GATE 2025 PH – Question 56
A neutral conducting sphere of radius $R$ in uniform field $E_0\hat z$ has outside potential $V(r,\theta)=V_0-E_0r(1-R^3/r^3)\cos\theta$. Which option(s) is/are correct?
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Correct answer: (C) The field is curl free for $r>R$; (D) The field is divergence free for $r>R$
Explanation
The sphere is neutral and conducting, in a uniform field $E_0\hat z$, with outside potential
$$V=V_0-E_0r\left(1-\frac{R^3}{r^3}\right)\cos\theta .$$
- **A. The surface charge density is proportional to $\sin\theta$.** Compute $\sigma=-\epsilon_0\dfrac{\partial V}{\partial r}\Big|_{r=R}$: this gives $\sigma=3\epsilon_0E_0\cos\theta$, so it is proportional to $\cos\theta$. ✗
- **B. As $r\to\infty$, $\mathbf E=E_0\cos\theta\,\hat r$.** At large distance the field tends to the uniform field $E_0\hat z=E_0(\cos\theta\,\hat r-\sin\theta\,\hat\theta)$, which also has a $\theta$ component. ✗
- **C. The field is curl-free for $r>R$.** True: $\mathbf E=-\nabla V$. ✓
- **D. The field is divergence-free for $r>R$.** True: there is no charge outside the sphere, so $\nabla^2V=0$. ✓
Answer **C and D**.