How to calculate energy stored in a parallel-plate capacitor?

[ajmCane22]

Homework Statement

A parallel-plate capacitor has plates with an area of 383 cm2 and an air-filled gap between the plates that is 1.51 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery.
(a) How much energy is stored in the capacitor?

(b) The separation between the plates is now increased to 4.53 mm. How much energy is stored in the capacitor now?

(c) How much work is required to increase the separation of the plates from 1.51 mm to 4.53 mm?
Please, explain your reasoning.

The Attempt at a Solution

This is what I did:

C = εA/d to find C and then used E = 1/2CV^2 to find E. Then for part (c) I subtracted the value for part (b) from the value for part (a). I got the following values:
(a)3.712e-5 J
(b)1.236e-5 J
(c)2.476e-5 J

(b) and (c) are incorrect and I'm not sure what I'm doing wrong. (a) and (b) should be the same equation, correct? I don't understand why I'm getting part (a) right and part (b) wrong. Please help.

 

[kuruman]

(b) and (c) are incorrect and I'm not sure what I'm doing wrong. (a) and (b) should be the same equation, correct? I don't understand why I'm getting part (a) right and part (b) wrong. Please help.

When the plates are separated by some additional distance, which of the quantities Q, C, V remain the same and which ones change?

 

[ajmCane22]

Oh! The voltage changes, so I need to calculate the second voltage, right?

 

[kuruman]

Yes, the voltage changes but the charge remains the same.

 

[rwisz]

To calculate the energy stored in a parallel-plate capacitor, you can use the formula E = 1/2CV^2 where C is the capacitance and V is the voltage across the capacitor.

(a) To find the capacitance, you can use the formula C = εA/d where ε is the permittivity of air, A is the area of the plates, and d is the distance between the plates. Plugging in the given values, we get:

C = (8.85x10^-12 F/m)(0.0383 m^2)/(0.00151 m) = 2.245x10^-10 F

Now, we can calculate the energy stored in the capacitor:

E = 1/2(2.245x10^-10 F)(575 V)^2 = 3.712x10^-5 J

(b) To calculate the energy stored in the capacitor with the increased separation between plates, we need to find the new capacitance. Using the same formula as above, we get:

C' = (8.85x10^-12 F/m)(0.0383 m^2)/(0.00453 m) = 7.483x10^-11 F

Now, we can calculate the new energy stored in the capacitor:

E' = 1/2(7.483x10^-11 F)(575 V)^2 = 1.236x10^-5 J

(c) To find the work required to increase the separation of the plates, we can use the formula W = 1/2CV^2 where C is the capacitance and V is the change in voltage. In this case, the change in voltage is 0 since the capacitor is disconnected from the battery. So, the work required is simply 1/2(C'-C)V^2. Plugging in the values calculated above, we get:

W = 1/2(7.483x10^-11 F - 2.245x10^-10 F)(0 V)^2 = 2.476x10^-5 J

I hope this helps clarify any confusion and shows the correct steps for calculating the energy stored in a parallel-plate capacitor.