Question 116: Electrostatics & Capacitance

Question

A $$600 ~\mathrm{pF}$$ capacitor is charged by $$200 \mathrm{~V}$$ supply. It is then disconnected from the supply and is connected to another uncharged $$600 ~\mathrm{pF}$$ capacitor. Electrostatic energy lost in the process is ____________ $$\mu \mathrm{J}$$

Accepted Answer

The energy stored in a capacitor can be calculated using the formula:
$ U = \frac{1}{2} C V^2 $
where:
(U) is the energy,
(C) is the capacitance,
(V) is the voltage.
Initially, the energy stored in the first capacitor is:
$ U_{	ext{initial}} = \frac{1}{2} C V^2 = \frac{1}{2} 	imes 600 	imes 10^{-12} \, 	ext{F} 	imes (200 \, 	ext{V})^2 = 0.012 \, 	ext{J} = 12 \, \mu	ext{J}. $
When the charged capacitor is connected to the uncharged capacitor, the charge will distribute equally between them because they have the same capacitance. Therefore, the final voltage across each capacitor is half of the initial voltage, i.e., 100 V.
The energy in each capacitor after the redistribution is:
$ U_{	ext{final each}} = \frac{1}{2} C \left(\frac{V}{2}ight)^2 = \frac{1}{2} 	imes 600 	imes 10^{-12} \, 	ext{F} 	imes (100 \, 	ext{V})^2 = 0.003 \, 	ext{J} = 3 \, \mu	ext{J}. $
As there are two capacitors, the total final energy is:
$ U_{	ext{final total}} = 2 	imes U_{	ext{final each}} = 2 	imes 3 \, \mu	ext{J} = 6 \, \mu	ext{J}. $
The energy loss is the difference between the initial energy and the final energy:
$ \Delta U = U_{	ext{initial}} - U_{	ext{final total}} = 12 \, \mu	ext{J} - 6 \, \mu	ext{J} = 6 \, \mu	ext{J}. $

Electrostatics & Capacitance

A $600 ~\mathrm{pF} capacitor is charged by 200 \mathrm{~V} supply. It is then disconnected from the supply and is connected to another uncharged 600 ~\mathrm{pF} capacitor. Electrostatic energy lost in the process is ____________ \mu \mathrm{J}

Integer / Numeric Answer Type