# Two infinite sheets of different charge density

1. May 19, 2012

### purplegreen

1. The problem statement, all variables and given/known data
From the picture attached it can be seen that there are two plates with charge density σa and -σb. We will chose → as the n direction.
What is that electric field everywhere for this system?

2. Relevant equations
∫∫E.dS = ∫∫σ/ε0 dA

3. The attempt at a solution

Using Gauss' law we can take three gaussian surfaces. One containing both of the plates and two more containing each individual plate.

Using the surface containing both plates, we have:
ESn + ESn = σ(a-b)/ε0 n

hence,
E = σ(a-b)/2ε0 n outside of the plates

Inside I think I should use superposition of the two other surfaces. However I'm not sure which direction the electric field goes in for these two surfaces?

Generally I'm not convinced by my answer and lack some understanding which I would like some help with. Thanks in advance!

#### Attached Files:

• ###### Twoinfinitesheets.jpg
File size:
23.9 KB
Views:
347
Last edited: May 19, 2012
2. May 19, 2012

### Dickfore

Using Gauss's Lw, you should be able to show that an infinite charged sheet creates a homogeneous electric field directed perpendicularly to the sheet:
$$\vec{E} = \frac{\sigma}{2 \epsilon_0} \, \hat{n}$$
where $\hat{n}$ is a unit normal directed from the sheet outwards (on each side).

Now you can use the principle of superposition to find the electric field due to two sheets of charge.

3. May 19, 2012

### purplegreen

Ok so would the answers for this question be:

on the left hand side of the 2 plates:
E= -σ(a-b)/2ε n

In the middle of the plates:
E= σ(a+b)/2ε n

Finally, on the right hand side of the plates:

E= σ(a-b)/2ε n

Thanks for the help.