A point charge $Q$ is placed inside a cavity within a solid isolated conducting sphere. Consider points $A, B$ and $C$ as shown in the figure, where the magnitudes of the electric fields are $E_A, E_B, E_C$, respectively. The points $B$ and $C$ are at the same distance from the center of the solid sphere. The correct option is :

![](https://cdn.mathpix.com/snip/images/NVNkW3nIOYaJw1IuwJOQatvE-WyFz7kSEix6bxoi8HU.original.fullsize.png)

Question

A point charge $Q$ is placed inside a cavity within a solid isolated conducting sphere. Consider points $A, B$ and $C$ as shown in the figure, where the magnitudes of the electric fields are $E_A, E_B, E_C$ , respectively. The points $B$ and $C$ are at the same distance from the center of the solid sphere. The correct option is :

![](https://cdn.mathpix.com/snip/images/NVNkW3nIOYaJw1IuwJOQatvE-WyFz7kSEix6bxoi8HU.original.fullsize.png)

Accepted Answer

C. 1. Point $A$ lies inside the cavity where the point charge $Q$ is located, so there **must** be an electric field there to satisfy Gauss's law inside the cavity:
$$E_A
eq0$$

2. Because the conductor is isolated and in electrostatic equilibrium, the induced charges rearrange themselves so that the charge on the **outer surface** distributes uniformly, regardless of where $Q$ sits inside the cavity. This is a consequence of **electrostatic shielding**.

3. Once the outer surface charge is uniform, the field outside the conductor depends only on the distance from the centre, exactly like a point charge placed at the centre. Since $B$ and $C$ are at the **same distance** from the centre:
$$E_B=E_C$$

4. Combining both results, $E_A
eq0$ and $E_B=E_C$.

Hence, the answer is **option C**.