Question

A conducting sphere of radius R and a concentric thick
spherical shell of inner radius 2R and outer radius 3R is shown
in figure. A charge + 10 Q is given to the shell and inner sphere
is earthed. Then charge on sphere is
A. -4Q
B. -10Q
C. zero
D. None​

Answer 1

Answer:

A) q=-4Q

Explanation:

Since the inner sphere is earthed, we get V =0

Note if we assume that charge on the sphere (surface) is q, we proceed as shown in figure

Hope this answer helped you ✌:)

Answer 2

Concept: When matter is held in an electric or magnetic field, it develops an electric charge, which causes it to experience a force. The flowing electric charge generates a magnetic field, which is coupled with an electric field. The electromagnetic field is the result of a combination of electric and magnetic fields. The electromagnetic force, which is the basis of physics, is created when charges interact.

Given: Radius of sphere = R

           Inner radius of shell = 2R

           Outer radius of shell = 3R

           Charge given to the shell = +10Q

To find: Charge on sphere

Solution:

Allow the conducting sphere to become charged q

Charge on the shell's inner surface = -q

Charge on the shell's outer surface = (+q+10Q)

Assume that the potential on the sphere is V, then

$V = \frac{kq}{R} + \frac{-kq}{2R} +\frac{k(q+10Q)}{3R}$

Because the sphere is grounded, therefore V=0

$0 = \frac{kq}{R} - \frac{kq}{2R} +\frac{kq+10kQ}{3R}$

$\frac{2kq-kq}{2R} + \frac{kq+10kQ}{3R} = 0\\\frac{kq}{2R} + \frac{kq+10kQ}{3R} = 0\\\frac{3kq + 2kq + 20kQ}{6R} = 0$

$5kq + 20kQ = 0\\k (5q + 20Q) = 0 (k=0)\\5q + 20Q = 0 \\5q = -20Q\\q = -4Q$

Hence, the correct option is (A).

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