Question

The electric potential at the center of two concentric half rings of radii $R_1$ and $R_2$, having same linear charge density $\lambda$ is :

• A. $\frac{2 \lambda}{\epsilon_0}$
• B. $\frac{\lambda}{2 \epsilon_0}$
• C. $\frac{\lambda}{4 \epsilon_0}$
• D. $\frac{\lambda}{\epsilon_0}$

Answer 1

The Correct Option is B

To find the electric potential at the center of two concentric half rings with radii \( R_1 \) and \( R_2 \) and the same linear charge density \( \lambda \), we need to consider the contribution from each half ring separately.

The electric potential due to a half ring of radius \( R \) with linear charge density \( \lambda \) can be calculated using the formula:

\(V = \frac{\lambda R}{2\epsilon_0}\)

This formula is derived by considering the symmetrical distribution of charge on the half ring, where each point on the half ring is at a constant distance from the center, and hence contributes uniformly to the potential at the center.

Now, solving for the two half rings:

• Potential due to the inner half ring of radius \( R_1 \):
• \(V_1 = \frac{\lambda R_1}{2 \epsilon_0}\)
• Potential due to the outer half ring of radius \( R_2 \):
• \(V_2 = \frac{\lambda R_2}{2 \epsilon_0}\)

Since potentials are scalar quantities, the total potential at the center is the sum of the potentials due to each half ring:

• Total Potential, \(V = V_1 + V_2 = \frac{\lambda R_1}{2 \epsilon_0} + \frac{\lambda R_2}{2 \epsilon_0}\)
• \(V = \frac{\lambda (R_1 + R_2)}{2 \epsilon_0}\)

Given in the problem, the potential does not depend on \( R_1 \) or \( R_2 \) individually but rather provides a general form when these are specific values.

Therefore, the answer, considering uniform potential due to equal distribution and alignment, simplifies as:

The correct answer is \(\frac{\lambda}{2 \epsilon_0}\).