Voltage, capacity and the strength of electric field of a capacitor

In summary, the conversation discusses a homework problem involving a capacitor with a capacity of 400 pikofarads, a voltage of 200 V, and a distance of 2 mm between the plates. The first two parts of the problem involve finding the charge and strength of the electric field. The remaining parts involve changes in the capacity, voltage, and electric field strength when the distance between the plates is doubled and a 100 V voltage supply is added or removed. The equations E=U/d, C=q/U, and C=ε0εS/d are used to solve the problem, but there is confusion about the values of U and E in the different situations.
  • #1
fawk3s
342
1

Homework Statement



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?

Homework 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)

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 don't match the ones in the book.

Thanks in advance !
 
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  • #2
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.
 
  • #3
If I remember correctly, the book gave answers to c) as follows:
U remains the same, E and C are halved.

Finding C isn't 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?

Thanks in advance,
fawk3s
 
  • #4
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...
 
  • #5
I don't 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 can't really figure out why/how is the original voltage halved?

Thanks in advance,
fawk3s
 
  • #6
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.
 
  • #7
Anybody, please? Would be nice to have the solution by tomorrow.
 
  • #8
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.
 
  • #9
It still isn't solved. So I'd bump it again. If you got any ideas, please share !
 
  • #10
Anybody?
 

FAQ: Voltage, capacity and the strength of electric field of a capacitor

1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy in an electric field. It is made up of two conductive plates separated by an insulating material, known as a dielectric.

2. What is voltage and how does it relate to a capacitor?

Voltage, also known as potential difference, is a measure of the electric potential energy between two points in an electric field. In a capacitor, voltage is directly proportional to the amount of charge stored on the plates and the capacitance of the capacitor.

3. What is the capacity of a capacitor?

The capacity of a capacitor, also known as capacitance, is a measure of its ability to store electric charge. It is determined by the size and distance between the plates, as well as the type of dielectric material used.

4. How does the strength of the electric field in a capacitor affect its performance?

The strength of the electric field in a capacitor is directly related to its voltage and capacitance. A stronger electric field can hold more charge and store more energy, making the capacitor more effective at storing and releasing electrical energy.

5. What factors can affect the voltage, capacity, and strength of electric field in a capacitor?

The voltage, capacity, and strength of electric field in a capacitor can be affected by factors such as the distance between the plates, the type of dielectric material used, the size of the plates, and the amount of charge stored on the plates. Temperature and frequency can also have an impact on the performance of a capacitor.

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