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Q.
Three capacitors of capacitance $1.0, 2.0$ and $5.0 \,\mu$F are connected in series to a $10\,V$ source. The potential difference across the $2.0 \,\mu F$ capacitor is
WBJEEWBJEE 2017Electrostatic Potential and Capacitance
Solution:
When the capacitors are connected in series, die resultant capacitance of combination
$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$
$=\frac{1}{1}+\frac{1}{2}+\frac{1}{5}=\frac{17}{10} \mu F $
$C=\frac{10}{17} \mu F$
The charge will be same on all the capacitors in series
$Q=C V=\frac{10}{17} \times 10=\frac{100}{17}$
The potential difference across $2.0\, \mu F$ capacitor
$V^{\prime}=\frac{Q}{C}$
$=\frac{100 / 17}{2}=\frac{50}{17} $ volt
