Question

The wire in the figure has linear charge density λ.

What is the electric potential at the center of the semicircle?

Give your answer in terms of λ, R, ε_0, and appropriate constants.

Answer 1

Potential is a scalar, so the potential due to this wire is the sum of the potential due to the curved part, \(V_{c}\), and twice the potential due to one straight part, \(V_{s}\).

Since every piece of the curved part is the same distance from the center, the potential due to this part is the same as that of a point particle with the same charge at the same distance.

$$ V_{c}=K \frac{Q_{c}}{R_{c}}=K \frac{\lambda L}{R}=K \frac{\lambda \pi R}{R}=K \lambda \pi $$

Each piece of the straight part, however, is a different distance from the center. Each small element of the straight part is like a point. Adding up the contributions of all these small elements of charge (taking an integral) yields the potential due to one entire straight part.

$$ V_{s}=\int d V_{s}=\int K \frac{d q}{r}=\int_{R}^{3 R} K \frac{\lambda d r}{r}=\left.K \lambda \ln r\right|_{R} ^{3 R}=K \lambda \ln \left(\frac{3 R}{R}\right)=K \lambda \ln 3 $$

So

$$ V=V_{c}+2 V_{s}=K \lambda \pi+2(K \lambda \ln 3)=K \lambda(\pi+2 \ln 3) \quad \text { Where } K=\frac{1}{4 \pi \varepsilon_{0}} $$