Question

Consider the figure.
The charge appearing on $C_2$ is

• A. $E\left(\frac{C_3{C}_4}{C_1+C_2}\right)$
• B. $E\left(\frac{C_1{C}_2}{C_1+C_2}\right)$
• C. $E\left(\frac{C_1{C}_2}{C_3+C_4}\right)$
• D. $E\left(\frac{C_3{C}_4}{C_3+C_4}\right)$

Answer 1

The correct option is B $E\left(\frac{C_1{C}_2}{C_1+C_2}\right)$

Let us mark the points across the whole combination of capacitors as C and D.From the above fiqure we can see that the voltage across C and D is to E.
$V_{C_1}=\frac{q}{C_1}$ (voltage across $C_1$)
IIy $V_{C_2}=\frac{q}{C_2}$
As $C_1$ and $C_2$ are in series, thus
$V_{C_1}+V_{C_2}=E$
$⟹\frac{q}{C_1}+\frac{q}{C_2}=E........\left(i\right)$
$⟹E=\frac{C_1+C_2}{C_1{C}_2}q$
$\therefore q=\frac{C_1{C}_2}{C_1+C_2}E$