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QPages 165-170: Q44: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q44 Solution: A parallel-plate capacitor having plate area and separation…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the energy absorbed by the battery with the work done by the external agent; remember that when plates are pulled apart while connected to a battery, charge flows back into the battery, so the energy absorbed is $W_{bat} = -\Delta Q \cdot V$, which is distinct from the mechanical work done by the person.

A parallel-plate capacitor having plate area 20 cm220 \text{ cm}^2 and separation between the plates 1.00 mm is connected to a battery of 12.0 V. The plates are pulled apart to increase the separation to 2.0 mm. a) Calculate the charge flown through the circuit during the process. b) How much energy is absorbed by the battery during the process? c) Calculate the stored energy in the electric field before and after the process. d) Using the expression for the force between the plates, find the work done by the person pulling the plates apart. e) Show and justify that no heat is produced during this transfer of charge as the separation is increased.

Step-by-Step Verified Solution

**Answer:** (a) Q=1.06×10−10Q = 1.06 \times 10^{-10} C (b) W=1.27×10−9W = 1.27 \times 10^{-9} J (c) Ui=6.37×10−10U_i = 6.37 \times 10^{-10} J, Uf=3.18×10−10U_f = 3.18 \times 10^{-10} J (d) W=3.18×10−10W = 3.18 \times 10^{-10} J (e) No heat. **Step-by-Step Explanation:** (a) Ci=ϵ0A/d=1.77×10−11C_i = \epsilon_0 A/d = 1.77 \times 10^{-11} F, Qi=CiV=2.13×10−10Q_i = C_i V = 2.13 \times 10^{-10} C. Cf=Ci/2C_f = C_i/2, Qf=CfV=1.06×10−10Q_f = C_f V = 1.06 \times 10^{-10} C. Charge flown = Qi−Qf=1.06×10−10Q_i - Q_f = 1.06 \times 10^{-10} C. (b) Energy absorbed by battery = Qflown×V=1.27×10−9Q_{flown} \times V = 1.27 \times 10^{-9} J. (c) Ui=12CiV2=1.27×10−9U_i = \frac{1}{2}C_iV^2 = 1.27 \times 10^{-9} J. Uf=12CfV2=6.37×10−10U_f = \frac{1}{2}C_fV^2 = 6.37 \times 10^{-10} J. (d) Work done = Ui−Uf=6.37×10−10U_i - U_f = 6.37 \times 10^{-10} J. (e) The process is reversible; no heat is produced .

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