
In the given figure, charge stored in the capacitor of capacitance 5\[\mu \]F is –

Answer
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Hint: We need to understand the type of combination of capacitors used in the given network to work out the possible relations between the voltages across each of the capacitors. We can use the relation between capacitance and voltage to find the charge.
Complete step-by-step answer:
We now know that the charge stored in a capacitor and the potential difference across the plates of the capacitor are directly proportional to each other. It is given as –
\[Q=CV\]
Where, Q is the charge stored in the capacitor,
C is the capacitance of the capacitor,
V is the potential difference across the plates.
To find the charge on the capacitors, we need to find the potential across the capacitor initially. For this, we need to find the arrangement of capacitors used in the circuit. We can redraw the given circuit by identifying the nodes joining the capacitors at same points. We will get the equivalent circuit as shown below.
From the circuit, we can see that the potential at points C and D are the same.
If the potentials across C and D are the same, then there will be no flow of charges as the potential differences across CD will be zero.
i.e.,
\[{{V}_{CD}}=0\]
From the relation between the charge and the potential difference, we can find that there will be no charge on the capacitors of capacitance \[5\mu F\text{ and }3\mu F\]will be zero.
The charge in the capacitor of capacitance of \[5\mu F\text{ }\]will be given as –
\[\begin{align}
& Q=C{{V}_{CD}} \\
& \text{but, } \\
& {{V}_{CD}}=0 \\
& \therefore Q=0 \\
\end{align}\]
This is the required solution.
Note: We should always keep in mind while solving the networks involving the combination of capacitors that the charge stored in the capacitors are dependent on the potential drop across its plates. No charges will be present for a zero potential difference.
Complete step-by-step answer:
We now know that the charge stored in a capacitor and the potential difference across the plates of the capacitor are directly proportional to each other. It is given as –
\[Q=CV\]
Where, Q is the charge stored in the capacitor,
C is the capacitance of the capacitor,
V is the potential difference across the plates.

To find the charge on the capacitors, we need to find the potential across the capacitor initially. For this, we need to find the arrangement of capacitors used in the circuit. We can redraw the given circuit by identifying the nodes joining the capacitors at same points. We will get the equivalent circuit as shown below.

From the circuit, we can see that the potential at points C and D are the same.
If the potentials across C and D are the same, then there will be no flow of charges as the potential differences across CD will be zero.
i.e.,
\[{{V}_{CD}}=0\]
From the relation between the charge and the potential difference, we can find that there will be no charge on the capacitors of capacitance \[5\mu F\text{ and }3\mu F\]will be zero.
The charge in the capacitor of capacitance of \[5\mu F\text{ }\]will be given as –
\[\begin{align}
& Q=C{{V}_{CD}} \\
& \text{but, } \\
& {{V}_{CD}}=0 \\
& \therefore Q=0 \\
\end{align}\]
This is the required solution.
Note: We should always keep in mind while solving the networks involving the combination of capacitors that the charge stored in the capacitors are dependent on the potential drop across its plates. No charges will be present for a zero potential difference.
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