Question:
Two concentric circular arcs of radii 2 m and 3 m have linear charge densities \( \alpha \lambda \) and \( \lambda \) respectively as shown in the figure. The condition for the electric potential to vanish at the common center is (electric potential = 0 at infinity):
Two concentric circular arcs of radii 2 m and 3 m have linear charge densities \( \alpha \lambda \) and \( \lambda \) respectively as shown in the figure. The condition for the electric potential to vanish at the common center is (electric potential = 0 at infinity):
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For circular arcs, potential at center depends only on charge density and angle: radius cancels out completely.
Updated On: Jun 19, 2026
- \( \alpha = -1 \)
- \( \alpha = -\frac{2}{3} \)
- \( \alpha = -\frac{3}{2} \)
- \( \alpha = -\pi \)
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The Correct Option is A
Solution and Explanation
Step 1: Expression for electric potential due to a charge element.
Electric potential at a point due to a small element of charge \(dq\) is: \[ dV = \frac{1}{4\pi \varepsilon_0}\frac{dq}{r} \] For a charged arc, all elements are at the same distance \(r\) from the center, so integration becomes straightforward.
Step 2: Express charge element for arc.
For a small arc element: \[ dq = \lambda \, dl \] and arc length: \[ dl = r\, d\theta \]
Step 3: Potential due to a full arc.
\[ V = \frac{1}{4\pi \varepsilon_0} \int \frac{\lambda r\, d\theta}{r} \] The radius cancels: \[ V = \frac{1}{4\pi \varepsilon_0} \lambda \int d\theta \] So, \[ V = \frac{1}{4\pi \varepsilon_0} \lambda \theta \] Hence, potential due to an arc depends only on charge density and angle, not on radius.
Step 4: Apply to both arcs.
Both arcs subtend the same angle (as shown in figure, total angle \( \theta = \frac{\pi}{3} \)).
Potential due to outer arc: \[ V_2 = k (\lambda \theta) \]
Potential due to inner arc: \[ V_1 = k (\alpha \lambda \theta) \]
Step 5: Net potential at center.
\[ V = k \lambda \theta (\alpha + 1) \]
Step 6: Condition for zero potential.
For potential to vanish: \[ \alpha + 1 = 0 \Rightarrow \alpha = -1 \]
Final Answer: \[ \boxed{\alpha = -1} \]
Electric potential at a point due to a small element of charge \(dq\) is: \[ dV = \frac{1}{4\pi \varepsilon_0}\frac{dq}{r} \] For a charged arc, all elements are at the same distance \(r\) from the center, so integration becomes straightforward.
Step 2: Express charge element for arc.
For a small arc element: \[ dq = \lambda \, dl \] and arc length: \[ dl = r\, d\theta \]
Step 3: Potential due to a full arc.
\[ V = \frac{1}{4\pi \varepsilon_0} \int \frac{\lambda r\, d\theta}{r} \] The radius cancels: \[ V = \frac{1}{4\pi \varepsilon_0} \lambda \int d\theta \] So, \[ V = \frac{1}{4\pi \varepsilon_0} \lambda \theta \] Hence, potential due to an arc depends only on charge density and angle, not on radius.
Step 4: Apply to both arcs.
Both arcs subtend the same angle (as shown in figure, total angle \( \theta = \frac{\pi}{3} \)).
Potential due to outer arc: \[ V_2 = k (\lambda \theta) \]
Potential due to inner arc: \[ V_1 = k (\alpha \lambda \theta) \]
Step 5: Net potential at center.
\[ V = k \lambda \theta (\alpha + 1) \]
Step 6: Condition for zero potential.
For potential to vanish: \[ \alpha + 1 = 0 \Rightarrow \alpha = -1 \]
Final Answer: \[ \boxed{\alpha = -1} \]
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