Consider a uniform spherical charge distribution of radius R 1 centred at the origin O. In this distribution…

1. $\overrightarrow{E}$ is uniform, its magnitude is independent of $R_2$ but its direction depends on $\overrightarrow{r}$
2. $\overrightarrow{E}$ is uniform, its magnitude is depends on $R_2$ and its direction depends on $\overrightarrow{r}$
3. $\overrightarrow{E}$ is uniform, its magnitude is independent of a but its direction depends on $\overrightarrow{a}$
4. $\overrightarrow{E}$ is uniform and both its magnitude and direction depend on $\overrightarrow{a}$

Solution

Considering any point A inside the cavity electric field can be calculated at A by using super position principle.
${\overrightarrow{E}}_A={\left({\overrightarrow{E}}_A\right)}_{Due\mathrm{ }to\mathrm{ }sphere\mathrm{ }without\mathrm{ }cavity}-{\left({\overrightarrow{E}}_A\right)}_{Due\mathrm{ }to\mathrm{ }sphere\mathrm{ }of\mathrm{ }radius\mathrm{ }{R}_2\mathrm{ }\&\mathrm{ }centre\mathrm{ }at\mathrm{ }P}$
${\overrightarrow{E}}_A=\frac{\rho }{3{ϵ}_0}O\overrightarrow{A}=\frac{\rho }{3{ϵ}_0}\left(P\overrightarrow{A}\right)$
But $O\overrightarrow{P}+P\overrightarrow{A}=O\overrightarrow{P}$
$\therefore O\overrightarrow{A}-P\overrightarrow{A}=O\overrightarrow{P}$
$\therefore {\overrightarrow{E}}_A=\frac{\rho }{3{ϵ}_0}\left(O\overrightarrow{P}\right)=\frac{\rho }{3{ϵ}_0}\left(\overrightarrow{a}\right)$
So E.F. is independent of location of A inside the cavity. `

Asked in: JEE Advanced 2015 (Paper 2)

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