As shown in the figure, a point charge $$Q$$ is placed at the centre of conducting spherical shell of inner radius $$a$$ and outer radius $$b$$. The electric field due to charge $$Q$$ in three different regions I, II and III is given by:
(I: $$r < a$$, II: $$a < r < b$$, III: $$r > b$$)

For a point charge Q placed at the centre of a conducting spherical shell:

Regions are:

• Region I:r<a(inside cavity)
• Region II: a<r<b (inside conducting material)
• Region III: r>b (outside shell)

We examine electric field in each region.

For Region I (r<a)

Using Gauss law, take a spherical Gaussian surface of radius r<ar

Only charge enclosed is Q.

$$\oint\vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_0}$$

Since symmetry makes E constant over the surface,

$$E(4\pi r^2)=\frac{Q}{\varepsilon_0}$$

$$E_I=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\ne0$$

For Region II (a<r<b):

This is the conducting material.

In electrostatic equilibrium, electric field inside a conductor is zero.

$$E_{II}=0$$

This happens because charge Q induces

• charge −Q on inner surface
• charge +Q on outer surface

so field inside the conductor cancels.

For Region III (r>b):

Take a Gaussian sphere of radius r>b.

Total enclosed charge is

$$Q+(-Q)+(+Q)=Q$$

Using Gauss law,

$$E(4\pi r^2)=\frac{Q}{\varepsilon_0}$$

$$E_{III}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\ne0$$

So field outside is not zero; the whole system behaves like a point charge Q at the centre.

Therefore,

$$EI\ne0,E_I\ne0,\quad\quad E_{III}\ne0$$

Correct option:

B