A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.
Question:

A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.

(a) A charge is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Solution:

(a) Charge placed at the centre of a shell is +q. Hence, a charge of magnitude −q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is −q.

Surface charge density at the inner surface of the shell is given by the relation

$\sigma_{1}=\frac{\text { Total charge }}{\text { Inner surface area }}=\frac{-q}{4 \pi r_{1}^{2}}$  ..(i)

A charge of +q is induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is Q + q. Surface charge density at the outer surface of the shell,

$\sigma_{2}=\frac{\text { Total charge }}{\text { Outer surface area }}=\frac{Q+q}{4 \pi r_{2}^{2}}$  ..(ii)

(b) Yes

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.

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