Two capacitors of capacitances 20.0 pF and 50.0 pF are connected in series with a 6.00 V battery. Calculate the energy stored in the 20.0 pF and 50.0 pF capacitors respectively.

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Quantrex Academy · Accepted Answer

Correct answer: D. 184 pJ and 73.5 pJ

Equivalent capacitance of the series combination is given by:

C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{20 \times 50}{20 + 50} = \dfrac{100}{7} \text{ pF}

The charge on each capacitor in series is the same and is equal to:

Q = C_{eq} V = \dfrac{100}{7} \times 6 = \dfrac{600}{7} \text{ pC}

Energy stored in the

20.0 \text{ pF}

capacitor is:

U_1 = \dfrac{Q^2}{2C_1} = \dfrac{\left(\dfrac{600}{7}\right)^2}{2 \times 20} = \dfrac{360000}{49 \times 40} = \dfrac{9000}{49} \approx 184 \text{ pJ}

Energy stored in the

50.0 \text{ pF}

capacitor is:

U_2 = \dfrac{Q^2}{2C_2} = \dfrac{\left(\dfrac{600}{7}\right)^2}{2 \times 50} = \dfrac{360000}{49 \times 100} = \dfrac{3600}{49} \approx 73.5 \text{ pJ}

Two capacitors of capacitances 20.0 pF and 50.0 pF are connected in...

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