# Electric Field between Three Plastic Sheets

1. Feb 13, 2016

### CARNiVORE

1. The problem statement, all variables and given/known data
Figure (a) shows three plastic sheets that are large, parallel, and uniformly charged. Figure (b) gives the component of the net electric field along an x axis through the sheets. The scale of the vertical axis is set by

Es = 8.4 × 105 N/C. What is the ratio of the charge density on sheet 3 to that on sheet 2?

2. Relevant equations
E = σ / 2ε0

3. The attempt at a solution
I know that the field between 1 and 2 is three times less strong as that between 2 and 3.

3E12 = E23

Not sure where to go from there. Thanks in advance!

2. Feb 13, 2016

### Staff: Mentor

Think superposition. Assume for the moment that all the charges are positive (we know that's unlikely, but it's a bit like assuming current directions in a circuit -- some may turn out to be negative). For each region sketch in field vectors associated with each sheet (there will be one vector for each sheet in each of the regions, so three vectors in each).

The directions of the vectors in each region will tell you how the fields add in those regions.

Does that give you any ideas?

3. Feb 13, 2016

### CARNiVORE

Not quite sure if I follow - I see that sheet 1 is pushed to the left by sheet 2 and slightly to the left by sheet 3, sheet 2 is pushed left by sheet 3 and right by sheet 1, and sheet 3 is pushed to the right by sheet 2 and slightly to the right by sheet 1. This, of course, is assuming that they are all positively charged, which might not be the case (like you said). I also know that, due to their positive charge, they all give an outward 'push.' I don't know what to get from this, tho.

4. Feb 13, 2016

### Staff: Mentor

"pushes" are irrelevant. Assume the sheets are rigid and fixed in place, then ignore them except as region separators.

For now, consider only the electric field directions in the regions. There are three fields, one per sheet. The field from each sheet occurs in every region. All that matters for now is the directions of the fields in each of the regions. It's really helpful to make a sketch, penciling in a short directional arrow for each field in each region.

So for example, in the leftmost region (to the left of sheet 1), all three fields would be pointing left. Can you fill in the field vectors for the other regions?

5. Feb 14, 2016

### CARNiVORE

Good idea - here's the sketch I drew:

Now, I know the lengths of the arrows I drew don't necessarily reflect their magnitude, because of differences in charge density and E-field. They're just convenient to display the distance between them. It's useful to see them out like this, but I'm still a bit confused how to approach even with this picture.

6. Feb 14, 2016

### Staff: Mentor

Okay. Now from your relevant equation you know that the individual field strengths are directly proportional to the charge densities on the sheets. If you can find the relative field strengths then you have the relative charge densities, too.

Assign a variable for each of the field strengths, say x1, x2, and x3. Then using the arrow directions in your figure to determine signs, write equations for the total field strength for each region. How many independent equations do you get?

7. Feb 14, 2016

### CARNiVORE

All right, so I ended up with four equations. One for the space to the left of 1, one between 1 and 2, one between 2 and 3, and one for the space to the right of 3. Here they are:

E1- = -E1 - E2 - E3
E12 = 1/3 * E23 = E1 - E2 - E3
E23 = E1 + E2 - E3
E3+ = E1 + E2 + E3

Combining equations 2 and 3, I got:
2/3 * E23 = E1 - E3
2 * E2 = E1 - E3

I know I can substitute my relevant equation, but I'm not sure if I have enough information to solve for the ratio yet.

8. Feb 14, 2016

### Staff: Mentor

You are given the values for the sums of the fields in each region. Set your expressions for the sums equal to the given values (just use the magnitude; no need to deal with the units at this point since you're looking for a ratio anyways so the units will disappear).

How many independent equations do you have?

9. Feb 14, 2016

### CARNiVORE

I see what you mean - in this case, I get 3 independent equations:

0 = E1 + E2 + E3
2.8 * 10^5 = E1 - E2 - E3
8.4 * 10^5 = E1 + E2 - E3

Using these three equations, I found that E2 = 2.8 * 10^5 and E3 = -4.2 * 10^5. Is this correct?

10. Feb 14, 2016

### Staff: Mentor

Yes. Or, since you're only interested in a ratio you could have simply written:

$0 = E_1 + E_2 + E_3$
$2 = E_1 - E_2 - E_3$
$6 = E_1 + E_2 - E_3$

Solve for the E's.

11. Feb 14, 2016

### CARNiVORE

Awesome! That was my last question, and now homework is finally over for the week... time to get started on next week's, lol. Thanks again, gneill!