Q: An electric dipole with dipole moment 5 10^ -6 Cm is aligned with the direction of a uniform electric field of magnitude 4 10^5 ~N / C The dipole is then rotated through an angle of 60^ with respect t

Given, aligned & p =5 10^ -6 Cm \\ & E =4 10^5 ~N / C \\ & _ i =0^ , _ f =60^ \\ & U = U _ f - U _ i \\ & =- pE _ f - (- pE _ i ) \\ & pE ( _ i - _ f ) \\ & 5 10^ -6 4 10^5 (1- 1 2 )=1 ~J aligned

Question 1

A point charge Q is placed inside a cavity within a solid isolated conducting sphere. Consider points A, B and C as shown in the figure, where the magnitudes of the electric fields are E_A, E_B and E_

Accepted Answer

Point A is located inside the cavity where a point charge Q is present. The electric field inside the cavity is the superposition of the field due to the charge Q and the induced charge on the inner surface of the cavity. This resultant field is non-zero, so E_A 0. The point charge Q inside the cavity induces a charge -Q on the inner surface of the cavity and a charge +Q on the outer surface of the isolated conducting sphere. Because the outer surface of the sphere is perfectly spherical and the

Question 2

Consider a fixed uniformly charged insulating sphere with radius R and total charge +Q. A point charge -q (q Q) with mass m is released from rest at a distance of 3R from the centre of the charged sph

Accepted Answer

Initial potential energy of the system when the charge -q is at a distance of 3R from the centre is: U_i = - 1 4 _0 Qq 3R Final potential energy of the system when the charge reaches the surface of the sphere (distance R from the centre) is: U_f = - 1 4 _0 Qq R According to the principle of conservation of mechanical energy: K_i + U_i = K_f + U_f 0 - Qq 12 _0 R = 1 2 mv^2 - Qq 4 _0 R 1 2 mv^2 = Qq 4 _0 R - Qq 12 _0 R 1 2 mv^2 = Qq 4 _0 R ( 1 - 1 3 ) 1 2 mv^2 = 2Qq 12 _0 R v^2 = Qq 3 _0 mR v = Qq

Question 3

A unit positive point charge is taken slowly through an infinitesimally thin tube that is inside a charged dielectric sphere of radius R, having uniform positive charge density , as shown in the figur

Accepted Answer

Since the electrostatic field is conservative, the work done in moving a charge depends only on the potential difference between the initial and final positions, regardless of the path taken. The electric potential V(r) at a distance r (r R) from the centre of a uniformly charged solid sphere of radius R and charge density is given by: V(r) = 1 4 _0 Q r The total charge Q of the sphere is: Q = 4 3 R^3 Substituting Q into the potential formula, we get: V(r) = 1 4 _0 4 3 R^3 r = R^3 3 _0 r The ini

Question 4

Which of the following statements are correct ? A. Inside a conductor, the electrostatic field is zero. B. Electric field at the surface of a charged conductor does not depend on its surface charge de

Accepted Answer

Statement A is correct because in electrostatics, the free charges inside a conductor redistribute themselves such that the net electric field inside is zero. Statement B is incorrect because the electric field at the surface of a charged conductor is given by E = _0 , which directly depends on the surface charge density . Statement C is correct because, by Gauss's law, since the electric field inside the conductor is zero, the net charge enclosed by any Gaussian surface inside the conductor mus

Question 5

Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharge

Accepted Answer

F = kQ ^2 r ^2 on touching; F^ = k(Q / 2)(3 Q / 4) r^2 F^ = 3 F 8 After touching ( A ) \&( C ) Total charge q +0= q Q _ A ^ = q 2 , Q _ C ^ = q 2 (B) & (C) aligned & Total charge =q+ q 2 = 3 2 q \ & Q_B^ = 3 4 q Q_C^ = 3 4 q aligned

Question 6

An electric dipole with dipole moment 5 10^ -6 Cm is aligned with the direction of a uniform electric field of magnitude 4 10^5 ~N / C The dipole is then rotated through an angle of 60^ with respect t

Accepted Answer

Given, aligned & p =5 10^ -6 Cm \ & E =4 10^5 ~N / C \ & _ i =0^ , _ f =60^ \ & U = U _ f - U _ i \ & =- pE _ f - (- pE _ i ) \ & pE ( _ i - _ f ) \ & 5 10^ -6 4 10^5 (1- 1 2 )=1 ~J aligned

Question 7

A metal cube of side 5 cm is charged with 6 C . The surface charge density on the cube is

Accepted Answer

In metal all the charge is present on surface. Q=6 C Total surface area S=6 a^2 =6 (5 10^ -2 )^2 =6 25 10^ -4 =150 10^ -4 ~m ^2 Surface charge density = Q S = 6 10^ -6 150 10^ -4 =0.4 10^ -3 C m ^ -2

Question 8

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R . Assertion A: The potential (V) at any axial point, at 2 m distance (r) from the centre of the dip

Accepted Answer

The potential V at any point, at distance r from centre of dipole = K P r^2 At axial point where =0^ , V= K P r^2 = 9 10^9 4 10^ -6 2^2 =9 10^3 ~V At axial point where =180^ , V= -K P r^2 =-9 10^3 ~V

Question 9

A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V ) shown in the figure is: (Take 1 4 _0 =9 10^9 SI units)

Accepted Answer

For uniformly charged spherical shell, aligned & V= k q R ( For r R) \ & V_C=V_P \ & \ & V_C-V_P= Zero aligned

Question 10

The value of electric potential at a distance of 9 cm from the point charge 4 10^ -7 C is [ . Given . 1 4 _0 =9 10^9 ~N ~m ^2 C ^ -2 ] :

Accepted Answer

V= k q r =9 10^9 4 10^ -7 9 10^ -2 =4 10^4 ~V

Electrostatics — Previous Year Questions & Solutions