Question

The energy sotred in parallel plate capacitor having charge Q and
capacitance C is:(a)1/2QC (b) Q^2C
(c)Q^2/2C
d) 2Q^2/C​

Answer 1

Let dQ be the charge b/w the parallel plate capacitor .

V be the potential difference b/w the plates .

Energy , dU gained from moving a charge dQ through the potential difference ,V is given by ,

$:\implies \sf dU=VdQ$

Integrating on both sides ,

$:\implies \displaystyle \sf \int dU=\int VdQ$

• Q = CV

→ V = Q / C

$:\implies \displaystyle \sf \int dU=\int \dfrac{Q}{C}\ dQ$

$\displaystyle :\implies \sf U=\dfrac{1}{C}\int QdQ$

$:\implies \sf U=\dfrac{1}{C}.\dfrac{Q^2}{2}$

$:\implies \sf \pink{U=\dfrac{Q^2}{2C}}\ \; \bigstar$

Option C