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 distance 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 R2n 0. The value of n is : ( 0 is the permittivity of vacuum) [figure]
- 18
- 2
- 6
- 9
Answer
(1) 18
Solution
The electrostatic field is conservative, so the work done depends only on the initial and final positions.
Total charge of the uniformly charged sphere:
Q = ( 43 R3 )
Potential at any external point (r R) is:
V(r) = Q4 0 r = R33 0 r
For initial point A (rA = 2R):
VA = R33 0 (2R) = R26 0
For final point B (rB = 3R):
VB = R33 0 (3R) = R29 0
Magnitude of work done on a unit positive charge (q = 1):
|W| = |VB - VA| = | R29 0 - R26 0 | = R2 0 ( 16 - 19 ) = R218 0
Comparing with R2n 0, we get n = 18.
Answer: (1)