Change in energy of a capacitor

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SUMMARY

The discussion centers on a capacitor with an initial capacitance of 5.0 μF charged to 5.0 μC, which is then modified to a capacitance of 10 μF by decreasing the separation between its plates. The work done by the agent is calculated using the change in energy formula, W = ΔU, where the energy stored in the capacitor is given by Uf = (1/2)C(Q^2). The key point is that while the charge remains constant, the increase in capacitance results from a reduction in plate separation, not an increase in plate area.

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MathewsMD
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A certain capacitor has a capacitance of 5.0 μF. After it is charged to 5.0 μC and isolated, the plates are brought closer together so its capacitance becomes 10μF. The work done by the agent is about:

I assumed Qfinal = Qinitial since I believe there is still some separation, right (it's still a capacitor, isn't it?)? So, I did W = ΔU and let Uf = (1/2C)Q2 and I did Ui = (1/2C)Q2 and I increased the capacitance by 2 since area is doubled, right? But how does the distance of separation change?

Any help please?
 
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MathewsMD said:
A certain capacitor has a capacitance of 5.0 μF. After it is charged to 5.0 μC and isolated, the plates are brought closer together so its capacitance becomes 10μF. The work done by the agent is about:

I assumed Qfinal = Qinitial since I believe there is still some separation, right (it's still a capacitor, isn't it?)? So, I did W = ΔU and let Uf = (1/2C)Q2 and I did Ui = (1/2C)Q2 and I increased the capacitance by 2 since area is doubled, right? But how does the distance of separation change?

Any help please?

The area of the capacitor plates is not changing. The increase in capacitance is due to the decrease in separation between them (What's the formula for the capacitance of a parallel plate capacitor?).

You're correct that the charge remains the same. So what's the change in energy stored on the capacitor between the initial configuration and final configuration?
 

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