"TS<\/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. The intensity of the electric field at a point midway between them is
    \nAnswer:
    \nCharge densities of infinitely long conductors = \u03bb and 2\u03bb
    \nDistance between conductors = r
    \nIntensity of electric field of mid point = ?
    \nFor mid point distance d = r\/2
    \nIntensity of electric field due to a long conductor.
    \n\"TS<\/p>\n

    Question 10.
    \nTwo infinitely long thin straight wires having uniform linear charge densities \\(\\ddot{\\mathrm{e}}\\) and 2\\(\\ddot{\\mathrm{e}}\\) are arranged parallel to each other at a distance r apart. The intensity of the electric field at a point midway between them is
    \nAnswer:
    \nLinear charge densities \u03c31<\/sub> = \u03c32<\/sub> and \u03c32<\/sub> = 2e.
    \nSeparation between two parallel conductors d = r
    \nFor mid-way between them d = r\/2
    \n\"TS
    \nAt midpoint intensities are in oppisite direction so resultant intensity ER<\/sub> = E1<\/sub> ~ E2<\/sub>
    \nIntensity of electric field from an infinitely long charged conductor E = \\(\\frac{\\lambda}{2 \\pi \\varepsilon_0r}\\)
    \n\"TS
    \nIntensity of electric field at mid point E = \\(\\frac{e}{\\pi \\varepsilon_0r}\\)<\/p>\n

    \"TS<\/p>\n

    Question 11.
    \nAn electron of mass m and chargee is fired perpendicular to a uniform electric field of intensity E with an initial velocity u. If the electron traverses a distance x in the field in the direction of firing. Find the transverse displacement y it suffers.
    \nAnswer:
    \nMass of electron = m ;
    \nCharge on electron = e
    \nIntensity of electric field = E
    \n\u2234 Force F = E . e
    \nInitial velocity of electron = u; acceleration of electron a = F\/m = E.e\/m<\/p>\n

    Distance travelled S = x. along x-axis
    \n\u21d2 time t = \\(\\frac{x}{4}\\) …………. (1)
    \nDistance travelled along y-axis = ?
    \nInitial velocity along y-axis = u = 0
    \nVertical displacement y = ut + \\(\\frac{1}{2}\\) at\u00b2 = \\(\\frac{1}{2}\\)at\u00b2
    \n\u21d2 y = \\(\\frac{1}{2}\\frac{Ee}{m}.\\frac{x}{u^2}\\)
    \n\u2234 Vertical displacement after travelling a eEx2 distance x is y = \\(\\frac{eEx^2}{2mu^2}\\)<\/p>\n

    Additional 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?
    \nAnswer:
    \nRepulsive force of magnitude 6 \u00d7 10-3<\/sup> N ;
    \nCharge on the first sphere, q: = 2 \u00d7 10-7<\/sup>
    \nC Charge on the second sphere, q\u00b2 = 3 \u00d7 10-7<\/sup> C ;
    \nDistance between the spheres, r = 30 cm = 0.3 m
    \nElectrostatic force between the spheres is given by the relation,
    \n\"TS
    \nHence, force between the two small charged spheres is 6 \u00d7 10-3<\/sup> N. The charges are of same nature. Hence, force between them will be repulsive.<\/p>\n

    Question 2.
    \nThe electrostatic force on a small sphere of charge 0.4 \u00b5C due to another small sphere of charge – 0.8 \u00b5C in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?
    \nAnswer:
    \na) Electrostatic force on the first sphere, F = 0.2 N
    \nCharge on this sphere, = 0.4 \u00b5C
    \n= 0.4 \u00d7 10-6<\/sup> C
    \nCharge on the second sphere,
    \nq2<\/sub> = -0.8 \u00b5C = -0.8 \u00d7 10-6<\/sup><\/p>\n

    Electrostatic force between the spheres
    \n\"TS<\/p>\n

    The distance between the two spheres is 0.12 m.<\/p>\n

    b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.<\/p>\n

    \"TS<\/p>\n

    Question 3.
    \nCheck that the ratio ke\u00b2\/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?
    \nAnswer:
    \n1) The given ratio is \\(\\frac{\\mathrm{ke^2}}{\\mathrm{Gm_em_p}}\\)
    \nWhere, G = Gravitational constant
    \nme<\/sub> and mp<\/sub> = Masses of electron and proton.; e = Electric charge is C.
    \nk = A constant. = \\(\\frac{1}{4 \\pi \\varepsilon_0}\\),
    \nWhere \u03b50<\/sub> = Permittivity of free space
    \nTherefore, unit of the given ratio
    \n\"TS
    \nHence, the given ratio is dimensionless,<\/p>\n

    ii) e = 1.6 \u00d7 10-19<\/sup>C; G = 6.67 \u00d7 10-11<\/sup> N m\u00b2kg-2<\/sup>; me<\/sub> = 9.1 \u00d7 10-31<\/sup> kg ; mp<\/sub> = 1.66 \u00d7 10-27<\/sup>kg
    \nHence, the numerical value of the given ratio is
    \n\"TS
    \nThis is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.<\/p>\n

    Question 4.
    \na) Explain the meaning of the statement \u2018electric charge of a body is quantised\u2019.
    \nb) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?
    \nAnswer:
    \na) Electric charge of a body is quantized. This means that only integral (1, 2, …………, n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.<\/p>\n

    b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.<\/p>\n

    Question 5.
    \nFour point charges qA<\/sub> = 2 \u00b5C, qB<\/sub> = – 5\u00b5C, qC<\/sub> = 2 \u00b5C, and qD<\/sub> = – 5 \u00b5C are located at thecomereofasquare ABCD of side 10cm. What is the force on a charge of 1 \u00b5C placed at the centre of the square?
    \nAnswer:
    \nThe given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.
    \n\"TS
    \nWhere, (Sides) AB = BC = CD = AD = 10 cm
    \n(Diagonals) AC = BD = 10\u221a2 cm
    \nAO = OC = DO = OB = 5\u221a2 cm
    \nA charge of amount 1 \u00b5C is placed at point O.<\/p>\n

    Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 \u00b5C charge at centre O is zero.<\/p>\n

    \"TS<\/p>\n

    Question6.
    \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 is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.<\/p>\n

    b) If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never crosr. each other.<\/p>\n

    Question 7.
    \nAn electric dipole with dipole moment 4 \u00d7 10-9<\/sup> C m is aligned at 30\u00b0 with the direction of a uniform electric field of magnitude 5 \u00d7 104<\/sup> N C-1<\/sup>. Calculate the magnitude of the torque acting on the dipole.
    \nAnswer:
    \nElectric dipole moment, p = 4 \u00d7 10-9<\/sup> C m ;
    \nElectric field, E = 5 \u00d7 104<\/sup> N C-1<\/sup>
    \nAngle made by p with a uniform electric field, \u03b8 = 30\u00b0<\/p>\n

    Torque acting on the dipole is given by the relation, \u03c4 = pE sin \u03b8
    \n= 4 \u00d7 10-9<\/sup> \u00d7 5 \u00d7 104<\/sup> \u00d7 sin 30
    \n= 20 \u00d7 10-5<\/sup> \u00d7 1\/2 = 10-4<\/sup> Nm<\/p>\n

    Therefore, the magnitude of the torque acting on the dipole is 10-4<\/sup> N m.<\/p>\n

    Question 8.
    \na) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 \u00d7 10-7<\/sup> C? The radii of A and B are negligible compared to the distance of separation.
    \nb) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
    \nAnswer:
    \na) Charge on sphere A, qA<\/sub> = Charge on sphere B, qB<\/sub> = 6.5 \u00d7 10-7<\/sup> C
    \nDistance between the spheres,
    \nr = 50 cm = 0.5 m<\/p>\n

    Force of repulsion between the two
    \n\"TS<\/p>\n

    Therefore, the force between the two spheres is 1.52 \u00d7 10-2<\/sup> N.<\/p>\n

    b) After doubling the charge, charge on sphere A, qA<\/sub> = Charge on sphere B,
    \nqB<\/sub> = 2 x 6.5 \u00d7 10-7<\/sup>C = 1.3 \u00d7 10-6<\/sup>C
    \nThe distance between the spheres is halved.
    \n\u2234 r = \\(\\frac{0.5}{2}\\) = 0.25 m
    \nForce of repulsion between the two spheres,
    \n\"TS
    \nTherefore, the force between the two spheres is 0.243 N.<\/p>\n

    Question 9.
    \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\"TS
    \nAnswer:
    \nOpposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively chargee plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.<\/p>\n

    The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.<\/p>\n

    Question 10.
    \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:
    \nAll the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.<\/p>\n

    \"TS<\/p>\n

    Question 11.
    \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 10\u00b3 N m\u00b2\/C.
    \na) What is the net charge inside the box?
    \nb) 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?
    \nAnswer:
    \na) Net outward flux through the surface of the box, \u03a6 = 8.0 \u00d7 10\u00b3 N m\u00b2\/C
    \nFor a body containing net charge q, flux is given by the relation, \u03a6 = \\(\\frac{q}{\\varepsilon_0}\\)
    \n\u03b50<\/sub> = Permittivity of free space = 8.854 \u00d7 10-12<\/sup> N-1<\/sup>C\u00b2m-2<\/sup>
    \nq = \u03b50<\/sub>\u03a6 = 8.854 \u00d7 10-12<\/sup> \u00d7 8.0 \u00d7 10\u00b3
    \n= 7.08 \u00d7 10-8<\/sup> = 0.07 \u00b5C
    \nTherefore, the net charge inside the box is 0.07 \u00b5C.<\/p>\n

    b) No
    \nNet flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.<\/p>\n

    Question 12.
    \nA point charge + 10 \u00b5C is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Figure. 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\"TS
    \nAnswer:
    \nThe square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss\u2019s theorem for a cube, total electric flux is through all its six faces.
    \nHence, electric flux through one face of the cube
    \n\u03a6total<\/sub> = \\(\\frac{q}{\\varepsilon_0}\\)
    \nHence, electric flux through one fact of the cube i.e., through the square,
    \n\"TS
    \nTherefore, electric flux through the square is 1.88 \u00d7 105<\/sup> Nm\u00b2C-1<\/sup><\/p>\n

    Question 13.
    \nA point charge of 2.0 \u00b5C is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
    \nAnswer:
    \nNet electric flux (\u03a6net<\/sub>) through the cubic surface is given by, \u03a6net<\/sub> = \\(\\frac{q}{\\varepsilon_0}\\)
    \nWhere, \u03b50<\/sub> = Permittivity of free space = 8.854 \u00d7 10-12<\/sup>N-1<\/sup>C\u00b2m-2<\/sup>
    \nq = Net charge contained inside the cube = 2.0 \u00b5C = 2 \u00d7 10-6<\/sup> C
    \n\"TS
    \nThe net electric flux through the surface is 2.26 \u00d7 105<\/sup> Nm\u00b2C-1<\/sup>.<\/p>\n

    Question 14.
    \nA point charge causes an electric flux of -1.0 \u00d7 10\u00b3 Nm\u00b2\/C to pass through a sphe\u00acrical Gaussian surface of 10.0 cm radius centered on the charge.
    \na) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
    \nb) What is the value of the point charge?
    \nAnswer:
    \na) Electric flux, \u03a6 = -1.0 \u00d7 10\u00b3 Nm\u00b2\/C ;
    \nRadius of the Gaussian surface, r = 10.0 cm
    \nElectric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., – 10\u00b3 N m\u00b2\/C.<\/p>\n

    b) Electric flux is given by the relation, \u03a6 = \\(\\frac{q}{\\varepsilon_0}\\)
    \nWhere, q = Net charge enclosed by the spherical surface
    \n\u03b50<\/sub> = Permittivity of free space
    \n= 8.854 \u00d7 10-12<\/sup> N-1<\/sup>C\u00b2m-2<\/sup>
    \n\u2234 q = \u03a6\u03b50<\/sub> = – 1.0 \u00d7 10\u00b3 \u00d7 8.854 \u00d7 10-12<\/sup>
    \n= – 8.854 \u00d7 10-9<\/sup> C = – 8.854 nC<\/p>\n

    Therefore, the value of the point charge is-8.854 nC.<\/p>\n

    Question 15.
    \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 10\u00b3 N\/C and points radially inward, what is the net charge on the sphere?
    \nAnswer:
    \nElectric field intensity (E) at a distance (d) from the centre of a sphere containing net charge q is given by the relation,
    \n\"TS
    \nTherefore, the net charge on the sphere is 6.67 nC.<\/p>\n

    Question 16.
    \nA uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 \u00b5C\/m\u00b2.
    \na) Find the charge on the sphere.
    \nb) What is the total electric flux leaving the surface of the sphere?
    \nAnswer:
    \na) Diameter of the sphere, d = 2.4 m ;
    \nRadius of the sphere, r = 1.2 m
    \nSurface charge density \u03c3 = 80.0 \u00b5C\/m\u00b2
    \n= 80 \u00d7 10-6<\/sup>C\/m\u00b2<\/p>\n

    Total charge on the surface of the sphere,
    \nQ = Charge density \u00d7 Surface area
    \nQ = \u03c3 \u00d7 4\u03c0r\u00b2 = 80 \u00d7 10-6<\/sup> \u00d7 4 \u00d7 3.14 \u00d7 (1.2)\u00b2
    \n= 1.447 \u00d7 10-3<\/sup>C
    \nTherefore, the charge on the sphere is 1.447 \u00d7 10-3<\/sup> C.<\/p>\n

    b) Total electric flux (\u03a6total<\/sub>) leaving out the surface of a sphere containing net charge
    \nQ is given by the relation, \u03a6total<\/sub> = \\(\\frac{Q}{\\varepsilon_0}\\)
    \n\"TS<\/p>\n

    Therefore, the total electric flux leaving the surface of the sphere is 1.63 \u00d7 10-8<\/sup>NC-1<\/sup>m\u00b2<\/p>\n

    \"TS<\/p>\n

    Question 17.
    \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.
    \nAnswer:
    \nElectric field produced by the infinite line charges at a distance d having linear charge density \u03bb is given by the relation,
    \n\"TS
    \nTherefore, the linear charge density is 10 \u00b5C\/m<\/p>\n","protected":false},"excerpt":{"rendered":"

    Telangana TSBIE\u00a0TS Inter 2nd Year Physics Study Material 4th Lesson Electric Charges and Fields Textbook Questions and Answers. TS Inter 2nd Year Physics Study Material 4th Lesson Electric Charges and Fields Very Short Answer Type Questions Question 1. What is meant by the statement ‘charge is quantized’? 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TS Inter 2nd Year Physics Study Material 4th Lesson Electric Charges and Fields Very Short Answer Type Questions Question 1. What is meant by the statement ‘charge is quantized’? Answer: The minimum charge that can be transferred ... 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