Calculate the charge on the second capacitor | vedclass

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

Question

(A) Before the switch is closed,the two capacitors (labeled $1$ and $2$) are in series with the battery $E$. Since both have capacitance $C$,the poten

Vedclass · Accepted Answer

$\frac{CE}{2}, CE$

Vedclass · Answer

(A) Before the switch is closed,the two capacitors (labeled $1$ and $2$) are in series with the battery $E$. Since both have capacitance $C$,the potential difference across each is $\frac{E}{2}$.Thus,the charge on the second capacitor is $Q = C 	imes \frac{E}{2} = \frac{CE}{2}$.After the switch is closed,the first capacitor is short-circuited. This means the potential difference across the first capacitor becomes $0$,and all the potential $E$ appears across the second capacitor.Therefore,the charge on the second capacitor becomes $Q' = C 	imes E = CE$.

Calculate the charge on the second capacitor before and after the switch in the circuit is closed.

Similar Questions

See the diagram. The area of each plate is $2.0 \,m^{2}$ and $d=2 \times 10^{-3} \,m$. $A$ charge of $8.85 \times 10^{-8} \,C$ is given to plate $Q$. Then the potential of $Q$ becomes (in $\,V$)

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

Initially $n$ identical capacitors are joined in parallel and are charged to potential $V$. Now they are separated and joined in series. Then

Twenty-seven drops of the same size are charged at $220 \, V$ each. They combine to form a bigger drop. Calculate the potential of the bigger drop (in $V$).

Find the potentials at points $A$ and $B$ in the circuit shown in the figure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module