The problem statement, all variables and given/known data 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?
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.