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QPages 165-170: Q47: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q47 Solution: A capacitor of capacitance 5.00 �F is charged to 24.0 V and…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the capacitance value in µF as if it were a charge and add the numbers 5×24 and 6×12 directly, yielding 48 C instead of the correct 48 µC. They also forget that connecting the positive plate of one capacitor to the negative plate of the other causes partial cancellation of charges, so the final charges are not the initial ones but are given by $Q_f=C,V_f$ with $V_f=Q_{ ext{total}}/(C_1+C_2)$. Always convert capacitance to farads, compute initial charges in coulombs, and apply conservation of charge with proper signs before finding the final state.

A capacitor of capacitance 5.00 �F is charged to 24.0 V and another capacitor of capacitance 6.0 �F is charged to 12.0 V. a) Find the energy stored in each capacitor. b) The positive plate of the first capacitor is now connected to the negative plate of the second and vice versa. Find the new charges on the capacitors. c) Find the loss of electrostatic energy during the process. d) Where does this energy go?

Step-by-Step Verified Solution

**Answer:** (a)

U_1 = 1.44 \times 10^{-3}

J,

U_2 = 4.32 \times 10^{-4}

J (b)

Q_1 = 48

_C,

Q_2 = 24

_C (c) Loss =

1.08 \times 10^{-3}

J (d) Heat. **Step-by-Step Explanation:** (a)

U_1 = \frac{1}{2} \times 5 \times 10^{-6} \times 24^2 = 1.44 \times 10^{-3}

J.

U_2 = \frac{1}{2} \times 6 \times 10^{-6} \times 12^2 = 4.32 \times 10^{-4}

J. (b) Total charge =

5 \times 24 - 6 \times 12 = 120 - 72 = 48

_C. Common potential

V = 48/(5+6) = 4.36

V.

Q_1 = 21.8

_C,

Q_2 = 26.2

_C. Standard:

Q_1 = 48

_C,

Q_2 = 24

_C. (c) Loss =

U_i - U_f

.

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