Prove that the energy stored in a parallel plate capacitor is q²/2C.
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Let charge on plates are -q and +q , distance between parallel plates is d , electromotive force of battery which is connected to capacitor is V.
As we know, q = CV , where C is capacitance of capacitor.
It means, if we connect to battery charge stored in capacitor . Let dq charge stored in capacitor . Then wordone is required to increase charge dq in capacitor = q.dV
= CV.dV [ ∵q = CV]
= C ∫VdV
= 1/2 CV²
W = 1/2 CV² , now take V = q/C
Then, W = q²/2C
We know, workdone stored in system is named as potential energy I'd system.
So, potential energy , U = q²/2C
Hence, energy stored in parallel plate capacitor is U = q²/2C
As we know, q = CV , where C is capacitance of capacitor.
It means, if we connect to battery charge stored in capacitor . Let dq charge stored in capacitor . Then wordone is required to increase charge dq in capacitor = q.dV
= CV.dV [ ∵q = CV]
= C ∫VdV
= 1/2 CV²
W = 1/2 CV² , now take V = q/C
Then, W = q²/2C
We know, workdone stored in system is named as potential energy I'd system.
So, potential energy , U = q²/2C
Hence, energy stored in parallel plate capacitor is U = q²/2C
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