Verified Solution • 2025–26 Session moderate

QPages 141-142: Q17: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q17 Solution: A non-conducting sheet of large surface area and thickness d…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently assume the electric field is constant throughout the sheet or incorrectly apply the infinite plane formula $E = \rho d / 2\epsilon_0$ inside the material. You must recognize that for a point inside the slab, the Gaussian pillbox extends only to distance $x$ from the center, enclosing charge $\rho A (2x)$, which results in a linear dependence $E \propto x$ rather than a constant value.

A non-conducting sheet of large surface area and thickness d contains a uniform charge distribution of density ρ\rho. Find the electric field at a point P inside the plate, at a distance x from the central plane. Draw a qualitative graph of E against x for 0<x<d0 < x < d.

Step-by-Step Verified Solution

**Answer:** E=ρxϵ0E = \frac{\rho x}{\epsilon_0}. **Step-by-Step Explanation:** For a non-conducting sheet of thickness d with uniform charge density _, at distance x from the central plane (x < d/2): Charge enclosed in a Gaussian surface of area A and thickness 2x: Qenc=ρA(2x)Q_{enc} = \rho A (2x). By Gauss's law: E×2A=ρA(2x)ϵ0E \times 2A = \frac{\rho A (2x)}{\epsilon_0}.
E=ρxϵ0E = \frac{\rho x}{\epsilon_0}
Graph: E increases linearly from 0 at the centre to ρd2ϵ0\frac{\rho d}{2\epsilon_0} at the surface, then remains constant outside .

All Questions in this Chapter

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