Question

A rod uniformly charged with charge \( - q\) is bent into a semicircular arc of radius\(b\), as shown in Figure 16.97. What is the potential relative to infinity at location\(A\), at the center of the arc?

Short answer

\(\frac{q}{{4\pi {\varepsilon _0}b}}\)

Step-by-step solution

Given data

Semicircular arc having charge \( - q\) and radius \(b\)

Concept/ Formula used

Electric potential is the work required to transport a unit charge from one location to another in the presence of an electric field.

\(V = \frac{{KQ}}{r}\)

Where,\(Q\)is charge and\(r\)is distance where potential to be calculated

Electric potential at point A

Semicircular arc having length\(\pi b\)carrying\( - q\)charge

So, arc having length\(dx\)carrying charge\(dq = \frac{{ - qdx}}{{\pi b}}\)

Potential at point A

\(\begin{aligned}{c}\int {d{V_1} &= \int {\frac{{Kdq}}{R}} } \\ &= \int {\frac{{ - Kqdx}}{{\pi b \times b}}} \\ &= \int\limits_0^{\pi b} {\frac{{ - Kqdx}}{{\pi b \times b}}} \\ &= - \frac{q}{{4\pi {\varepsilon _0}b}}\end{aligned}\)

So potential at point A with reference to infinity is\(\frac{q}{{4\pi {\varepsilon _0}b}}\)