How does the energy change in a capacitor with variable plate distance?

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SUMMARY

The discussion focuses on the energy change in a capacitor with variable plate distance while maintaining a constant charge Q. When the distance between the plates increases by 20%, the capacitance C decreases according to the formula C = εA/d, where ε is the permittivity and A is the plate area. Consequently, the energy stored in the capacitor, given by E = (1/2)QV or E = Q²/2C, decreases as capacitance diminishes. Thus, the energy in the electric field between the plates changes inversely with the increase in distance.

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Homework Statement


A capacitor has variable distance between the plates . It charges with a constant charge Q and the distance between the plates increase with 20%. How many percent does the energy change in the electric field between the plates?


Homework Equations


E = (1/2)QV = (1/2)CV^2 = Q^2/2C


The Attempt at a Solution


I remember my teacher said that the energy only depends on e. So the energy is the same.
 
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chopticks said:

Homework Equations


E = (1/2)QV = (1/2)CV^2 = Q^2/2C


Here Q is constant.because distance between the plates changes,capacitance also changes.

[tex]C=\epsilon\frac{A}{d}[/tex]

[tex]\epsilon[/tex] and A are also constants.
 
ok. If I want to find the maximum energy/volume then the distance doesn't matter, right?
 

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