Shown in the figure is a shell made of a cond | vedclass

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(B) The total charge of the dipole is $0$,so the net charge induced on the inner surface is $0$. Since the shell is a conductor,the total charge $Q$ r

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Electric field outside the shell is the same as that of a point charge $Q$ at the centre of the shell.

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(B) The total charge of the dipole is $0$,so the net charge induced on the inner surface is $0$. Since the shell is a conductor,the total charge $Q$ resides on the outer surface.Because the dipole is at the centre,it creates a non-uniform electric field inside the cavity. This induces a non-uniform charge distribution on the inner surface of the shell.However,for any point outside the shell $(r > b)$,the electric field produced by the induced charges on the inner surface and the dipole itself cancels out due to the principle of superposition and the properties of a conductor in electrostatic equilibrium.Thus,the electric field outside the shell is solely due to the total charge $Q$ distributed uniformly on the outer surface,which is equivalent to a point charge $Q$ at the centre: $E = \frac{kQ}{r^2}$.

Shown in the figure is a shell made of a conductor. It has inner radius $a$ and outer radius $b$,and carries a total charge $Q$. At its centre is a dipole $\vec{P}$ as shown. In this case,

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