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Condenser Ahas 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.......$V$
$400$
$800$
$1200$
$1600$
Solution
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 $V_{c}=\frac{Q_{1}+Q_{2}}{C_{1}+C_{2}}=\frac{1500+100}{1+1}=800 \mathrm{V}$
