Verified Solution • 2025–26 Session moderate

QPages 165-170: Q49: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q49 Solution: The two square faces of a rectangular dielectric slab…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the dimensions in centimeters or millimeters directly into $C=\kappa\epsilon_0 A/d$, forgetting to convert to meters; this makes $A$ too large and $d$ too small, inflating the capacitance by orders of magnitude. Always convert $20\,\text{cm}=0.20\,\text{m}$ and $1.0\,\text{mm}=1\times10^{-3}\,\text{m}$ before using the formula.

The two square faces of a rectangular dielectric slab dielectric constant 4.0) of dimensions 20 cm×20 cm×1.0 mm20 \text{ cm} \times 20 \text{ cm} \times 1.0 \text{ mm} are metal-coated. Find the capacitance between the coated surfaces.

Step-by-Step Verified Solution

**Answer:** C=1.42×10−9C = 1.42 \times 10^{-9} F. **Step-by-Step Explanation:** A=0.2×0.2=0.04A = 0.2 \times 0.2 = 0.04 m_, d=1.0d = 1.0 mm =10−3= 10^{-3} m, K=4.0K = 4.0.
C=Kϵ0Ad=4×8.854×10−12×0.0410−3=1.42×10−9 FC = \frac{K\epsilon_0 A}{d} = \frac{4 \times 8.854 \times 10^{-12} \times 0.04}{10^{-3}} = 1.42 \times 10^{-9} \text{ F}

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