Question

The electric potential at the center of two concentric half rings of radii \(R_1\) and \(R_2\), having same linear charge density \(λ\) is (\(ε_0\) = permittivity of free space)

• A. \(2λ/ε_0\)
• B. \(λ/2ε_0\)
• C. \(λ/4ε_0\)
• D. \(λ/ε_0\)

Hint:

Each half ring is at the same distance from the centre, so its potential is kQ/R with Q = lambda times pi R. Add the two.

Answer 1

The Correct Option is B

Step 1: Understand the figure
Two concentric half rings with centre O have radii \(R_1\) and \(R_2\). Both carry the same positive linear charge density \(\lambda\).

Step 2: Potential of a half ring at its centre
Every bit of charge on a half ring is at the same distance \(R\) from the centre. So \(V=\frac{1}{4\pi\epsilon_0}\frac{Q}{R}\). The length is \(\pi R\), so \(Q=\lambda\pi R\).

Step 3: Calculate
\[ V=\frac{1}{4\pi\epsilon_0}\frac{\lambda\pi R}{R}=\frac{\lambda}{4\epsilon_0} \] The radius cancels, so each half ring gives the same value.

Step 4: Add the two
Potential is a scalar, so we add directly: \[ V_{net}=\frac{\lambda}{4\epsilon_0}+\frac{\lambda}{4\epsilon_0}=\frac{\lambda}{2\epsilon_0} \]

Step 5: Other options
\(\frac{2\lambda}{\epsilon_0}\) and \(\frac{\lambda}{\epsilon_0}\) are too large, and \(\frac{\lambda}{4\epsilon_0}\) counts only one ring.

Final Answer:
Net potential is \(\frac{\lambda}{2\epsilon_0}\), option (B). \[ \boxed{\dfrac{\lambda}{2\epsilon_0}} \]