Question 392: Electrostats & Capacitance

Question

A capacitor of capacitance $$900 \mu \mathrm{F}$$ is charged by a $$100 \mathrm{~V}$$ battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as $$x 	imes 10^{-} { }^{2} \mathrm{~J}$$. The value of $$x$$ is _____________.

Accepted Answer

$${U_i} = {1 \over 2}C{V^2} = {1 \over 2} 	imes 900 	imes {10^{ - 6}} 	imes {100^2} = 4.5$$ J
As the other capacitor is identical therefore charge is equally divided and potential difference across the capacitors becomes half. So
$${U_f} = {1 \over 2}2C{\left( {{V \over 2}} ight)^2} = {1 \over 2} 	imes 2 	imes 900 	imes {10^{ - 6}}{\left( {{{100} \over 2}} ight)^2}$$
$$ = {9 \over 4}$$ J = 2.25 J
So, loss in energy $$\Delta {U_{loss}} = {U_i} - {U_f}$$
= 2.25 J
= 225 $$	imes$$ 10$$^{-2}$$ J

Electrostats & Capacitance

Choose the correct answer: