On the surface of a spherical shell enclosing a charge free region, the electrostatic
potential values are as follows: One quarter of the area has potential 𝜙_0, another
quarter has potential 2𝜙_0 and the rest has potential 4𝜙_0. The potential at the
centre of the shell is
(You can use a property of the solution of Laplace's equation.)
Options:
(A) $\frac{11}{4} \phi_0$
(B) $\frac{11}{2} \phi_0$
(C) $\frac{7}{3} \phi_0$
(D) $\frac{7}{4} \phi_0$
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.15)**
**1. Electrostatics Principles & Laplace's Equation**:
- In a charge-free region inside a spherical shell, the electrostatic potential $V(\vec{r})$ satisfies Laplace's equation:
$\nabla^2 V = 0$
- **Mean Value Theorem for Harmonic Functions**:
The potential at the center of a spherical shell of radius $R$ enclosing a charge-free volume is strictly equal to the average of the potential over the spherical surface:
$V(\text{center}) = \frac{1}{4\pi R^2} \iint_{\text{surface}} V(\theta, \phi)\, dA$
**2. Step-by-Step Derivation**:
- The spherical surface area $A = 4\pi R^2$ is partitioned into three regions:
- Region 1: Area $A_1 = \frac{1}{4} A$, potential $V_1 = \phi_0$.
- Region 2: Area $A_2 = \frac{1}{4} A$, potential $V_2 = 2\phi_0$.
- Region 3: Remaining Area $A_3 = A - (A_1 + A_2) = \frac{1}{2} A$, potential $V_3 = 4\phi_0$.
- Computing the surface average:
$V(\text{center}) = \frac{1}{A} \left[ A_1 V_1 + A_2 V_2 + A_3 V_3 \right]$
$V(\text{center}) = \frac{1}{4}\phi_0 + \frac{1}{4}(2\phi_0) + \frac{1}{2}(4\phi_0)$
$V(\text{center}) = \left(\frac{1}{4} + \frac{2}{4} + 2\right)\phi_0 = \left(\frac{3}{4} + \frac{8}{4}\right)\phi_0 = \frac{11}{4}\phi_0$
**3. Official Verification**:
The verified official IIT Master Answer is **Option (A)** (`11/4 \phi_0`).
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