Energy of parallel plate capacitor problem

In summary, before insertion of dielectric: U = (1/2)CV^2After insertion of dielectric: U = (1/2)KCV^2
  • #1
meteorologist1
100
0
Hi, I have trouble on the following problem:

Given a parallel plate capacitor, fixed area A, and fixed separation d. Find the energy stored, before and after insertion of a slab of dielectric, which completely fills the space between plates, for each of the two cases:

a) Plates are connected to a battery which maintains constant potential difference

b) Plates are charged with fixed charges +Q and -Q, and battery disconnected.

Explain differences in the two cases.

For part a, I think the answers would be just:
Before insertion of dielectric: U = (1/2)CV^2 where C = A(epsilon)/d and
after insertion of dielectric: U = (1/2)KCV^2 where C = A(epsilon)/d and K is the dielectric constant.

But I'm not sure about part b. Thanks.
 
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  • #2
How about [itex] U=\frac{Q^{2}}{2C} [/itex],where C is the electric capacity before and after inserting the diellectric...?

Daniel.
 
  • #3
meteorologist1 said:
For part a, I think the answers would be just:
Before insertion of dielectric: U = (1/2)CV^2 where C = A(epsilon)/d and
after insertion of dielectric: U = (1/2)KCV^2 where C = A(epsilon)/d and K is the dielectric constant.
You've got the right idea, but be careful how you express it. The capacitance changes when you insert the dielectric: before insertion, C = A(epsilon)/d; after insertion, C = KA(epsilon)/d.

U = (1/2)CV^2 is always true, but C changes, so U changes from (1/2)[A(epsilon)/d]V^2 to (1/2)K[A(epsilon)/d]V^2. (Don't write U = (1/2)KCV^2; that's not true!)

But I'm not sure about part b.
If the charge is fixed, what happens to the potential difference when the dielectric is inserted. (What happens to the electric field within the plates?)
In this case, both C and V change.

You can also write the stored energy directly in terms of Q and C. (Figure that out.) Then you'd only have to worry about C changing.
 
  • #4
Ok, I understand now. Thank you very much.
 

1. What is the formula for calculating the energy of a parallel plate capacitor?

The formula for calculating the energy of a parallel plate capacitor is U = 1/2 CV², where U is the energy in joules, C is the capacitance in farads, and V is the potential difference in volts.

2. How is the energy of a parallel plate capacitor related to its capacitance and voltage?

The energy of a parallel plate capacitor is directly proportional to its capacitance and the square of its voltage. This means that as the capacitance or voltage increases, the energy stored in the capacitor also increases.

3. Can the energy of a parallel plate capacitor be negative?

No, the energy of a parallel plate capacitor cannot be negative. It is always a positive value, representing the amount of potential energy stored in the electric field between the capacitor plates.

4. How does the spacing between the plates affect the energy of a parallel plate capacitor?

The spacing between the plates of a parallel plate capacitor is directly proportional to its capacitance. This means that as the spacing decreases, the capacitance and energy of the capacitor increase. However, the voltage also increases, so the energy may not change significantly.

5. Is the energy of a parallel plate capacitor affected by the material of the plates?

Yes, the energy of a parallel plate capacitor can be affected by the material of the plates. This is because the dielectric constant of the material affects the capacitance of the capacitor, which in turn affects the energy stored. Different materials will have different dielectric constants, thus resulting in different energies for the same capacitor.

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