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

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To calculate the energy stored in a parallel-plate capacitor, the capacitance is determined using the formula C = εA/d, followed by applying E = 1/2CV^2 for energy calculations. The user correctly calculated the energy for part (a) but is confused about discrepancies in parts (b) and (c). When the plate separation increases, the charge remains constant while the voltage changes, which affects the energy stored. The user needs to recalculate the voltage for part (b) to find the correct energy value. Understanding these relationships is crucial for accurate calculations in capacitor problems.
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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.
 
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ajmCane22 said:
(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?
 
Oh! The voltage changes, so I need to calculate the second voltage, right?
 
Yes, the voltage changes but the charge remains the same.
 
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