Question

A semicircular arc of radius '\(R\)' is charged uniformly and has charge per unit length as \(\lambda\). The electric potential \(V\) at the centre of curvature is [\(\varepsilon _{0}\) = permittivity of vacuum]

• A. \(\frac{\lambda }{4\varepsilon _{0}}\)
• B. \(\frac{\lambda }{2\varepsilon _{0}}\)
• C. \(\frac{\lambda }{8\varepsilon _{0}}\)
• D. \(\frac{\lambda }{\varepsilon _{0}}\)

Hint:

Every element of the arc is at the same distance R from the centre, so add up the charge and use kq over R.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Potential is a scalar, and every charge element on the arc lies at the same distance \(R\) from the centre.

Step 2: Total Charge
Arc length \(=\pi R\), so \(q=\lambda\pi R\).

Step 3: Calculation
\(V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\lambda\pi R}{R}=\dfrac{\lambda}{4\varepsilon_0}\)

Step 4: Compare the options
This is the first option.

Final Answer:
The potential at the centre is lambda divided by 4 epsilon naught. \[ \boxed{\text{(A) }\dfrac{\lambda}{4\varepsilon_{0}}} \]