Question

A neutral conducting sphere of radius \( R \) is placed in a uniform electric field of magnitude \( E_0 \), that points along the \( z \)-axis. The electrostatic potential at any point \( r \) outside the sphere is given by: \[ V(r, \theta) = V_0 - E_0 r \left(1 - \frac{R^3}{r^3} \right) \cos \theta \] where \( V_0 \) is the constant potential of the sphere. Which of the following option(s) is/are correct?

• A. The induced surface charge density on the sphere is proportional to \( \sin \theta \)
• B. As \( r \to \infty \), \( \vec{E} = E_0 \cos \theta \hat{r} \)
• C. The electric field at any point is curl-free for \( r>R \)
• D. The electric field at any point is divergence-free for \( r>R \)

Hint:

The electric field outside a conducting sphere in a uniform electric field is both curl-free and divergence-free. These properties hold for any electrostatic field.

Answer 1

The Correct Option is C, D

1. Induced Surface Charge Density:
The potential outside the conducting sphere is given by the expression \( V(r, \theta) \). The induced surface charge density \( \sigma \) on the surface of the sphere is related to the electric field just outside the surface by the boundary condition:
\[ \sigma = \epsilon_0 \mathbf{E} \cdot \hat{n} \]
where \( \hat{n} \) is the unit normal to the surface, and \( \mathbf{E} \) is the electric field. From the given potential expression, we can derive that the induced surface charge density depends on \( \sin \theta \). Therefore, the induced surface charge density on the sphere is proportional to \( \sin \theta \), which suggests option (A). However, the correct answer is (C) and (D).

2. Electric Field as \( r \to \infty \):
As \( r \to \infty \), the potential becomes dominated by the term \( -E_0 r \cos \theta \), and the electric field is derived as the gradient of the potential:
\[ \mathbf{E} = -\nabla V(r, \theta) = E_0 \cos \theta \hat{r} \]
This matches option (B). Thus, the electric field at large distances is \( \mathbf{E} = E_0 \cos \theta \hat{r} \), and this result is consistent with the behavior outside the sphere.

3. Curl-Free Electric Field:
Since the electric field is derived from a scalar potential, it is conservative, which means the electric field is curl-free. Therefore, the electric field at any point for \( r > R \) is curl-free, matching option (C).

4. Divergence-Free Electric Field:
In electrostatics, the electric field satisfies Gauss's law, which states that the divergence of the electric field is zero in regions where there are no charges. For \( r > R \), there are no charges, so the electric field is divergence-free, matching option (D).

Thus, the correct answers are (C) and (D).