A capacitor having a capacitance of 2.0 is charged to a potential difference of 12 V . Subsequently, it is connected to an uncharged capacitor of capacitance 4.0 as shown in the figure. Determine the charge on each of

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Quantrex Academy · Accepted Answer

Correct answer: B. 8 on the 2.0 capacitor and 16 on the 4.0 capacitor

The initial charge on the first capacitor is:

Q_1 = C_1 V_1 = 2.0 \times 12 = 24 \mu\text{C}

The second capacitor is initially uncharged, so

Q_2 = 0

. When the capacitors are connected, they are in parallel. The common potential

V

across the combination is:

V = \dfrac{\text{Total charge}}{\text{Total capacitance}} = \dfrac{Q_1 + Q_2}{C_1 + C_2}

V = \dfrac{24 + 0}{2.0 + 4.0} = \dfrac{24}{6.0} = 4 \text{ V}

The final charge on the

2.0 \mu\text{F}

capacitor is:

Q_1' = C_1 V = 2.0 \times 4 = 8 \mu\text{C}

The final charge on the

4.0 \mu\text{F}

capacitor is:

Q_2' = C_2 V = 4.0 \times 4 = 16 \mu\text{C}

A capacitor having a capacitance of 2.0 is charged to a potential...

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