QJEEAdv15Phc513: JEE Advanced 2015 Physics - Class 12 JEEAdv15Phc513 Solution: Consider a uniform spherical charge distribution of radius…

Let $\rho$ be the charge density of the spherical charge distribution of radius r₁ centred at the origin O. A spherical cavity of radius r₂ centred at P with distance OP = a = r₁ $-$ r₂ is made in the spherical charge distribution.

The sphere with cavity is equivalent to a sphere of uniform charge density $-$ $\rho$ and radius r₂ centred at P embedded in the original sphere. Thus, the electric field at a point Q in the cavity is superposition of (i) electric field at Q due to the sphere of charge density $\rho$ and radius r₁ centred at O (say $\overrightarrow E$ ₁), and (ii) electric field at Q due to the sphere of charge density $-$ $\rho$ and radius r₂ centred at P (say $\overrightarrow E$ ₂). Let $\overrightarrow a$ , $\overrightarrow r$ , and $\overrightarrow r$ $-$ $\overrightarrow a$ be the vectors as shown in the figure. The electric fields $\overrightarrow E$ ₁, $\overrightarrow E$ ₂, and their superposition $\overrightarrow E$ ₁₂ are given by

${\overrightarrow E _1} = {1 \over {4\pi { \in _0}}}{{{4 \over 3}\pi |\overrightarrow r {|^3}\rho } \over {|\overrightarrow r {|^2}}}\widehat r = {\rho \over {3{ \in _0}}}\overrightarrow r$ ,

${\overrightarrow E _2} = - {\rho \over {3{ \in _0}}}(\overrightarrow r - \overrightarrow a )$ ,

${\overrightarrow E _{12}} = {\overrightarrow E _1} + {\overrightarrow E _2} = {\rho \over {3{ \in _0}}}\overrightarrow a$ .

Thus, the electric field at a point within the cavity is uniform and its magnitude and direction both depend on $\overrightarrow a$ .