# Capacitor with uniform space charge between them

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1. Mar 3, 2017

### rohanlol7

1. The problem statement, all variables and given/known data

2 large plates are seperated by a distance d and a space charge of uniform charge density p is placed between them and a potential difference V is applied accross the plates. Find the electric field stength at a distance x fromt the positive plate
The answer is -V/d +p(x-2d)/2e ( e = epsiolon0)
2. Relevant equations
gauss law

3. The attempt at a solution
using gauss law:
EA = Qenc/e, Qenc = A*g + charge inside my box, which i can't seem to find, since i have no idea how all the charge will redistribute, i'm guessing some will go onto the plates, or maybe they will redistrubute so as to create an opposing field of equal magnitude to the external field....

2. Mar 3, 2017

### kuruman

Suppose the charges are fixed and do not redistribute. What would your answer be in that case?

3. Mar 3, 2017

-V/d + px/e

4. Mar 3, 2017

### kuruman

How did you find this? DId you consider the electric flux through both sides of the Gaussian surface?

5. Mar 3, 2017

### rohanlol7

no i didn't. So if i try and consider this, does is go like this ? the E field going from the positive plate will be -V/2d + px/e and through the left that should give -V/2d -p(d-x)/e and adding those two would give -V/d + pd/e ?

6. Mar 3, 2017

### kuruman

You have to be careful here. Suppose you only have the space charge. The electric field at a point equidistant from the two ends must be zero because you have as much charge on the left side as on the right. Therefore, whatever expression you find for the space charge electric field contribution alone must give zero at x = d/2. Do you agree?

7. Mar 3, 2017

### rohanlol7

yes i definitely do agree, i did realise that if its at a point x then a further distance of x on the other side will cancel that E field, however i'm not sure how to go around formulating that properly mathematically properly

8. Mar 3, 2017

### kuruman

Consider a Gaussian surface with one edge at d/2. Call that x = 0 (temporarily). The other edge of the surface is at x to its right. Use Gauss's law to find the field through the surface at x. Move the origin to the zero that the problem has defined by transforming x → x - d/2.