Two identical capacitors of 12 pF each are connected in series across a 50 V battery. Calculate the electrostatic energy stored in the combination. If these were connected in parallel across the same battery, find out the value of the energy stored in this combination.

The energy stored in capacitor, when connected in series combination;

We have given, $C_{S}=6{P}F=6\times 10^{-12}F$

Hence, the energy stored $(E_{S})$ in capacitor ;-

$E_{S}=\frac{1}{2}C_{S}V^{2}$

$=\frac{1}{2}\times 6\times 10^{-12}\times 50\times 50$  (Given; V=50)

$=7500\times 10^{-12}J$

$E_{S}=7.5\times 10^{-9}J$

Now, when capacitors are connected in parallel,

the value of the capacitor is given as,

$C_{P}=-24{P}F=24\times 10^{-12}F$

there, energy in the capacitor in parallel combination is given as ;

$E_{P}=\frac{1}{2}C_{P}V^{2}$

$E_{P}=\frac{1}{2}\times 24\times 10^{-12}\times 50\times 50$

$E_{P}=3\times 10^{-8}J$

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