# What is electric field problem?

## Homework Statement

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?

## Homework Equations

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

## 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.

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.

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

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

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

E = (1.13 x 1011)(10) = 1.13x1012 Coulombs/m2
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.
b) V = EL = (1.13x1012)(.0002m) = 2.26x108Volts
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?

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

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

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)?

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

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 ...

So then the final answer is 3.6Coulombs?