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QPages 165-170: Q53: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q53 Solution: A parallel-plate capacitor of capacitance 5 �F is connected to…

⚠️ Common Exam Traps & Conceptual Pitfalls

When a dielectric slab of thickness $d_2$ and constant $K$ fills part of the gap, the effective separation in the series formula $C = \frac{\epsilon_0 A}{d_1 + d_2/K}$ is not simply the physical sum of distances; students often forget to divide the dielectric thickness by $K$ or incorrectly treat the layers as parallel.

A parallel-plate capacitor of capacitance 5 �F is connected to a battery of emf 6 V. The separation between the plates is 2 mm. a) Find the charge on the positive plate. b) Find the electric field between the plates. c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination. d) How much charge has flown through the battery after the slab is inserted?

Step-by-Step Verified Solution

**Answer:** (a) Q=30Q = 30 _C (b) E=3000E = 3000 V/m (c) C=8.33C = 8.33 _F (d) ΔQ=20\Delta Q = 20 _C. **Step-by-Step Explanation:** (a) Q=CV=5×6=30Q = CV = 5 \times 6 = 30 _C. (b) E=V/d=6/0.002=3000E = V/d = 6/0.002 = 3000 V/m. (c) With dielectric of thickness 1 mm and K=5 in half the gap, the capacitor is equivalent to two capacitors in series: one with air (1 mm) and one with dielectric (1 mm, K=5).
C=ϵ0Ad1+d2/K=ϵ0A0.001+0.001/5=ϵ0A0.0012C = \frac{\epsilon_0 A}{d_1 + d_2/K} = \frac{\epsilon_0 A}{0.001 + 0.001/5} = \frac{\epsilon_0 A}{0.0012}
With ϵ0A=C0d=5×10−6×0.002=10−8\epsilon_0 A = C_0 d = 5 \times 10^{-6} \times 0.002 = 10^{-8}.
C=10−80.0012=8.33×10−6 FC = \frac{10^{-8}}{0.0012} = 8.33 \times 10^{-6} \text{ F}
(d) New charge = 8.33×6=508.33 \times 6 = 50 _C. Extra charge = 20 _C .

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