Consider a uniform spherical charge distribution of radius R 1 centred at the origin O. In this distribution, a spherical cavity of radius R 2 , centred at P with distance O P = a = R 1 - R 2 (see fig

Question

Consider a uniform spherical charge distribution of radius $R_{1}$ centred at the origin O. In this distribution, a spherical cavity of radius $R_{2}$ , centred at $P$ with distance $O P = a = R_{1} - R_{2}$ (see figure) is made. If the electric field inside the cavity at position $\vec{r}$ is $\vec{E} (\vec{r})$ , then the correct statement(s) is (are)

TestHub · Accepted Answer

Considering any point A inside the cavity electric field can be calculated at A by using super position principle.
${\vec{E}}_{A} = {({\vec{E}}_{A})}_{D u e t o s p h e r e w i t h o u t c a v i t y} - {({\vec{E}}_{A})}_{D u e t o s p h e r e o f r a d i u s R_{2} & c e n t r e a t P}$
${\vec{E}}_{A} = \frac{ρ}{3 ϵ_{0}} O \vec{A} = \frac{ρ}{3 ϵ_{0}} (P \vec{A})$
But $O \vec{P} + P \vec{A} = O \vec{P}$
$∴ O \vec{A} - P \vec{A} = O \vec{P}$
$∴ {\vec{E}}_{A} = \frac{ρ}{3 ϵ_{0}} (O \vec{P}) = \frac{ρ}{3 ϵ_{0}} (\vec{a})$
So E.F. is independent of location of A inside the cavity.

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