# Voltage, capacity and the strength of electric field of a capacitor (1 Viewer)

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#### fawk3s

1. The problem statement, all variables and given/known data

So basically we're given a capacitor, or 2 metal plates, which has the capacity of 400 pikofarads (4*10^-10 F), a voltage of 200 V, and the distance between the two plates is 2 mm (2*10^-3 m).

a) Find the charge of a plate.
b) The strength of the electric field.
c) How does the capacity, voltage and the strength of the electric field change, when the distance between the 2 plates is doubled and the system is given a 100 V voltage supply?
d) How do they change, when the 100 V supply is now removed and the distance is yet once again doubled?

2. Relevant equations

E=U/d - (strength of the electric field = voltage / distance)
C=q/U - (capacity = charge / voltage)
C=ε0εS/d - (capacity = epsilon-0*epsilon*area / distance)

3. The attempt at a solution

a) q=U*C => 8*10^-8 coulombs
b) E=U/d => 10^5 V/m

But points c) and d) are which confuse me. My answers dont match the ones in the book.

Thanks in advance !

#### Tusike

Use your equations. If d is doubled, then C=eo*S/d is halved. q = U*C, where both U and C have been halved, so it's one-fourth of the original value (I'm assuming you miswrote c) and it asked for the charge, not the given 100V voltage Similarly you should get the rest of the answers. Hope this helped you start out.

#### fawk3s

If I remember correctly, the book gave answers to c) as follows:
U remains the same, E and C are halved.

Finding C isnt a problem, as you stated above. But finding U confuses me. As it is given the extra 100 V supply, it could remain the same when the original value halves. But how do we know its halves? And why does the charge change?
E=U/d, so it is halved when U stays the same (with the extra voltage supply). But again, why is the original value of U halved?

fawk3s

#### Tusike

In the first situation the capacitor had some charge on it, and so it's voltage happened to be 200V. In the second, you put it on 100V, and it's charge changed according to that. However I think I'm misunderstanding something, because you said the answer to c) was that U remains the same, 200V. I don't see how that can be if you put it on 100V...

#### fawk3s

I dont think its meant that you put it on 100 V. I think it means that after doubling the distance, the 100 volts are added extra to the voltage. IF I understood correctly.

So that would mean that the original voltage would have to be 100 V. And when adding the 100 extra volts, it becomes 200 V, aka the same as in the first situation.
But the problem is, I cant really figure out why/how is the original voltage halved?

fawk3s

#### fawk3s

Anybody got any ideas? Since the exercise seems really easy, its bugging me now.
If anybody could point out the flaw in my logic, I'd really appreciate it.

#### fawk3s

Anybody, please? Would be nice to have the solution by tomorrow.

#### fawk3s

I hate bumping this, but I really need it in a couple of hours. So if anybody's got any ideas, I'd be grateful if you'd share them.

#### fawk3s

It still isnt solved. So I'd bump it again. If you got any ideas, please share !

Anybody?

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