JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Electrostatics

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 $OP=a=R_1-R_2$ (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)

1. (A)$\vec{E}$ is uniform, its magnitude is independent of $R_2$ but its direction depends on $\vec{r}$
2. (B)$\vec{E}$ is uniform, its magnitude depends on $R_2$ and its direction depends on $\vec{r}$
3. (C)$\vec{E}$ is uniform, its magnitude is independent of a but its direction depends on $\vec{a}$
4. (D)$\vec{E}$ is uniform, and both its magnitude and direction depend on $\vec{a}$✓ Correct

Answer: (D)

Solution

Using the superposition principle, the field inside the cavity can be found by considering:

1. A uniform sphere of radius $R_1$ with charge density $\rho$ (positive contribution)
2. A uniform sphere of radius $R_2$ centered at P with charge density $-\rho$ (negative contribution, representing the cavity)

For a point inside a uniform sphere at distance $r$ from center:
$\vec{E}=\frac{\rho \vec{r}}{3{\epsilon }_0}$

At a point inside the cavity:

• Field due to large sphere: ${\vec{E}}_1=\frac{\rho \vec{r}}{3{\epsilon }_0}$ (from O)
• Field due to removed sphere: ${\vec{E}}_2=-\frac{\rho (\vec{r}-\vec{a})}{3{\epsilon }_0}$ (from P)

Net field:
$\vec{E}={\vec{E}}_1+{\vec{E}}_2=\frac{\rho \vec{r}}{3{\epsilon }_0}-\frac{\rho (\vec{r}-\vec{a})}{3{\epsilon }_0}=\frac{\rho \vec{a}}{3{\epsilon }_0}$

Since $\vec{a}=\overrightarrow{OP}$ is constant:

• $\vec{E}$ is uniform (independent of position $\vec{r}$ inside cavity)
• Its magnitude depends on $|\vec{a}|=a=R_1-R_2$ (depends on both $R_1$ and $R_2$)
• Its direction is along $\vec{a}$ (depends on $\vec{a}$)

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options