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QPages 141-142: Q23: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q23 Solution: Two conducting plates X and Y, each with a large surface area A …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume a neutral plate has zero charge on all surfaces, ignoring electrostatic induction. Remember that the neutral plate Y will develop induced charges $+Q/2$ and $-Q/2$ on its inner and outer surfaces respectively, which is essential for correctly calculating the field cancellation outside the plates.

Two conducting plates X and Y, each with a large surface area A on one side), are placed parallel to each other, as shown in the following figure. Plate X is given a charge Q, whereas the other is kept neutral. Find a) the surface charge density at the inner surface of plate X b) the electric field at a point to the left of the plates c) the electric field at a point in between the plates and d) the electric field at a point to the right of the plates.

Step-by-Step Verified Solution

**Answer:** (a)

\sigma_{inner} = -Q/(2A)

(b)

E_{left} = 0

(c)

E_{between} = Q/(2A\epsilon_0)

(d)

E_{right} = 0

. **Step-by-Step Explanation:** Plate X has charge Q, plate Y is neutral. (a) Charge on inner surface of X =

-Q/2

(shared equally with Y's inner surface). (b) Left of plates: fields cancel. E = 0. (c) Between:

E = \frac{Q}{2A\epsilon_0}

. (d) Right: fields cancel. E = 0 .

All Questions in this Chapter

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