Answer

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**Hint:**To solve this question, we need to use the formula for the equivalent capacitance in a parallel combination. Then we have to use the basic formula of the capacitance to calculate the total charge supplied by the battery. Using the same formula for each of the individual capacitances, we can calculate the charge on each capacitor.

**Formula used:**The formulae used for solving this question are given by

$C = {C_1} + {C_2} + {C_3} + ........$, here $C$ is the equivalent capacitance for the parallel combination of the capacitances ${C_1}$, ${C_2}$, ${C_3}$,….

$Q = CV$, here $Q$ is the charge stored by a capacitor of capacitance $C$ at the voltage of $V$.

**Complete step-by-step solution:**

We know that the equivalent capacitance of a combination of capacitors connected in parallel is given by

$C = {C_1} + {C_2}$

According to the question, the capacitors of capacitances \[2\mu F\] and \[4\mu F\] are connected in parallel. Therefore we substitute ${C_1} = 2\mu F$ and ${C_2} = 4\mu F$ in ….(1) to get the equivalent capacitance of the given combination as

$C = 2\mu F + 4\mu F$

$ \Rightarrow C = 6\mu F$.............(2)

So the equivalent capacity of the combination is equal to $6\mu F$.

Now, this combination is connected across a $9{\text{V}}$ battery. We know that the potential difference across each element connected in parallel is the same and is equal to the emf of the battery, so the potential difference across each of the capacitors is equal to $9{\text{V}}$.

Now, we know that the charge stored by a capacitor is given by

$Q = CV$..............(3)

Since the emf of the battery is given as $9{\text{V}}$, so we substitute $V = 9{\text{V}}$ above to get

$Q = 9C$

Substituting (2) above, we get

$Q = 9 \times 6\mu C$

$ \Rightarrow Q = 54\mu C$..............(4)

So the total charge supplied by the battery is equal to $54\mu C$.

Now, let ${Q_1}$ be the charge on the \[2\mu F\] capacitor, and ${Q_2}$ be the charge on the \[4\mu F\] capacitor.

From (3) we have

$Q = CV$

$ \Rightarrow V = \dfrac{Q}{C}$

Since the voltage across each capacitor is the same, so we have

$\dfrac{{{Q_1}}}{{{C_1}}} = \dfrac{{{Q_2}}}{{{C_2}}}$

Substituting ${C_1} = 2\mu F$ and ${C_2} = 4\mu F$, we have

$\dfrac{{{Q_1}}}{2} = \dfrac{{{Q_2}}}{4}$

$ \Rightarrow {Q_2} = 2{Q_1}$ ………………………….(5)

Now, from the conservation of charge, the total charge on both the capacitors is equal to the charge supplied by the battery. So we have

${Q_1} + {Q_2} = Q$

Substituting (4) and (5) in the above equation, we have

${Q_1} + 2{Q_1} = 54\mu C$

$ \Rightarrow 3{Q_1} = 54\mu C$

Dividing by $3$ both sides, we get

\[{Q_1} = 18\mu C\]..............................(6)

Putting (6) in (5) we get

${Q_2} = 2 \times 18\mu C$

$ \Rightarrow {Q_2} = 36\mu C$

**Thus, the charge on the \[2\mu F\] capacitor is equal to $18\mu C$, and the charge on the \[4\mu F\] capacitor is equal to $36\mu C$.**

**Note:**We should not use the inverse relation which is used for calculating the equivalent resistance of a parallel combination. The capacitances are added in a parallel combination.

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