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QPages 141-142: Q21: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q21 Solution: Two large conducting plates are placed parallel to each other…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often insert the plate separation as 2.00 (cm) and the travel time as 2.00 (µs) directly into the SI equation $d=\frac12 a t^2$, forgetting to convert to meters (0.02 m) and seconds (2×10⁻⁶ s). Always express all quantities in base SI units before applying $a=eE/m=e\sigma/(\epsilon_0 m)$ to avoid orders‑of‑magnitude errors.

Two large conducting plates are placed parallel to each other with a separation of 2.00 cm between them. An electron starting from rest near one of the plates reaches the other plate in 2.00 microseconds. Find the surface charge density on the inner surfaces.

Step-by-Step Verified Solution

**Answer:** σ=2.02×10−12\sigma = 2.02 \times 10^{-12} C/m_. **Step-BY-STEP Explanation:** Given: d=2.00d = 2.00 cm, t=2.00t = 2.00 _s, electron starting from rest. a=eEm=eσmϵ0a = \frac{eE}{m} = \frac{e\sigma}{m\epsilon_0}. d=12at2⇒σ=2dmϵ0et2=2×0.02×9.1×10−31×8.85×10−121.6×10−19×(2×10−6)2=2.02×10−12d = \frac{1}{2}at^2 \Rightarrow \sigma = \frac{2dm\epsilon_0}{et^2} = \frac{2 \times 0.02 \times 9.1 \times 10^{-31} \times 8.85 \times 10^{-12}}{1.6 \times 10^{-19} \times (2 \times 10^{-6})^2} = 2.02 \times 10^{-12} C/m_ .

All Questions in this Chapter

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