Question

Condenser A has a capacity of $15\mu F$ when it is filled with a medium of dielectric constant $15$. Another condenser $B$ has a capacity $1\mu F$ with air between the plates. Both are charged separately by a battery of $100V$. After charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is

• A. 400 V
• B. 800 V
• C. 1200 V
• D. 1600 V

Answer 1

Answer: (B)

As battery is disconnected so the charges on two capacitor remain constant .
$Q_1=15\times 10^{-6}\times 100=1500\mu C$
and $Q_2=10^{-6}\times 100=100\mu C$
As dielectric is removed so the capacitance
$15\mu F$ becomes $C_1=\frac{15}{15}=1\mu F$
and capacitance $1\mu F$ will be same $C_2=1\mu F$
Common potential
$V_c=\frac{Q_1+Q_2}{C_1+C_2}=\frac{1500+100}{1+1}=800V$