# UY1: Electric Potential Of A Line Of Charge

Positive electric charge Q is distributed uniformly along a line (you could imagine it as a very thin rod) with length 2a, lying along the y-axis between y = -a and y = +a. A point p lies at x along x-axis. Find the electric potential at point P.

Linear charge density:

$$\lambda = \frac{Q}{2a}$$

Small element of charge:

$$dQ = \lambda \, dy$$

dV in terms of linear charge density:

\begin{aligned} dV &= \frac{dQ}{4 \pi \epsilon_{0} r} \\ &= \frac{\lambda \, dy}{4 \pi \epsilon_{0} \sqrt{x^{2} + y^{2}}} \end{aligned}

Integrate from -a to a by using the integral in integration table, specifically $\int \frac{dx}{\sqrt{a^{2} +x^{2}}} = \text{ln} \, \left(x + \sqrt{a^{2} + x^{2}} \right)$,

\begin{aligned} V &= \frac{\lambda}{4 \pi \epsilon_{0}} \int\limits_{-a}^{a} \frac{dy}{\sqrt{x^{2}+y^{2}}} \\ &= \frac{\lambda}{4 \pi \epsilon_{0}} \text{ln} \left( \frac{\sqrt{a^{2}+x^{2}}+a}{\sqrt{a^{2} + x^{2}} – a} \right) \end{aligned}

We can check the expression for V with the expression for electric field derived in Electric Field Of A Line Of Charge.

\begin{aligned} E &= \, – \frac{\partial V}{\partial x} \\ &= \frac{Q}{4 \pi \epsilon_{0} \sqrt{x^{2} + a^{2}}} \end{aligned}

Next: Electric Potential Of An Infinite Line Charge

Previous: Electric Potential Of A Ring Of Charge

Back To Electromagnetism (UY1)

##### Mini Physics

As the Administrator of Mini Physics, I possess a BSc. (Hons) in Physics. I am committed to ensuring the accuracy and quality of the content on this site. If you encounter any inaccuracies or have suggestions for enhancements, I encourage you to contact us. Your support and feedback are invaluable to us. If you appreciate the resources available on this site, kindly consider recommending Mini Physics to your friends. Together, we can foster a community passionate about Physics and continuous learning.

This site uses Akismet to reduce spam. Learn how your comment data is processed.