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 $$\overrightarrow a $$.