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QPages 165-170: Q5: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q5 Solution: A parallel-plate capacitor has plate area and a separation of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume the charge $Q$ remains constant when the plate separation changes, leading them to calculate the new voltage instead of the new charge. Since the capacitor remains connected to the battery, the potential difference $V$ is fixed, so the extra charge must be calculated as $\Delta Q = \Delta C \cdot V$.

A parallel-plate capacitor has plate area 25.0 cm225.0 \text{ cm}^2 and a separation of 2.00 mm between the plates. The capacitor is connected to a battery of 12.0 V. a) Find the charge on the capacitor. b) The plate separation is decreased to 1.00 mm. Find the extra charge given by the battery to the positive plate.

Step-by-Step Verified Solution

**Answer:** (a) Q=1.33×10−10Q = 1.33 \times 10^{-10} C (b) Extra charge =1.33×10−10= 1.33 \times 10^{-10} C. **Step-by-Step Explanation:** (a) A=25.0A = 25.0 cm_ =2.5×10−3= 2.5 \times 10^{-3} m_, d=2.00d = 2.00 mm =2×10−3= 2 \times 10^{-3} m, V=12.0V = 12.0 V.
C=ϵ0Ad=8.854×10−12×2.5×10−32×10−3=1.107×10−11 FC = \frac{\epsilon_0 A}{d} = \frac{8.854 \times 10^{-12} \times 2.5 \times 10^{-3}}{2 \times 10^{-3}} = 1.107 \times 10^{-11} \text{ F}
Q=CV=1.107×10−11×12=1.33×10−10 CQ = CV = 1.107 \times 10^{-11} \times 12 = 1.33 \times 10^{-10} \text{ C}
(b) When separation is halved, CC doubles, so QQ doubles. Extra charge =1.33×10−10= 1.33 \times 10^{-10} C .

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