Verified Solution • 2025–26 Session moderate

QPages 165-170: Q36: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q36 Solution: Two capacitors of capacitance 20.0 pF and 50.0 pF are connected …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently treat series capacitors like resistors, splitting the battery voltage proportionally to the capacitance values. In reality the charge $Q$ is common, so $V_i=Q/C_i$; thus the smaller capacitor gets the larger voltage. Remember to compute the series charge first, then use $V_i=Q/C_i$ to avoid the wrong voltage distribution.

Two capacitors of capacitance 20.0 pF and 50.0 pF are connected in series with a 6.00 V battery. Find a) the potential difference across each capacitor and b) the energy stored in each capacitor.

Step-by-Step Verified Solution

**Answer:** (a) V1=4.29V_1 = 4.29 V, V2=1.71V_2 = 1.71 V (b) U1=1.84×10−10U_1 = 1.84 \times 10^{-10} J, U2=7.35×10−11U_2 = 7.35 \times 10^{-11} J. **Step-by-Step Explanation:** Equivalent capacitance: Ceq=20×5070=14.29C_{eq} = \frac{20 \times 50}{70} = 14.29 pF. Charge Q=CeqV=14.29×10−12×6=8.57×10−11Q = C_{eq}V = 14.29 \times 10^{-12} \times 6 = 8.57 \times 10^{-11} C.
V1=Q/C1=4.29 V,V2=Q/C2=1.71 VV_1 = Q/C_1 = 4.29 \text{ V}, \quad V_2 = Q/C_2 = 1.71 \text{ V}
U1=12C1V12=1.84×10−10 J,U2=12C2V22=7.35×10−11 JU_1 = \frac{1}{2}C_1V_1^2 = 1.84 \times 10^{-10} \text{ J}, \quad U_2 = \frac{1}{2}C_2V_2^2 = 7.35 \times 10^{-11} \text{ J}

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