(A) Initial charges on the capacitors are $Q_1 = CV$ and $Q_2 = (2C)(2V) = 4CV$.Since they are connected with opposite polarities,the net charge on th

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) Initial charges on the capacitors are $Q_1 = CV$ and $Q_2 = (2C)(2V) = 4CV$.Since they are connected with opposite polarities,the net charge on the system is $Q_{net} = |Q_2 - Q_1| = |4CV - CV| = 3CV$.The equivalent capacitance of the system in parallel is $C_{eq} = C + 2C = 3C$.The common potential is $V_{com} = \frac{Q_{net}}{C_{eq}} = \frac{3CV}{3C} = V$.The final energy stored in the system is $U = \frac{1}{2} C_{eq} V_{com}^2 = \frac{1}{2} (3C) (V)^2 = \frac{3}{2} CV^2$.

Class 12 Physics Electric Potential and Capacitance Sharing of Charge in Capacitor Circuit

Difficult

$A$ parallel plate capacitor of capacitance $C$ is charged to a potential $V$ by a battery. Another capacitor of capacitance $2C$ is charged to a potential $2V$ by another battery. After removing the batteries,they are connected in parallel such that the positive plate of one is connected to the negative plate of the other. Calculate the final energy of the system.

A
$3/2 \ CV^2$
B
$5/2 \ CV^2$
C
$7/3 \ CV^2$
D
$4/5 \ CV^2$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Electric Potential and Capacitance

Subtopic

Sharing of Charge in Capacitor Circuit

Two metal spheres of capacitance $C_1$ and $C_2$ carry some charges. They are put in contact and then separated. The final charges $Q_1$ and $Q_2$ on them will satisfy

Easy

View Solution

Two spheres of radii $1 \, cm$ and $2 \, cm$ are charged with $1.5 \times 10^{-8} \, C$ and $0.3 \times 10^{-7} \, C$ of positive charge,respectively. When they are connected by a wire,the charge will flow:

Easy

View Solution

In the circuit shown here $C_1 = 6\,\mu F$,$C_2 = 3\,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge finally on $C_2$ in $\mu C$?

Difficult

View Solution

Two capacitors of capacitances $C$ and $2C$ are charged to potential differences $V$ and $2V$,respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is $.....CV^2$.

DifficultJEE MAIN 2020

View Solution

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

MediumNEET 2017

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

Two metal spheres of capacitance $C_1$ and $C_2$ carry some charges. They are put in contact and then separated. The final charges $Q_1$ and $Q_2$ on them will satisfy

Two spheres of radii $1 \, cm$ and $2 \, cm$ are charged with $1.5 \times 10^{-8} \, C$ and $0.3 \times 10^{-7} \, C$ of positive charge,respectively. When they are connected by a wire,the charge will flow:

In the circuit shown here $C_1 = 6\,\mu F$,$C_2 = 3\,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge finally on $C_2$ in $\mu C$?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module