Q: Six infinitely large and thin non-conducting sheets are fixed in configurations I and II. As shown in the figure, the sheets carry uniform surface charge densities which are indicated in terms of _0.

aligned ( E _4 )_ I = & _0 2 _0 [1-1+1-1-1+1]=0 \\ ( ~V _ First )_ I & = - _0 2 _0 [-1+2-3+4-5] d \\ & = - _0 2 _0 [-3] d = _0 3 ~d 2 _0 aligned aligned & aligned (V_ Last )_I & = - _0 2 _0 [1-2+3-4+5] \\ & = _0 2 _0 [-3 ~d ] aligned \\ & aligned (V_ First -V_ Last )_I & = 3 _0 ~d _0 \\ & = 3 9 10^ -6 1 10^ -6 9 10^ -12 =3 volt aligned aligned aligned & aligned ( E _3 )_ II = & _0 2 _0 [ 1 2 -1+1+1-1+ 1 2 ]= _0 2 _0 \\ & = - _0 2 _0 [-1+2-3+4- 5 2 ] d \\ & = - _0 2 _0 [2-2.5] d = _0 ~d 4 _0 alig

Q: A small electric dipole p _0, having a moment of inertia I about its center, is kept at a distance r from the center of a spherical shell of radius R. The surface charge density is uniformly distribut

$ aligned & =| p E | \\ & I =p_0 E \\ & = p I ( 1 4 _0 4 R^2 r^2 ) \\ & = ( p 0 R^2 I _0 r^2 ) \\ & = p 0 R^2 I _0 r^2 aligned $ For $r=2 R$ $ = p_0 4 / _0 ( C $ is incorrect $)$ Also, for $r=10 R$ $ = p_0 4 l(100) $ (D is correct) It will oscillate for any finite value of $r>R$. (B is correct)

Q: An infinitely long thin wire, having a uniform charge density per unit length of 5 nC / m , is passing through a spherical shell of radius 1 ~m , as shown in the figure. A 10 nC charge is distributed

due to wire aligned & dV =- E dx \\ & _ v _ P ^ v _ R dV =- _ 0.5 ^2 2 k x dx \\ & v _ R - v _ P =-2 k 2 0.5 \\ & =-2 9 10^9 3 10^ -9 2 0.7=-126 ~V aligned due to sphere aligned & v _ R - v _P= kQ 2 - kQ 1 =- kQ 2 = -9 10^9 10 10^ -9 2 \\ &=-45 ~V \\ & v _ R - v _ P =-126-45=-171 ~V \\ & v _ P - v _ R =171 ~V aligned

Q: Two beads, each with charge q and mass m, are on a horizontal, frictionless, non-conducting, circular hoop of radius R. One of the beads is glued to the hoop at some point, while the other one perform

Restoring force = qE ( 2 ) aligned & = qE ( 2 ) R = I \\ & E = Kq (2 R 2 )^2 = 1 4 _0 q 4 R ^2 ^2 ( 2 ) \\ & 1 4 _0 qR 4 R ^2 ^2 ( 2 ) ( 2 ) q = mR ^2 aligned For very small, aligned & - q ^2 32 _0 R ^3 ~m = \\ & ^2= q ^2 32 _0 mR ^3 aligned Hence option (2)

Question 1

Two charges Q_1 = q and Q_2 = mq are placed at the points P_1(a, b) and P_2(ma, mb), respectively, in the XY plane, where a, b 0 and m 0, 1. If V_1 is the potential at a point in the XY plane due to c

Accepted Answer

The potential at a point (x, y) due to charge Q_1 is V_1 = 1 4 _0 q (x-a)^2 + (y-b)^2 . The potential at (x, y) due to charge Q_2 is V_2 = 1 4 _0 mq (x-ma)^2 + (y-mb)^2 . Given |V_1| = |V_2|, we have: 1 (x-a)^2 + (y-b)^2 = m^2 (x-ma)^2 + (y-mb)^2 Cross-multiplying and expanding both sides: (x-ma)^2 + (y-mb)^2 = m^2[(x-a)^2 + (y-b)^2] x^2 + m^2a^2 - 2max + y^2 + m^2b^2 - 2mby = m^2(x^2 + a^2 - 2ax + y^2 + b^2 - 2by) x^2 + y^2 - 2m(ax+by) + m^2(a^2+b^2) = m^2(x^2+y^2) - 2m^2(ax+by) + m^2(a^2+b^2)

Question 2

Consider an electric dipole comprising two charges +q and -q each with mass m, separated by a fixed distance d and initially at rest with its dipole moment pointing along i . A uniform electric field

Accepted Answer

The net force on the dipole in a uniform electric field is zero, as the forces on the two charges are equal and opposite (qE j and -qE j ). Therefore, the center of mass remains at rest and is not deflected. Option (A) is incorrect. The moment of inertia of the dipole about its center of mass is I = m ( d 2 )^2 + m ( d 2 )^2 = md^2 2 . The torque on the dipole is = p E . The angle between the dipole moment p and the electric field E is 90^ - . The magnitude of the torque is = pE (90^ - ) = qdE .

Question 3

List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole has a dipole moment of magnitude p, oriented as marked by arrows in the figures. In all the configura

Accepted Answer

E_ net = -2 k P r^3 j E_ n e t = -P j 2 _0 r^3 E _ net =0 2 P i 4 _0 r ^3 - P j 4 _0 r ^3 E _ net = 4 kP i r ^3 P 2, Q 1, R 4, ~S 5 Ans. (C)

Question 4

Two co-axial conducting cylinders of same length with radii 2 R and 2 R are kept, as shown in Fig. 1. The charge on the inner cylinder is Q and the outer cylinder is grounded. The annular region betwe

Accepted Answer

Here we are assuming that " " is very large just for the sake of symmetry. Outside cylinder will have zero electric field inside, so the flux generated on the plate will be due to inner cylinder only in sections AB and CD , as section be will be at that place where electric field is zero. flux through element will be aligned d & = E dS \ d & = 2 k dy ..... (1) aligned from figure we can say that aligned & = R r r = R \ & = y R y = R \ & dy = R ^2 ~d \ & ~d = 2 k R R ^2 ~d \ & ~d =2 k dq \

Question 5

A positive point charge of 10^ -8 C is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm . The sphere is then grounded and the charge on the sphere is measured

Accepted Answer

Before grounding aligned & V _ sphere = ( V _ C )_ net = ( V _ C )_ q + ( V _ C )_ ind \ & V _ sphere = kq +0= 9 10^9 10^ -8 0.2 = 90 0.2 = 900 2 =450 volt aligned After grounding aligned & kQ + kq _ S R = V _ sphere ^ =0 \ & q _ s =- R q =- 1 2 10^ -8 =-5 10^ -9 \ & q _ s =-5 10^ -9 Coulomb aligned Charge flower from sphere to ground =5 10^ -9 Coulomb After grounding is removed aligned & ( V _ sphere )_ final = kq ^ + kq _ s R \ & = 9 10^9 10^2 10^ -8 30 - 9 10^9 5 10^ -9 10^2 10 \ & = 9 1

Question 6

Six infinitely large and thin non-conducting sheets are fixed in configurations I and II. As shown in the figure, the sheets carry uniform surface charge densities which are indicated in terms of _0.

Accepted Answer

aligned ( E _4 )_ I = & _0 2 _0 [1-1+1-1-1+1]=0 \ ( ~V _ First )_ I & = - _0 2 _0 [-1+2-3+4-5] d \ & = - _0 2 _0 [-3] d = _0 3 ~d 2 _0 aligned aligned & aligned (V_ Last )_I & = - _0 2 _0 [1-2+3-4+5] \ & = _0 2 _0 [-3 ~d ] aligned \ & aligned (V_ First -V_ Last )_I & = 3 _0 ~d _0 \ & = 3 9 10^ -6 1 10^ -6 9 10^ -12 =3 volt aligned aligned aligned & aligned ( E _3 )_ II = & _0 2 _0 [ 1 2 -1+1+1-1+ 1 2 ]= _0 2 _0 \ & = - _0 2 _0 [-1+2-3+4- 5 2 ] d \ & = - _0 2 _0 [2-2.5] d = _0 ~d 4 _0 alig

Question 7

A small electric dipole p _0, having a moment of inertia I about its center, is kept at a distance r from the center of a spherical shell of radius R. The surface charge density is uniformly distribut

Accepted Answer

$ aligned & =| p E | \ & I =p_0 E \ & = p I ( 1 4 _0 4 R^2 r^2 ) \ & = ( p 0 R^2 I _0 r^2 ) \ & = p 0 R^2 I _0 r^2 aligned $ For $r=2 R$ $ = p_0 4 / _0 ( C $ is incorrect $)$ Also, for $r=10 R$ $ = p_0 4 l(100) $ (D is correct) It will oscillate for any finite value of $r>R$. (B is correct)

Question 8

A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle at P , as shown in the figure. When =30^ , then the electric flux through the curved surface of the

Accepted Answer

Solid angle made by plane surfaces =2 2 (1- ) =4 -4 So solid angle made by curved surface =4 - aligned & =4 -(4 -4 )=4 \ & _ 30^ = = 4 30^ 4 Q _0 = 30^ Q _0 \ & _ 60 = 4 60^ 4 Q _0 = 60^ Q _0 \ & _ 30 _ 60 = 30^ 60^ = 3 \ & _ 60 = 3 \ & _ 60 = 3 n =3 aligned

Question 9

An infinitely long thin wire, having a uniform charge density per unit length of 5 nC / m , is passing through a spherical shell of radius 1 ~m , as shown in the figure. A 10 nC charge is distributed

Accepted Answer

due to wire aligned & dV =- E dx \ & _ v _ P ^ v _ R dV =- _ 0.5 ^2 2 k x dx \ & v _ R - v _ P =-2 k 2 0.5 \ & =-2 9 10^9 3 10^ -9 2 0.7=-126 ~V aligned due to sphere aligned & v _ R - v _P= kQ 2 - kQ 1 =- kQ 2 = -9 10^9 10 10^ -9 2 \ &=-45 ~V \ & v _ R - v _ P =-126-45=-171 ~V \ & v _ P - v _ R =171 ~V aligned

Question 10

Two beads, each with charge q and mass m, are on a horizontal, frictionless, non-conducting, circular hoop of radius R. One of the beads is glued to the hoop at some point, while the other one perform

Accepted Answer

Restoring force = qE ( 2 ) aligned & = qE ( 2 ) R = I \ & E = Kq (2 R 2 )^2 = 1 4 _0 q 4 R ^2 ^2 ( 2 ) \ & 1 4 _0 qR 4 R ^2 ^2 ( 2 ) ( 2 ) q = mR ^2 aligned For very small, aligned & - q ^2 32 _0 R ^3 ~m = \ & ^2= q ^2 32 _0 mR ^3 aligned Hence option (2)

Question 11

An electric dipole is formed by two charges  + q and - q located in x y -plane at ( 0 , &#160; 2 ) &#160; mm and ( 0 , &#160; &#8722; 2 ) &#160; mm , respectively, as shown in the figure. The ele

Accepted Answer

Electric potential at any general point for vertical dipole is written as,  V 1 &#8733; p 1 cos &#952; 1 r 1 2 . Electric potential at any general point for  V 2 &#8733; p 2 cos &#952; 2 r 2 2 . For first case:  tan &#952; 1 = 100 100 = 1 &#8658; &#952; 1 = 45 &#176; For second case: slope in this is  = - 2 - 2 1 - - 1 = - 2 which is equal to  tan 45 &#176; + &#952; 2 = - 2 &#8658; 1 + tan &#952; 2 1 - tan &#952; 2 = - 2 &#8658; 1 + tan &#952; 2 = - 2 + 2 tan &#952; 2 &#

Question 12

A charge q  is surrounded by a closed surface consisting of an inverted cone of height h  and base radius R , and a hemisphere of radius R as shown in the figure. The electric flux through t

Accepted Answer

According to the Gauss law, the total flux linked with a closed surface is 1 &#949; 0  times the charge enclosed by the closed surface.  As we can see from the diagram, the given shape encloses the charge  q , therefore from Gauss law, we can write total flux  = q &#949; 0 . Now, the plane connecting hemispherical shape & conical shape cuts the charge in two equal parts, therefore flux through the conical part will be exactly half of the total flux value. Hence, flux thro

Question 13

In the figure, the inner (shaded) region  A represents a sphere of radius r A = 1 , within which the electrostatic charge density varies with the radial distance r from the center as &rho; A = k

Accepted Answer

Since both the densities are positive, the total charge can not be zero. Hence, option A is incorrect. For total charge, Q Total = &#8747; 0 r A k r 4 &#960; r 2 d r + &#8747; r A r B 2 k r 4 &#960; r 2 d r = 4 &#960; k 4 r A 4 + 8 &#960; k 2 r B 2 - r A 2 = &#960; k + 4 &#960; k r B 2 - r A 2 If  r B = 3 2 Q Total &#160; = &#960; k + 4 &#960; k 9 4 - 1 = &#960; k + 4 &#960; k 5 4 = 6 &#960; k The potential just outside  B  will be, V = 1 4 &#960; &#949; 0 Q total r B = 1 4 &#960;

Question 14

A disk of radius R  with uniform positive charge density &#963;  is placed on the x y  plane with its center at the origin. The Coulomb potential along the z -axis is V z = &#963; 2

Accepted Answer

For the particle to reach at origin, work done by the force should be greater than the increase in the potential energy of the charged particle. Therefore, c z &#8805; &#963; q 2 &#1013; 0 R - R 2 + z 2 - z &#8658; c 2 &#1013; 0 &#963; q &#8805; R - R 2 + z 2 - z z &#8658; &#946; &#8805; R - R 2 + z 2 - z z For option A R - R 2 + z 2 - z z = R - R 1 + 625 49 - 25 7 R 25 7 &#8771; 0 . 242 Since,  0 . 25 &#62; 0 . 242 Option A is correct. For option B R - R 2 + z 2 - z z = R - R 1 + 9 49 - 3

Question 15

Six charges are placed around a regular hexagon of side length a  as shown in the figure. Five of them have charge q , and the remaining one has charge x . The perpendicular from each charge to t

Accepted Answer

​​​​​​​ The given arrangements of the charges also form a regular hexagon with side length equal to  3 a When x = q , the situation is symmetric &rArr; Electric field at O  would be zero. &rArr; A is correct. When x = - q , we can think of x  as q + - 2 q . Therefore, the magnitude of the electric field will be equivalent to that if only charge of - 2 q  is kept at the the bottom point. Hence, E O = 1 4 &pi; ϵ 0 2 q 3 a 2 = 1 4 &pi; ϵ 0 2 q 3 a 2 = q 6 &pi; ϵ 0 a 2 &rArr

Question 16

Two point charges -Q and +Q / 3 are placed in the x y-plane at the origin (0,0) and a point (2,0), respectively, as shown in the figure. This results in an equipotential circle of radius R and potenti

Accepted Answer

Let  P h , k  be be a general point with potential to be zero. V 1 + V 2 = 0    (as potential given  = 0 ) K &#8722; Q h 2 + k 2 + K Q 3 h &#8722; 2 2 + k 2 = 0 1 3 h &#8722; 2 2 + k 2 = 1 h 2 + k 2 3 h &#8722; 2 2 + k 2 = h 2 + k 2 3 h 2 &#8722; 4 h + 4 + k 2 = h 2 + k 2 2 h 2 + 2 k 2 &#8722; 12 h + 12 = 0 h 2 + k 2 &#8722; 6 h + 6 = 0 h 2 &#8722; 6 h + 6 + k 2 = 0 h 2 &#8722; 6 h + 9 + k 2 = 3 h &#8722; 3 2 + k 2 = 3 Standard equation of circle is given by x &#8722; x

Question 17

Two point charges -Q and +Q / 3 are placed in the x y-plane at the origin (0,0) and a point (2,0), respectively, as shown in the figure. This results in an equipotential circle of radius R and potenti

Accepted Answer

Let  P h , k  be be a general point with potential to be zero. V 1 + V 2 = 0    (as potential given  = 0 ) K &#8722; Q h 2 + k 2 + K Q 3 h &#8722; 2 2 + k 2 = 0 1 3 h &#8722; 2 2 + k 2 = 1 h 2 + k 2 3 h &#8722; 2 2 + k 2 = h 2 + k 2 3 h 2 &#8722; 4 h + 4 + k 2 = h 2 + k 2 2 h 2 + 2 k 2 &#8722; 12 h + 12 = 0 h 2 + k 2 &#8722; 6 h + 6 = 0 h 2 &#8722; 6 h + 6 + k 2 = 0 h 2 &#8722; 6 h + 9 + k 2 = 3 h &#8722; 3 2 + k 2 = 3 Standard equation of circle is given by x &#8722; x

Question 18

Two large circular discs separated by a distance of 0 . 01 m are connected to a battery via a switch as shown in the figure. Charged oil drops of density 900 kg m - 3 are released through a tiny hole

Accepted Answer

q E = m g q = m g E = 900 &#215; 4 3 &#960; 8 &#215; 10 - 7 3 &#215; 10 200 0 . 01 = 900 &#215; 4 3 &#960; &#215; 512 &#215; 10 - 21 &#215; 10 2 &#215; 10 4 N = 900 &#215; 4 3 &#960; &#215; 512 &#215; 10 - 21 &#215; 10 2 &#215; 10 4 &#215; 1 . 6 &#215; 10 - 19 N &#8776; 6

Question 19

Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting, massless strings of same length. At equilibrium, the angle between th

Accepted Answer

T sin θ = F ; T cos θ = m g Divide, tan θ = F m g . . . 1 As force between two charged bodies doesn't depend upon medium. Hence, the force between them remains same because the distance spheres is same. When it is submerged in liquid then T ' sin θ = F ε r T ' cos θ = m g 1 - ρ ℓ ρ s Divide, tan θ = F mg 1 - ρ l ρ S ε r . . . 2 Equating (1) & (2) 1 - ρ ℓ ρ s = 1 ε r ; 1 - 800 ρ s = 1 21 ρ s = 840 kg m - 3 T ' = T ε r = T 21

Question 20

One end of a spring of negligible unstretched length and spring constant k is fixed at the origin ( 0 , 0 ) . A point particle of mass m carrying a positive charge q is attached at its other end. The

Accepted Answer

Original frequency f = 1 2 π k m If dipole appears, At equilibrium 2 K p ℓ + x 0 3 · q = k x 0 . . . i Now disturbed 2 K p ℓ + x 0 + x 3 · q - k x 0 + x = m a ⇒ m a = 2 K pq ℓ + x 0 3 1 + x ℓ + x 0 - 3 - k x 0 + x ⇒ m a = 2 K p q ℓ + x 0 3 1 - 3 x ℓ + x 0 - 3 - k x 0 + x = - 6 K p q x ℓ + x 0 4 - k x = - 3 x ℓ + x 0 k x 0 - k x = - k x 3 x 0 ℓ + x 0 + 1 As ℓ is negative, m a = - 4 k x ; ⇒ a = 4 k m x New frequency : f ' = 1 2 π 4 k m = 2 f = 1 δ k m ∴ δ = π = 3 . 14

Electrostatics — Previous Year Questions & Solutions