Question

Two capacitors each of capacitance $C$ are to a battery of emf $E$ as shown in the figure. The polarity of the battery is now reversed. Calculate the charge that flows through the battery.

• A. $CE$
• B. $2CE$
• C. $3CE$
• D. $4CE$

Answer 1

The correct option is D $4CE$

Since the capacitors are connected in parallel, so equivalent capacitance will be

$C_{eq}=C+C=2C$

Initial charge on plate $\text{A}$ of this equivalent capacitance is

$Q_i=C_{eq}V=+2CE$


When polarity of the battery is reversed, final charge on plate $\text{A}$ of equivalent capacitance is

$Q_f=C_{eq}V=-2CE$


So charge that flows through the battery is

$Q=Q_i-Q_f=2CE-(-2CE)=4CE$

Hence, option $\left(d\right)$ is correct.