Consider a uniform spherical charge distribution of radius $R_{1}$ centered at the origin O. In this distribution, a spherical cavity of radius $R_{2}$ , centered 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 $\vec{E} \overset{⇀}{(r)}$ , then the correct statement(s) is (are)

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$\vec{E}$ is uniform its magnitude is independent of $R_{2}$ but its direction depends on $\vec{r}$

$\vec{E}$ is uniform , its magnitude is independent of a but its direction depends on $\vec{a}$

$\vec{E}$ is uniform and both its magnitude and direction depend on $\vec{a}$

$\vec{E}$ is uniform , its magnitude depends on $R_{2}$ and its direction depends on $\vec{r}$

answer is D.

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Detailed Solution

Let us consider a point M inside the cavity where electric field has to be calculated . Assume the cavity to contain similar charge distribution of positive and negative charge as the rest of sphere electicfiled at M due to uniformly distributed charge of the whole sphere of radius $R_{1} {\vec{E}}_{1} = \frac{ρ}{3 ε} \vec{r}$
Electric field at M due to the negative charge distribution in the cavity

${\vec{E}}_{2} = \frac{ρ}{3 ε} \overset{⇀}{M P}$
The total electric field at M inside the cavity,
$\vec{E} = \vec{E} 1 + {\vec{E}}_{2} = \frac{ρ}{3 ε} \vec{r} + \frac{ρ}{3 ε} \overset{⇀}{M P} ∴ \vec{E} = \frac{ρ}{3 ε} \vec{r} + \frac{ρ}{3 ε} (\vec{a} - \vec{r}) [∴ \vec{r} + \overset{⇀}{M P} = \vec{a}] ∴ \vec{E} = \frac{ρ}{3 ε} \vec{a}$