UY1: Electric Field Of A Uniformly Charged Sphere



Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R. Find the magnitude of the electric field at a point P, a distance r from the center of the sphere.

Using Gauss’s Law for $r \geq R$,

$$\begin{aligned} EA &= \frac{Q}{\epsilon_{0}} \\  E (4 \pi r^{2}) &= \frac{Q}{\epsilon_{0}} \\ E &= \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{r^{2}} \end{aligned}$$

 

For $r < R$,

$$\begin{aligned} EA &= \frac{q}{\epsilon_{0}} \\ E \times 4 \pi r^{2} &= \frac{q}{\epsilon_{0}} \end{aligned}$$

q is just the net charge enclosed by a spherical Gaussian surface at radius r. Hence, we can find out q from volume charge density, $\rho$

$$ \rho = \frac{Q}{\frac{4}{3} \pi R^{3}}$$

$$\begin{aligned} q &= \rho \times \frac{4}{3} \pi r^{3} \\ &= Q \frac{\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R^{3}} \\ &= Q \frac{r^{3}}{R^{3}}\end{aligned}$$

Hence, sub. q into the expression for E to get:

$$E = \frac{Q}{4 \pi \epsilon_{0}} \frac{r}{R^{3}}$$

 

Next: Using Gauss’s Law For Common Charge Distributions

Previous: Electric Field And Potential Of Charged Conducting Sphere

Back To Electromagnetism (UY1)

 

Back To University Year 1 Physics Notes



Mini Physics

Administrator of Mini Physics. If you spot any errors or want to suggest improvements, please contact us. If you like the content in this site, please recommend this site to your friends!



Leave a Comment

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