Verified Solution • 2025–26 Session moderate

QPages 165-170: Q58: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q58 Solution: Figure shows two identical parallel plate capacitors connected…

⚠️ Common Exam Traps & Conceptual Pitfalls

When the switch is opened, the system is isolated, so charge $Q$ remains constant; students must use $U = \frac{Q^2}{2C}$ to show energy decreases by the dielectric constant, rather than $U = \frac{1}{2}CV^2$ which incorrectly assumes constant voltage.

Figure shows two identical parallel plate capacitors connected to a battery through a switch S. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.

Step-by-Step Verified Solution

**Answer:** Ratio = 3:1. **Step-by-Step Explanation:** Initial energy (both capacitors charged):

U_i = 2 \times \frac{1}{2}CV^2 = CV^2

. After switch opened and dielectric inserted in both:

C' = 3C

for each. Charge is constant, so

U_f = \frac{Q^2}{2C'} \times 2 = \frac{(CV)^2}{2 \times 3C} \times 2 = \frac{CV^2}{3}

. Ratio

U_i/U_f = 3

.

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