Electric field is measured along the axis of a uniformly charged disc of radius 25
cm. At a distance d from the centre, the field differs by 10% from that of an
infinite plane having the same charge density. The value of d is _______cm.
(Round off to one decimal place)
Official Answer Key:
2.4 to 2.6
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.31)**
**1. Electrostatics Formulation**:
- Electric field along the symmetry axis of a uniformly charged disc of radius $R$ at distance $d$:
$E_{\text{disc}}(d) = \frac{\sigma}{2\epsilon_0} \left[1 - \frac{d}{\sqrt{d^2 + R^2}}\right]$
- Electric field of an infinite plane with the same surface charge density:
$E_\infty = \frac{\sigma}{2\epsilon_0}$
**2. Detailed Step-by-Step Derivation**:
- Given that $E_{\text{disc}}(d)$ differs by $10\%$ ($0.10$) from $E_\infty$:
$\frac{E_\infty - E_{\text{disc}}(d)}{E_\infty} = 0.10$
$\frac{d}{\sqrt{d^2 + R^2}} = 0.10$
- Square both sides:
$\frac{d^2}{d^2 + R^2} = (0.10)^2 = 0.01$
$d^2 = 0.01(d^2 + R^2) = 0.01d^2 + 0.01R^2$
$0.99 d^2 = 0.01 R^2$
$d^2 = \frac{1}{99} R^2 \implies d = \frac{R}{\sqrt{99}}$
- Given $R = 25$ cm:
$d = \frac{25}{\sqrt{99}} = \frac{25}{9.94987} \approx 2.5126 \text{ cm}$
- Rounding to one decimal place yields $2.5$ cm.
**3. Official Verification**:
The verified official IIT Master Answer range is **2.4 to 2.6**.
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