Understand the configuration

We have:

• A solid metal sphere of radius $R$ carrying charge $q$.
• It is surrounded by a concentric conducting spherical shell with:
  • inner radius $a$
  • outer radius $b$

Since both are conductors and concentric, the electric field depends only on the radial distance $r$.

Use electrostatic properties of conductors

In electrostatic equilibrium:

• Electric field inside a conductor is zero.
• Charge resides on the outer surface of a conductor.
• For the shell, induced charge appears as:
  • $-q$ on the inner surface $r=a$
  • $+q$ on the outer surface $r=b$

so that the net charge enclosed in the conducting material of the shell gives zero field inside the metal of the shell.

Find $E(r)$ in different regions

Region I: $0 \le r < R$

This is inside the solid metal sphere.

Therefore, $E=0$

Region II: $R < r < a$

Take a Gaussian sphere of radius $r$ in this region. It encloses only charge $q$.

By Gauss's law, $E(4\pi r^2)=\frac{q}{\varepsilon_0}$ so $E=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$

Thus in this region, $E$ decreases as $1/r^2$.

Region III: $a < r < b$

This lies inside the conducting shell material.

Hence, $E=0$

Region IV: $r > b$

Now the Gaussian surface encloses:

• central sphere charge $q$
• induced charge $-q$ on inner surface
• induced charge $+q$ on outer surface

Total enclosed charge: $q + (-q) + q = q$

Therefore, $E=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$

Again field decreases as $1/r^2$.

Nature of the graph

So the variation of $E$ vs $r$ is:

• $E=0$ for $r<R$
• sudden jump at $r=R$ to $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R^2}$
• then decreases as $1/r^2$ for $R<r<a$
• drops suddenly to $0$ at $r=a$
• remains $0$ for $a<r<b$
• jumps again at $r=b$ to $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{b^2}$
• then decreases as $1/r^2$ for $r>b$

Match with the option

The correct graph must show exactly this piecewise behavior:

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

Hence, they agree.