Three capacitors of \(2 \mu \mathrm{F}, 3 \mu\) F and \(6 \mu\) F are connected in series to a \(10…
Three capacitors of \(2 \mu \mathrm{F}, 3 \mu\) F and \(6 \mu\) F are connected in series to a \(10 \mathrm{~V}\) source. The charge on the middle capacitor is
\(10 \mu \mathrm{C}\)
\(5 \mu \mathrm{C}\)
\(1 \mu \mathrm{C}\)
Zero
Solution
In a series connection of capacitors, the charge on each capacitor is the same. This is because the charge that leaves one plate of a capacitor must go to the other plate of the next capacitor.
The total charge $Q$ supplied by the source is given by the formula $Q=CV$, where $C$ is the total capacitance and $V$ is the voltage of the source.
In a series connection of capacitors, the total capacitance $C$ is given by the formula $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$, where $C_1$, $C_2$, $C_3$ are the capacitances of the individual capacitors.
Substituting the given values, we get $\frac{1}{C}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{6}{6}=1$. Hence, $C=1 \mu \mathrm{F}$.
Substituting $C=1 \mu \mathrm{F}$ and $V=10 \mathrm{V}$ in the formula $Q=CV$, we get $Q=1 \mu \mathrm{F} \times 10 \mathrm{V} = 10 \mu \mathrm{C}$.
Therefore, the charge on the middle capacitor (or any capacitor in the series) is $10 \mu \mathrm{C}$, which is option A.