A spherical conducting shell of inner radius $r_1$ and outer radius $r_2$ has a charge $Q. $
$(a)$ A charge $q$ 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.
A spherical conducting shell of inner radius $r_1$ and outer radius $r_2$ has a charge $Q. $
$(a)$ A charge $q$ 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.
$(a)$ 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,
$\sigma_{1}=\frac{\text { Total charge }}{\text { Inner surface area }}=\frac{-q}{4 \pi r_{1}^{2}}$
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,
$\sigma_{2}=\frac{\text { Toter surface of the shell, }}{\text { Outer surface area }}=\frac{Q+q}{4 \pi r_{2}^{2}}$
$(b)$ 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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