Question 344: Electrostatics & Capacitance

Question

In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is :
![](https://imagex.cdn.examgoal.net/vnww7VDht40I7wh5x/fkAjNc8NXJwAcIqLgoufLEhOdM1SQ/huB2wqxLgSNbpJbzhtGrsc/image.png)

Accepted Answer

V
i
= ${1 \over 2}$CE
2
V
f
= ${{{{\left( {CE} ight)}^2}} \over {2 	imes 4c}}$ = ${1 \over 2}{{C{E^2}} \over 4}$
$\Delta $E = ${1 \over 2}$CE
2
$ 	imes $ ${3 \over 4}$ = ${3 \over 8}$ CE
2

Answer

${1 \over 8}{{{Q^2}} \over C}$

Answer

${5 \over 8}{{{Q^2}} \over C}$

Answer

${3 \over 4}{{{Q^2}} \over C}$

Answer

${3 \over 8}{{{Q^2}} \over C}$

Electrostatics & Capacitance

In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is :

Choose the correct answer: