A capacitor with a capacitance of 100 is charged to a potential difference of 24 V . After disconnecting the charging battery, the capacitor is connected to a different battery of emf 12 V such that the positive plate

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 21.6 mJ

Initial energy of the capacitor is given by:

U_i = \dfrac{1}{2}CV_1^2

Substituting the given values,

C = 100 \mu\text{F}

and

V_1 = 24 \text{ V}

:

U_i = \dfrac{1}{2} \times 100 \times 10^{-6} \times (24)^2 = 28.8 \times 10^{-3} \text{ J} = 28.8 \text{ mJ}

When the capacitor is connected to a

12 \text{ V}

battery, the final potential difference across it becomes

V_2 = 12 \text{ V}

.

Final energy of the capacitor is:

U_f = \dfrac{1}{2}CV_2^2

U_f = \dfrac{1}{2} \times 100 \times 10^{-6} \times (12)^2 = 7.2 \times 10^{-3} \text{ J} = 7.2 \text{ mJ}

The decrease in the electrostatic field energy of the capacitor is:

\Delta U = U_i - U_f

\Delta U = 28.8 \text{ mJ} - 7.2 \text{ mJ} = 21.6 \text{ mJ}

A capacitor with a capacitance of 100 is charged to a potential...

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