Question

An arc of radius \(r\) carries charge. The linear density of charge is \(\lambda\) and the arc subtends an angle \(\dfrac{\pi}{3}\) at the centre. What is electric potential at the centre?

• A. \( \dfrac{\lambda}{4\varepsilon_0} \)
• B. \( \dfrac{\lambda}{8\varepsilon_0} \)
• C. \( \dfrac{\lambda}{12\varepsilon_0} \)
• D. \( \dfrac{\lambda}{16\varepsilon_0} \)

Hint:

For an arc, all elements are at same distance from centre. So potential is \(V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}\) with \(Q=\lambda r\theta\).

Answer 1

The Correct Option is C

Step 1: Potential at centre due to small element.
For an element \(dq\) at distance \(r\):
\[ dV = \frac{1}{4\pi\varepsilon_0}\frac{dq}{r} \] Step 2: Total charge on arc.
Arc length:
\[ L = r\theta = r\left(\frac{\pi}{3}\right) \] Charge:
\[ Q = \lambda L = \lambda r\frac{\pi}{3} \] Step 3: Potential at centre.
Since every element is at same distance \(r\),
\[ V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r} \] Substitute \(Q\):
\[ V = \frac{1}{4\pi\varepsilon_0}\frac{\lambda r\frac{\pi}{3}}{r} = \frac{1}{4\pi\varepsilon_0}\cdot \frac{\lambda\pi}{3} = \frac{\lambda}{12\varepsilon_0} \] Final Answer: \[ \boxed{\dfrac{\lambda}{12\varepsilon_0}} \]