Verified Solution • 2025–26 Session moderate

QPages 139-140: Q1: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 139-140: Q1 Solution: A charge Q is uniformly distributed over a large plastic plate. …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume that a conductor will shield the external field or alter its magnitude. For a large flat plate, the field just outside depends only on the surface charge density $sigma= rac{Q}{A}$, so replacing a plastic plate with a copper plate of the same charge and area leaves $E$ unchanged at $10, ext{V,m}^{-1}$.

A charge Q is uniformly distributed over a large plastic plate. The electric field at a point P close to the centre of the plate is 10 V m−110 \text{ V m}^{-1}. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at the point P will become

Multiple Choice Options

  • A zero
  • B 5 V m−15 \text{ V m}^{-1}
  • C 10 V m−110 \text{ V m}^{-1}
  • D 20 V m−120 \text{ V m}^{-1}

Step-by-Step Verified Solution

**Answer:** (c) 10 V m__. **Step-by-Step Explanation:** The electric field near a large charged plate is E = _/(2__), independent of the material. So it remains 10 V/m.

All Questions in this Chapter

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