Potential

Electricity and Magnetism Level 2

A rod is bent into semi circular arc of radius $R$ . The rod has uniform linear charge density $\lambda$ . What is the potential at the center of the arc?

\frac{\lambda}{2 \pi \epsilon_0 R}

\frac{\lambda}{\epsilon_0}

\frac{\lambda}{4 \epsilon_0}

2 solutions

Steven Chase
Jul 14, 2017

Potential is a scalar quantity rather than a vector quantity. So the main factor is the distance of the charge from the test point. Since all of the charge is the same distance from the origin, it can be treated as an equivalent point charge. The potential expression is:

$U = \frac{1}{4 \pi \epsilon_0} \frac{Q_{total}}{R} = \frac{1}{4 \pi \epsilon_0} \frac{\pi R \lambda}{R} = \frac{\lambda}{4 \epsilon_0}$

Adharsh M
Jul 14, 2017

please anyone solve this

Potential

2 solutions

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