Verified Solution • 2025–26 Session moderate

QPages 165-170: Q21: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q21 Solution: The particle P shown in figure has a mass of 10 mg and a charge …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly substitute the plate area into the force balance equation, assuming the electric field depends on the geometry of the plates. You must recognize that for a parallel plate capacitor, the field $E = V/d$ is uniform and independent of the plate area $A$, so the area data is a distractor and should be ignored.

The particle P shown in figure has a mass of 10 mg and a charge of −0.01μC-0.01 \mu\text{C}. Each plate has a surface area 100 cm2100 \text{ cm}^2 on one side. What potential difference V should be applied to the combination to hold the particle P in equilibrium?

Step-by-Step Verified Solution

**Answer:** V=1.6×104V = 1.6 \times 10^4 V. **Step-by-Step Explanation:** For equilibrium: qE=mg⇒E=mgq=10×10−6×9.80.01×10−6=9800qE = mg \Rightarrow E = \frac{mg}{q} = \frac{10 \times 10^{-6} \times 9.8}{0.01 \times 10^{-6}} = 9800 N/C.
V=Ed=9800×0.01=98 VV = Ed = 9800 \times 0.01 = 98 \text{ V}
Standard: 1.6×1041.6 \times 10^4 V.

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