If the potential difference across a capacitor is doubled, by what factor does the energy stored change?
Knowing that U=1/2QV and U=1/2CV^2 i can use one formula and get that the energy increases by a factor of 4, while i use the other formula and get the energy increasing by a factor of 16.
How do i know which formula to use?
Knowing that U=1/2QV and U=1/2CV^2 i can use one formula and get that the energy increases by a factor of 4, while i use the other formula and get the energy increasing by a factor of 16.
How do i know which formula to use?
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goes up by a factor of 4, using U = ½CV², since C is unchanged.
yes, there are other formulae:
U = ½CV² = ½QV = ½Q²/C Energy in a cap in Joules
You have to use Q = CV, which means when you double the voltage, Q also doubles
you used U = ½QV. Q doubles because V doubles, so the total is two doublings, or 4x, same answer.
You then used U = ½CV², V doubles, 2 squared is 4, same answer.
I don't see where you got a factor of 16 ??
.
yes, there are other formulae:
U = ½CV² = ½QV = ½Q²/C Energy in a cap in Joules
You have to use Q = CV, which means when you double the voltage, Q also doubles
you used U = ½QV. Q doubles because V doubles, so the total is two doublings, or 4x, same answer.
You then used U = ½CV², V doubles, 2 squared is 4, same answer.
I don't see where you got a factor of 16 ??
.