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Q.
Two capacitors of $10\, PF$ and $20\, PF$ are connected to $200 \,V$ and $100\, V$ sources respectively. If they are connected by the wire, what is the common potential of the capacitors ?
KCETKCET 2014Electrostatic Potential and Capacitance
Solution:
Given, $C_{1}=10 pF =10 \times 10^{-12} F$
$C_{2}=20 pF =20 \times 10^{-12} F$
$V_{1}=200\, V ,\, V_{2}=100\, V$
$C_{1}=$ Capacitance of Ist capacitor
$C _{2}=$ Capacitance of IInd capacitor
$V_{1}=$ Voltage across Ist capacitor
$V_{2}=$ Voltage across IInd capacitor
We know that $V_{1}=\frac{q_{1}}{C_{1}}$
and $V_{2}=\frac{q_{2}}{C_{2}}$
$\Rightarrow q_{1} =V_{1} C_{1}$...(i)
$q_{2} = V_{2} C_{2}$...(ii)
So, common potential of capacitors
$V=\frac{q_{1}+q_{2}}{C_{1}+C_{2}}=\frac{V_{1} C_{1}+V_{2} C_{2}}{C_{1}+C_{2}}$
$=\frac{200 \times 10 \times 10^{-12}+100 \times 20 \times 10^{-12}}{10 \times 10^{-12}+20 \times 10^{-12}}$
$=\frac{200 \times 10+100 \times 20}{10+20}$
$=\frac{2000+2000}{30}=\frac{4000}{30}=133.3\, V$