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Old Feb 28th 2018, 07:20 PM   #1
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What is the electrostatic energy between two charged plates?


Electrostatic energy = kq1q2/r
Or U = qE?
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Old Mar 1st 2018, 07:00 AM   #2
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The energy stored in an electric field is 1/2 QV

Can you do your questions now or do you need more help?

Last edited by studiot; Mar 1st 2018 at 07:14 AM.
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Old Mar 1st 2018, 08:00 AM   #3
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Why the formula that one?
Not kq1q2/r ?
Or Ue = qE

What change is V
Is V = $$\Delta$$ kq/r

Last edited by helly123; Mar 1st 2018 at 08:08 AM.
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Old Mar 1st 2018, 08:29 AM   #4
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Not kq1q2/r ?
Well that is the formula for the force exterted between two charges and you asked about energy.

and this one is the potential energy of a single charge in an existing electric field.
You do not have such charges.

Or Ue = qE
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Old Mar 1st 2018, 08:32 AM   #5
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Thank you.
Then, P = 1/2 QV
What can influence the V?

Last edited by helly123; Mar 1st 2018 at 08:34 AM.
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Old Mar 1st 2018, 09:32 AM   #6
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Originally Posted by helly123 View Post

Electrostatic energy = kq1q2/r
Or U = qE?
The forum rules require that you show us what you've tried before we can help you.

Until then let me ask you what other problems that you've done in this portion of your course, such as other problems with capacitors and the energy stored in them.

Do you know the expression for the energy stored in a capacity or the energy density in an electric field? Do you know the difference between the potential energy associated with two charged particles and the energy stored in an electric field?
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Old Mar 1st 2018, 04:17 PM   #7
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Energy stored in capasitor = 1/2 QV
V is amount of energy per charge
V = E/q = Ed
d is distance traveled parallel to electric field. I am not sure which is d..
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Old Mar 3rd 2018, 07:41 AM   #8
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I found out electrical potential (V) for capacitor usually V = Ed
Can i substitute that V to 1/2QV
So ratio of Electrostatic energy 2 and 1
Is QEd2/QEd1 = d/2d = 1/2

Is it right?
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Old Mar 5th 2018, 08:57 AM   #9
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I don't think that's right. You're comparing a situation where the total distance between the plates is d in one case and 2d in the other, which is not quite the situation you have.

I think it would help you if you find the answer to this question:

Q: what happens to electrons in the conductor when being placed under the electric field?
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