Question 635: Electrostatics & Capacitance

Question

A capacitor is charged by a battery. The battery is removed and another identical unchanged capacitor is connected in parallel. The total electrostatic energy of resulting system

Accepted Answer

When the capacitor is charged by a battery of potential V, then energy stored in the capacitor,
$${U_i} = {1 \over 2}C{V^2}$$   ...(i)
When the battery is removed and another identical uncharged capacitor is connected in parallel
Common potential, $$V' = {{CV} \over {C + C}} = {V \over 2}$$
Then the energy stored in the capacitor,
$${U_f} = {1 \over 2}\left( {2C} ight){\left( {{V \over 2}} ight)^2} = {1 \over 4}C{V^2}$$   ...(ii)
$$ 	herefore $$ From eqns. (i) and (ii)
$${U_f} = {{{U_i}} \over 2}$$
that means the total electrostatic energy of resulting system will decreases by a factor of 2.

Answer

decreases by a factor of 2

Answer

remains the same

Answer

increases by a factor of 2

Answer

increases by a factor of 4

Electrostatics & Capacitance

A capacitor is charged by a battery. The battery is removed and another identical unchanged capacitor is connected in parallel. The total electrostatic energy of resulting system

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