Question

A capacitor of capacitance 800pf is charged by a source 100v.What is the energy stored by the capacitor??

Answer 1

Hello ❣️

Initially, C = 800 pF = 8*10-10 F, V (battery) = 100 V

Therefore, charge (q) on capacitor = CV = 8*10-8 C

Now when the battery is disconnected and the capacitor is connected to another capacitor of same capacitance, the quantity that is conserved is the charge. Total charge remains unchanged as there is no external agency (battery has been removed) to supply further charge. But this charge is now equally divided between the two capacitors in the ratio of their capacitances, i.e. qnew = 4*10-8 C on each of the two capacitors. So electrostatic energy now stored is 2*(qnew2/2C) =

2*(4*10-8)2/(2*8*10-10) = 2*10-6 J

Thus, total electrostatic energy now stored = 2 µJ.

Answer 2

Answer:

A capacitor of capacitance 800pf is charged by a source 100v, the energy stored by the capacitor is $4*10^{-6}J$

Explanation:

• A capacitor is used to store charge

• The electrical potential energy is stored in a capacitor

• For a capacitor with capacitance (C) and potential difference across the battery (V), the energy stored is

        $U=\frac{1}{2}CV^{2}$

From the question, we have

the capacitance of the capacitor    $C=800pF=800*10^{-12}F$

the voltage across the battery $V=100v$

put the values given,

the energy stored in the capacitor is

$U=\frac{1}{2}*800*10^{-12} *(100^{2} )$

⇒  $U=4*10^{-6} J$