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Homework Help: What is electric field problem?

  1. Jul 1, 2010 #1
    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

    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. jcsd
  3. Jul 1, 2010 #2
    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.
  4. Jul 1, 2010 #3
    a) Well it gives 10 Coulombs/m2 which equals Q/A, so how do I relate this to E=1.13x1011 Q/A?

    b) V=(answer in A)(.00002m)

    c) I still don't quite understand
  5. Jul 1, 2010 #4
    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?
  6. Jul 1, 2010 #5
    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:


    27.8 coulombs = Q

    ...This seems wrong
  7. Jul 1, 2010 #6
    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?
  8. Jul 1, 2010 #7
    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
  9. Jul 1, 2010 #8
    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)?
  10. Jul 1, 2010 #9
    I posted it above ^ you didn't say if its correct or not lol.
  11. Jul 1, 2010 #10
    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 ...
  12. Jul 1, 2010 #11
    So then the final answer is 3.6Coulombs?
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