2. Electric Potential and Capacitance
hard

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

A

$6$

B

$5$

C

$4$

D

$3$

(JEE MAIN-2023)

Solution

$Q = CV =600 \times 10^{-12} \times 200=12 \times 10^{-8} C$

$\text { Initial energy }=\frac{1}{2} CV ^2$

$=\frac{1}{2} \times 600 \times 10^{-12} \times(200)^2=12\,\mu J$

When connected to another uncharged capacitor

Charge will be equally distributed on identical capacitor

$Q ^{\prime}=\frac{ Q }{2}=6 \times 10^{-8}$

Final energy $=2 \times \frac{ Q ^{\prime 2}}{2 C }=\frac{ Q ^{\prime 2}}{ C }$

$\frac{\left(6 \times 10^{-8}\right)^2}{600 \times 10^{-12}}=6\,\mu J$

Energy lost = Initial energy – Final energy

$=(12-6)\,\mu\, J =6\, \mu\, J$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.