HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q59 Solution: A parallel-plate capacitor of plate area A and plate separation …

QPages 165-170: Q59: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q59 Solution: A parallel-plate capacitor of plate area A and plate separation …

⚠️ Common Exam Traps & Conceptual Pitfalls

When the dielectric is inserted with the battery disconnected, the capacitance increases and the stored electrostatic energy drops; students often assume the work done on the system equals the energy increase and thus miss the negative sign. Remember that the work done by the external agent is the negative of the change in stored energy ($W = U_i - U_f$), not $U_f-U_i$.

A parallel-plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.

Step-by-Step Verified Solution

**Answer:**

W = \frac{Q^2}{2C}\left(1 - \frac{1}{K}\right)

. **Step-BY-Step Explanation:** Initial energy:

U_i = \frac{Q^2}{2C}

. Final energy (with dielectric):

U_f = \frac{Q^2}{2KC}

. Work done =

U_i - U_f = \frac{Q^2}{2C}\left(1 - \frac{1}{K}\right)

QPages 165-170: Q59: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q59 Solution: A parallel-plate capacitor of plate area A and plate separation …

Step-by-Step Verified Solution

All Questions in this Chapter

Related Learning Resources & Exam Preparation