- #1

Guillem_dlc

- 188

- 17

- Homework Statement
- We have a sphere of radius ##R=6\, \textrm{cm}## charged with a volume density of charge ##\rho =10\, \mu \textrm{C}/\textrm{m}^3## centered at the origin of coordinates. An infinite plane located in the ##z=0##-plane will be charged with a surface charge density ##\sigma =0,2\, \mu \textrm{C}/\textrm{m}^2##. What is the modulus of the electric field at its point ##(0,3,4)\, \textrm{cm}?

Answer: ##2,87\cdot 10^4\, \textrmV}/\textrm{m}##

- Relevant Equations
- ##E_\sigma=\dfrac{\sigma}{2\varepsilon_0}##

At point ##P(0,0'03,0'04)## the field caused by the sphere is added to the field caused by the plane.

First, ##E_\sigma##

$$E_\sigma=\dfrac{\sigma}{2\varepsilon_0}=\dfrac{0,2\cdot 10^{-6}}{2\varepsilon_0}=11299,44\, \textrm{V}/\textrm{m}$$

Then, ##E_0##: Because ##r<R##:

$$E_0=\dfrac{\rho}{3\varepsilon_0}r=\dfrac{10^{-6}\cdot 10}{3\varepsilon_0}$$

We see that ##r=|\vec{r}|=\sqrt{0,03^2+0,04^2}=0,05\, \textrm{m}## because ##\vec{r}=(0,0'03,0'04)-(0,0,0)=(0,0'03,0'04)##. Then,

$$E_0=\dfrac{10\cdot 10^{-6}}{3\varepsilon_0}\cdot 0,05=18832,39\, \textrm{V}/\textrm{m}$$

Finally,

$$E_T=E_\sigma +E_0=18832,39+11299,44=30131,83\, \textrm{V}/\textrm{m}$$

I don't get the solution I should give, should I have done modulus instead of direct summation?