Energy of a Capacitor in the Presence of a Dielectric

In summary: So, the total energy U_2 = (1/2)*(C1+C2)*V^2 = (1/2)*epsilon_0*((A/2/d) + K*(A/2/d))*V^2 = (1/2)*epsilon_0*K*(A/d)*V^2.
  • #1
odd_concept
2
0
An dielectric-filled parallel-plate capacitor has plate area A and plate separation d. The capacitor is connected to a battery that creates a constant voltage V. The dielectric constant is K.

A. Find the energy U_1 of the dielectric-filled capacitor. The capacitor remains connected to the battery

I was able to figure out this one:
(epsilon_0*K*A*V^2)/(2d)

But I cannot figure out the rest...

B.The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U_2 of the capacitor at the moment when the capacitor is half filled with the dielectric. Express your answer in terms of A,d,V,K,epsilon_0.

C.The capactor is now disconnected from the battery, and the dielectric plate is then slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U_s.Express your answer in terms of A,d,V,K,epsilon_0

D. In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric.Express your answer in terms of A,d,V,K,epsilon_0

ANY help would be greatly appreciated... Thanks in advance
 
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  • #2
When the capacitor is half filled with the dielectric, you can imagine two capacitors having area A/2 each and connected in parallel. One of them is with dielectric and other without dielectric. Calculate the equivalent capacitance and charge on it. By this hint you can solve C and D
 
  • #3
I tried this for C and it was wrong?
epsilon_0*K*(A/2)*V^2/(2d)

what am I doing wrong?
 
  • #4
C1 = epsilon_0*A/2*/d and C2 = epsilon_0*K*(A/2)/d. Since the battery is still connected the potential difference is same in C1 and C2.
 

1. What is the energy of a capacitor in the presence of a dielectric?

The energy of a capacitor in the presence of a dielectric is the amount of electrical potential energy stored in the capacitor's electric field when a dielectric material is placed between the capacitor's plates. This energy is equal to the work done in moving charges from one plate to the other, and is given by the equation:
W = 1/2 * C * V^2

2. How does a dielectric affect the energy of a capacitor?

A dielectric material, when placed between the plates of a capacitor, reduces the electric field strength and increases the capacitance of the capacitor. This results in an increase in the energy stored in the capacitor, as given by the equation:
W = 1/2 * C * V^2

3. Can the energy of a capacitor in the presence of a dielectric be negative?

No, the energy of a capacitor in the presence of a dielectric cannot be negative. This is because the energy is a measure of the potential energy stored in the electric field, and potential energy cannot be negative.

4. What happens to the energy of a capacitor if the dielectric material is removed?

If the dielectric material is removed from between the plates of a capacitor, the electric field strength increases and the capacitance decreases. This results in a decrease in the energy stored in the capacitor, as given by the equation:
W = 1/2 * C * V^2

5. How does the distance between the plates of a capacitor affect its energy in the presence of a dielectric?

The distance between the plates of a capacitor does not directly affect the energy stored in the capacitor in the presence of a dielectric. However, it does affect the capacitance of the capacitor, which in turn affects the energy stored. The closer the plates are, the higher the capacitance and the greater the energy stored, and vice versa.

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