A capacitor of capacitance 5.00 is charged to 24.0 V , while a second capacitor of capacitance 6.0 is charged to 12.0 V . The positive plate of the first capacitor is subsequently connected to the negative plate of the

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 1.77 mJ

When the positive plate of the first capacitor is connected to the negative plate of the second, the common potential

V

is given by:

V = \dfrac{C_1 V_1 - C_2 V_2}{C_1 + C_2}

The loss of electrostatic energy during this process is given by the formula:

\Delta U = \dfrac{1}{2} \dfrac{C_1 C_2}{C_1 + C_2} (V_1 + V_2)^2

Substituting the given values

C_1 = 5.00 \mu\text{F}

,

C_2 = 6.0 \mu\text{F}

,

V_1 = 24.0 \text{ V}

, and

V_2 = 12.0 \text{ V}

:

\Delta U = \dfrac{1}{2} \times \dfrac{5.00 \times 6.0}{5.00 + 6.0} \times (24.0 + 12.0)^2 \mu\text{J}

\Delta U = \dfrac{1}{2} \times \dfrac{30.0}{11.0} \times (36.0)^2 \mu\text{J}

\Delta U = \dfrac{15}{11} \times 1296 \mu\text{J}

\Delta U = \dfrac{19440}{11} \mu\text{J} \approx 1767.27 \mu\text{J}

\Delta U \approx 1.77 \text{ mJ}

A capacitor of capacitance 5.00 is charged to 24.0 V , while a...

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