An air capacitor of capacity \(\mathrm{C}=10 \mu \mathrm{F}\) is connected to a constant voltage battery of…
An air capacitor of capacity \(\mathrm{C}=10 \mu \mathrm{F}\) is connected to a constant voltage battery of \(12 \mathrm{~V}\). Now, the space between the plates is filled with a liquid of dielectric constant \((n)=5\). The charge that now flows from battery to the capacitor in micro coulomb is
Solution
Initial charge \(=10 \mu \mathrm{F} \times 12 \mathrm{~V}=120 \mu \mathrm{C}\) New capacitance \(\mathrm{C}^{\prime}=\mathrm{nC}\)
\(\&=5 \times 10 \mu \mathrm{F}=50 \mu \mathrm{F}\)
New charge \(=\mathrm{VC}^{\prime}\)
\(\&=50 \mu \mathrm{F} \times 12 \mathrm{~V}=600 \mu \mathrm{C}\)
So, charge flowing now from battery to the capacitor \(\mathrm{Q}^{\prime}=\) Final charge - initial charge
\(=(600-120) \mu \mathrm{C}=480 \mu \mathrm{C}\)