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

A capacitor of capacitance

2.0 \mu\text{F}

is charged to a potential difference of

12 \text{ V}

. It is then connected to an uncharged capacitor with a capacitance of

4.0 \mu\text{F}

. Calculate the electrostatic energy stored in each of the two capacitors after the connection is established.
A capacitor of capacitance 2.0 is charged to a potential difference of 12 V . It is then connected to an uncharged

A capacitor of capacitance 2.0 is charged to a potential difference of 12 V . It is then connected to an uncharged

Options

A $24 \mu\text{J}$ in both capacitors
B $16 \mu\text{J}$ in the $2.0 \mu\text{F}$ capacitor and $32 \mu\text{J}$ in the $4.0 \mu\text{F}$ capacitor
C $48 \mu\text{J}$ in the $2.0 \mu\text{F}$ capacitor and $96 \mu\text{J}$ in the $4.0 \mu\text{F}$ capacitor
D $32 \mu\text{J}$ in the $2.0 \mu\text{F}$ capacitor and $16 \mu\text{J}$ in the $4.0 \mu\text{F}$ capacitor

Correct answer

B. $16 \mu\text{J}$ in the $2.0 \mu\text{F}$ capacitor and $32 \mu\text{J}$ in the $4.0 \mu\text{F}$ capacitor

Step-by-step solution

Initial charge on the

2.0 \mu\text{F}

capacitor is:

Q = C_1 V_1 = 2.0 \mu\text{F} \times 12 \text{ V} = 24 \mu\text{C}

When it is connected to the uncharged

4.0 \mu\text{F}

capacitor, the total charge is conserved and they reach a common potential

V

V = \dfrac{\text{Total Charge}}{\text{Total Capacitance}} = \dfrac{Q}{C_1 + C_2}

V = \dfrac{24 \mu\text{C}}{2.0 \mu\text{F} + 4.0 \mu\text{F}} = \dfrac{24}{6.0} \text{ V} = 4 \text{ V}

The electrostatic energy stored in the

2.0 \mu\text{F}

capacitor is:

U_1 = \dfrac{1}{2} C_1 V^2 = \dfrac{1}{2} \times 2.0 \mu\text{F} \times (4 \text{ V})^2 = 16 \mu\text{J}

The electrostatic energy stored in the

4.0 \mu\text{F}

capacitor is:

U_2 = \dfrac{1}{2} C_2 V^2 = \dfrac{1}{2} \times 4.0 \mu\text{F} \times (4 \text{ V})^2 = 32 \mu\text{J}

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