Electrostatics — NEET UG practice

1. Point $A$ lies inside the cavity where the point charge $Q$ is located, so there must be an electric field there to satisfy Gauss's law inside the cavity: $E_A\neq0$

2. Because the conductor is isolated and in electrostatic equilibrium, the induced charges rearrange themselves so that the charge on the outer surface distributes uniformly, regardless of where $Q$ sits inside the cavity. This is a consequence of electrostatic shielding.

3. Once the outer surface charge is uniform, the field outside the conductor depends only on the distance from the centre, exactly like a point charge placed at the centre. Since $B$ and $C$ are at the same distance from the centre: $E_B=E_C$

4. Combining both results, $E_A\neq0$ and $E_B=E_C$.

Hence, the answer is option C.

1. Statement A: Inside a conductor in electrostatic equilibrium, free charges rearrange themselves so that the net electric field inside becomes zero. So statement A is true.

2. Statement B: The electric field just outside the surface of a charged conductor is given by $E=\frac{\sigma}{\varepsilon_0}\hat{n}$ This clearly depends on the surface charge density $\sigma$, so statement B is false.

3. Statement C: In the static situation, all excess charge on a conductor resides only on its outer surface; the interior has no excess charge. So statement C is true.

4. Statement D: If the field at the surface had a component along the surface, charges would keep moving until equilibrium is reached, so in equilibrium the field must be normal to the surface at every point. Statement D is true.

5. Statement E: Since the field inside a conductor is zero, the potential inside is constant, but this constant value need not be zero — it equals the potential of the surface. So statement E is false.

6. The correct statements are A, C and D.

Hence, the answer is option C.

1. Both points $A$ (at $2R$) and $B$ (at $3R$) lie outside the sphere (radius $R$), so the sphere behaves like a point charge $Q$ placed at its centre when calculating the potential at these points. The potential outside a uniformly charged sphere is $V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$

2. The total charge $Q$ of the sphere is found by multiplying the uniform charge density $\rho$ by the sphere's volume: $Q=\rho\times\frac{4}{3}\pi R^3$

3. Since the charge is moved slowly, all the work done by the external agent goes into changing potential energy, so the work done equals the (unit) charge times the potential difference between the final and initial points: $W=q(V_B-V_A)$

4. Substituting the potentials at $r=3R$ and $r=2R$ (with $q=1$ for a unit charge): $W=\frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{3R}-\frac{1}{2R}\right)=\frac{Q}{4\pi\varepsilon_0 R}\left(-\frac{1}{6}\right)$

5. Taking the magnitude and substituting $Q=\rho\times\frac{4}{3}\pi R^3$: $|W|=\frac{1}{4\pi\varepsilon_0}\times\frac{1}{R}\times\frac{1}{6}\times\rho\times\frac{4}{3}\pi R^3=\frac{\rho R^2}{18\varepsilon_0}$

6. Comparing this with the given expression $\frac{\rho R^2}{n\varepsilon_0}$, we get $n=18$.

Hence, the answer is option A.