# Fields through an insulating slab

1. Mar 6, 2013

### Fluxthroughme

After writing this, I realised that I really have no idea what I'm going on about. Mainly, I don't understand the significance of it being an insulating object. I gather this means the charge is distributed uniformly throughout the object, rather than at the surfaces, but I don't know how I can use this to answer the question(s). Any guidance towards getting the right answer is appreciated. Thank you.

2. Mar 6, 2013

### rude man

It says E_x is a function of x but then doesn't give the function? (Only gives it at x = 0.)

3. Mar 6, 2013

### Fluxthroughme

Yes. Everything I know about the problem is included in the image.

4. Mar 6, 2013

### SammyS

Staff Emeritus
In part c, you have calculated the total flux exiting the rectangular solid by using Gauss's Law.

I part a, you have the flux exiting the 'front' surface. In part b, you have the flux exiting the 'sides'.

From that you should be able to come up with the flux exiting the 'back' surface. From this, find E at the back surface.

5. Mar 6, 2013

### Fluxthroughme

So the total flux from the charge is -2710. As it states it's part of a very large slab, we treat it as part of an infinite one, so the field has a component only in the x direction. Thus, half of the flux goes from the front, half through the back. From the E field from the start, there is a flux of 750 from each surface, giving a total flux as -2710+1500=-1210. As there are two surfaces, the flux through the back one is -605. Using $\phi=EA$, we get a value for E as -100.83N/C.

I'm pretty sure that's wrong, but it's the best I can do from what you've told me. Could you help further?

EDIT:

How about this: At x=0, there is a flux of 750 in the positive x direction. The flux from the charge is half of the total flux (By above reasoning), which means the total flux from the original electric field would have been (2710/2) + 750 = 2105. Thus, the flux through the back would be this plus the flux from the charge, which are both towards the positive x direction, giving 2105+ (2710/2) = 3460. To find the value of E at this surface, divide by the area to give 3460/6 = 576.6N/C

This seems better to me?

Last edited: Mar 6, 2013
6. Mar 6, 2013

### SammyS

Staff Emeritus
You can do the problem by superposition, which is what it appears you are doing -- without actually stating that.

If there were no external field, then yes, half the flux due to the charged slab would exit (or enter) the front surface (x=0), and half would exit (or enter) the back surface (x=-1).

The flux through each of the front and back of the rectangular solid (due only to the slab) would be -2710/2 = -1355 Nm2/C . Check your units. BTW: 1 Nm2/C = 1Vm. (That's Volt∙meter)

From the flux and the area of each face, you can find the E-field strength.

Superimpose upon that a uniform field extending throughout all space, which when added to the field at the front surface gives a resulting field of 125 N/C .