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QPages 165-170: Q60: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q60 Solution: A capacitor having a capacitance of 100 �F is charged to a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mistakenly equate the induced surface charge on the dielectric with the total charge on the capacitor plates. You must calculate the induced charge as the difference between the original free charge ($Q$) and the charge required to produce the new potential without the dielectric ($Q' = CV/K$), i.e., $Q_{induced} = Q(1 - 1/K)$.

A capacitor having a capacitance of 100 �F is charged to a potential difference of 50 V. a) What is the magnitude of the charge on each plate? b) The charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted. Calculate the new potential difference between the plates. c) What charge would have produced this potential difference in absence of the dielectric slab. d) Find the charge induced at a surface of the dielectric slab.

Step-by-Step Verified Solution

**Answer:** (a) Q=5×10−3Q = 5 \times 10^{-3} C (b) V=20V = 20 V (c) Q=2×10−3Q = 2 \times 10^{-3} C (d) Q′=3×10−3Q' = 3 \times 10^{-3} C. **Step-by-Step Explanation:** (a) Q=CV=100×10−6×50=5×10−3Q = CV = 100 \times 10^{-6} \times 50 = 5 \times 10^{-3} C. (b) V′=V/K=50/2.5=20V' = V/K = 50/2.5 = 20 V. (c) Without dielectric, charge for same potential = CV′=100×10−6×20=2×10−3CV' = 100 \times 10^{-6} \times 20 = 2 \times 10^{-3} C. (d) Induced charge = Q−Q′=3×10−3Q - Q' = 3 \times 10^{-3} C .

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