Affect on stored energy if charge is doubled

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When the charge on a parallel plate capacitor is doubled while maintaining a constant potential difference, the stored energy increases. The energy stored in a capacitor is given by the formula U = 0.5QV, and if the charge (Q) is doubled, the energy calculation shows it actually quadruples due to the relationship E = Q^2 / 2C. The voltage must also adjust to keep the potential difference constant, leading to a new voltage of 2V when Q is doubled. Therefore, the conclusion is that the stored energy in the capacitor is quadrupled when the charge is doubled. This aligns with the formulas provided, confirming that the stored energy increases by a factor of four.
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Homework Statement

A parallel plate capacitor is air filled and assumed to be ideal. It is connected to a battery so it has constant potential difference. By what factor does stored energy change if the charge is doubled..

The attempt at a solution

Using the formula U = 0.5QV..

If charge is doubled, then U is doubled correct?

I say the answer is stored energy is doubled, but the answer at the back reckons quadrupled. Who is right?
 
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E = Q^2 / 2C is another formula. That would yield a 4 times greater energy being stored.

Also, if E = (0.5)QV, then if Q = 2Q

Q = CV
C = Q/V

C = 2Q/2V, as C cannot change, so V must change in order to keep it constant.

E = (0.5).2Q.2V

I think.
 
well I'm given two formulas, U = (0.5)QV and U = 1/2CV^2

The second formula is not relevant as charge is not a variable, hence my above reasoning...
 
In case the charge doubles, the voltage would also get doubled. Hence stored energy would get quadrupled. All the formulas would yield the same result.
 
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