Verified Solution • 2025–26 Session moderate

QPages 165-170: Q15: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q15 Solution: Take the potential of the point B in figure to be zero. a) Find …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often apply the resistor voltage‑division formula directly to series capacitors, using $V_i=V_{total}\frac{C_i}{C_{total}}$. In reality, the voltage across a capacitor in series is proportional to the *other* capacitance: $V_i=V_{total}\frac{C_{j}}{C_i+C_j}$. Remember to use the inverse‑capacitance relationship to avoid a wrong potential and consequently an incorrect charge prediction.

Take the potential of the point B in figure to be zero. a) Find the potentials at the points C and D. b) If a capacitor is connected between C and D, what charge will appear on this capacitor?

Step-by-Step Verified Solution

**Answer:** (a) VC=VD=50/3V_C = V_D = 50/3 V (b) Charge = 0. **Step-by-Step Explanation:** (a) The bridge is balanced (ratio of capacitances in the two branches is equal). So VC=VDV_C = V_D. Using voltage division: VC=VD=24+2×50=50/3V_C = V_D = \frac{2}{4+2} \times 50 = 50/3 V. (b) Since VC=VDV_C = V_D, no charge flows when a capacitor is connected between C and D .

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