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 figure. The initial and final positions of the charge are marked by A and B at distances 2R and 3R respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is R^2 n _0 . The value of n is : ( _0 is the permittivity of vacuum)
A. 18
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 initial position A is at a distance r_A = 2R from the centre. The potential at A is: V_A = R^3 3 _0 (2R) = R^2 6 _0 The final position B is at a distance r_B = 3R from the centre. The potential at B is: V_B = R^3 3 _0 (3R) = R^2 9 _0 The work done by the external agent in moving a unit positive charge (q = 1) slowly from A to B is equal to the change in its potential energy: W = q(V_B - V_A) = 1 ( R^2 9 _0 - R^2 6 _0 ) W = R^2 _0 ( 2 - 3 18 ) = - R^2 18 _0 The magnitude of the total work done is: |W| = R^2 18 _0 Comparing this with the given expression R^2 n _0 , we find: n = 18 Answer: 18
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