Verified Solution • 2025–26 Session moderate

QPages 141-142: Q22: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q22 Solution: Two large conducting plates are placed parallel to each other…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often add the magnitudes $\sigma/2\epsilon_0$ from each plate without considering direction, concluding a non‑zero field outside. Remember that the field of a uniformly charged infinite plate points away from positive and toward negative charge; for opposite plates the vectors are opposite on the outer sides, so they cancel, leaving $E=0$ there.

Two large conducting plates are placed parallel to each other and they carry equal and opposite charges with surface density σ\sigma as shown in the figure. Find the electric field a) at the left of the plates b) in between the plates and c) at the right of the plates.

Step-by-Step Verified Solution

**Answer:** (a) E=0E = 0 (left) (b) E=σ/ϵ0E = \sigma/\epsilon_0 (between) (c) E=0E = 0 (right). **Step-by-Step Explanation:** Two large conducting plates with equal and opposite charges ±σ\pm\sigma. (a) To the left: fields from both plates cancel. E = 0. (b) Between: fields add up. E=σ2ϵ0+σ2ϵ0=σϵ0E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}. (c) To the right: fields cancel again. E = 0 .

All Questions in this Chapter

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