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):

• A. \( \alpha = -1 \)
• B. \( \alpha = -\frac{2}{3} \)
• C. \( \alpha = -\frac{3}{2} \)
• D. \( \alpha = -\pi \)

Hint:

For circular arcs, potential at center depends only on charge density and angle: radius cancels out completely.

Answer 1

The Correct Option is A

Step 1: Potential from a charge element.
dV = (1/4πε₀) dq/r; for an arc, all elements are equidistant from the center.

Step 2: Charge element on arc.
dq = λ dl = λ r dθ.

Step 3: Integrating for one arc.
V = (1/4πε₀) ∫ λ r dθ / r = (1/4πε₀) λ θ; independent of radius.

Step 4: Contributions from both arcs.
Outer arc: V₂ = k λ θ; inner arc: V₁ = k α λ θ.

Step 5: Net potential at center.
V_net = k λ θ (α + 1).

Step 6: Zero potential condition.
α + 1 = 0 → α = -1.