# What is electric field problem?

1. Jul 1, 2010

### physics(L)10

1. The problem statement, all variables and given/known data
Two uniform plates have a uniform charge density of 10 Coulombs/m2. The plates have dimensions of 0.6x0.6 m2 and are seperated by 2mm

a) The electric field (V/m) between the two plates is given by E=1.13x1011 Q/A where Q is the charge in Coulombs on one plate and A is the area in m2. Evaluate the field.

b) What is the voltage between the two plates?

c) What is the charge on one plate?

2. Relevant equations

E=F/q
V=W/q=EL,
where L=distance,E=electric field, F=force, q=charge, V=voltage, W=work (J)

3. The attempt at a solution

This is really confusing me. There's so much information, I just don't know which info I can ignore. I think I got all the equations needed for this question though. Help pleaseeee.

2. Jul 1, 2010

### dulrich

Part (a). Evaluate the field means calculate its value. You are given the formula in the problem statement, but you need to plug in Q/A. What are the units of Q/A?

Part (b). Use V = EL. Plug in the value for E that you get from part (a).

Part (c). If you did part (a), you must have figured out what Q/A is. Use this to figure out Q.

3. Jul 1, 2010

### physics(L)10

a) Well it gives 10 Coulombs/m2 which equals Q/A, so how do I relate this to E=1.13x1011 Q/A?

c) I still don't quite understand

4. Jul 1, 2010

### dulrich

Maybe if I rewrite this in symbols you will see:

Q/A = 10 columbs/m2

E = (1.13 x 1011)(Q/A)

What is E?

5. Jul 1, 2010

### physics(L)10

Ooooooh I see, thanks, you made that really clear. So,

E = (1.13 x 1011)(10) = 1.13x1012 Coulombs/m2

b) V = EL = (1.13x1012)(.0002m) = 2.26x108Volts

c) I'm gonna attempt this one and say that since you said E = (1.13 x 1011)(Q/A), then:

1.13x1012=(1.13x1011)(Q/.36)

27.8 coulombs = Q

...This seems wrong

6. Jul 1, 2010

### dulrich

Yep ... except the units are wrong. They should be V/m. This really isn't your fault because in the original equation the units for the 1.13 x 1011 are actually Vm/C.
Close, 2 mm = 0.002 m.

For part (c), you know Q/A = 10 and you want Q. Do you know A?

BTW, these seem like really big numbers. Are you sure that Q/A = 10?

7. Jul 1, 2010

### physics(L)10

Don't you divide 2mm by 100 twice to get to m?

Yes, A is 0.36m. They give the dimensions of the plates as .6mx.6m

Yup, that's what they give in the question. It says charge density is 10 Coulombs/m2

8. Jul 1, 2010

### dulrich

No, there are 1000 millimeters in one meter so 1 mm = 0.001 m. You might be thinking of converting cm2 to m2. The square means you have to convert twice (and a cube would mean convert three times).

Did you get the answer to part (c)?

9. Jul 1, 2010

### physics(L)10

I posted it above ^ you didn't say if its correct or not lol.

10. Jul 1, 2010

### dulrich

Oh, I see. Sorry that one's wrong too.

But it's even easier to do this:

Q/A = 10, and A = 0.36 so, 10 = Q/.36 ...

11. Jul 1, 2010

### physics(L)10

So then the final answer is 3.6Coulombs?