Verified Solution • 2025–26 Session moderate

QPages 165-170: Q34: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q34 Solution: The capacitance between the adjacent plates shown in figure is…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently split the middle‑plate charge as $Q/2$ for each gap and then use $V=Q/(2C)$, overlooking that each individual capacitor in the series carries the full charge $Q$ and the equivalent capacitance is $C/2$; thus the voltage between the upper and middle plates is $V=Q/C$, not $Q/(2C)$.

The capacitance between the adjacent plates shown in figure is 50 nF. A charge of 1.0 �C is placed on the middle plate. a) What will be the charge on the outer surface of the upper plate? b) Find the potential difference developed between the upper and the middle plates.

Step-by-Step Verified Solution

**Answer:** (a) Q=0.5Q = 0.5 _C (b) V=10V = 10 V. **Step-by-Step Explanation:** (a) Charge on outer surface of upper plate = Q/2=0.5Q/2 = 0.5 _C. (b) The middle plate has charge 1.0 _C. It forms two capacitors in series with the upper and lower plates.
V=Q2C=1.0×10−62×50×10−9=10 VV = \frac{Q}{2C} = \frac{1.0 \times 10^{-6}}{2 \times 50 \times 10^{-9}} = 10 \text{ V}

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