Question

A potential difference of 300 V is applied to a combination of 2.0$\mu$F and 8.0$\mu$F capacitors connected in series. The charge on the 2.0$\mu$F capacitor is

• A. 2.4 x 10^-4 C
• B. 4.8 x 10^-4 C
• C. $7.2\times 10^{-4}C$
• D. $9.6\times 10^{-4}C$

Answer 1

Answer: (B)

V = 300V, $C_1=2.0\mu F,C_2=8.0\mu F$
Not Capacitance, ${\frac{1}{C}}_s={\frac{1}{C}}_1+{\frac{1}{C}}_2$
$\Rightarrow C_8=\frac{C_1{C}_2}{C_1+C_2}$
$\Rightarrow C_s=\frac{2\times 8}{2+8}=\frac{16}{10}=1.6\mu F.$
Now total charge,
$Q=V_s\times C_s=300\times 1.6\times 10^{-6}=4.8\times 10^{-4}C.$
$\because$ In series charge is same on capacitors
$\Rightarrow$ Charge on $2\mu F$ capacitor is $4.8X10^{-4}C$