# UY1: Electric Field Of 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 field at point P.

Linear charge density, $\lambda$:

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

To find the dE, we will need dQ. dQ:

\begin{aligned}dQ &= \lambda \, dy \\ &= \frac{Q}{2a} \, dy \end{aligned}

Notice the symmetry in the problem. The vertical (y-axis) component of the electric field at P will vanish due to cancellation. Hence, only the horizontal (x-axis) component survives.

\begin{aligned} dE_{x} &= \frac{dQ}{4 \pi \epsilon_{0} \left(x^{2} + y^{2} \right)} \, \text{cos} \, \alpha \\ &= \frac{x}{4 \pi \epsilon_{0} (x^{2} + y^{2})^{\frac{3}{2}}} \, dQ \\ &= \frac{xQ}{8 \pi a \epsilon_{0} (x^{2} + y^{2})^{\frac{3}{2}}} \, dy \end{aligned}

You can use matlab or mathematica to evaluate the integral above from -a to a. Or you can use the integration result: $\int \frac{dx}{ \left( a^{2} + x^{2} \right)^{\frac{3}{2}}} = \frac{1}{a^{2}} \frac{x}{\sqrt{a^{2} +x^{2}}}$ from integration table.

You will obtain:

$$\vec{E} = \frac{Q}{4 \pi \epsilon_{0} x \sqrt{x^{2} + a^{2}}} \, \vec{i}$$

Next: Electric Field Of Uniformly Charged Disk

Previous: Electric Field 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.