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 $100\, V$. 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
$V _{ c }=\frac{ Q _{1}+ Q _{2}}{ C _{1}+ C _{2}}=\frac{1500+100}{1+1}=800 \,V$