Question

Space between the plates of a parallel plate capacitor is filled with a dielectric slab. The capacitor is charged and then the supply is disconnected to it. If the slab is now taken out , then:

• A. work is not done to take out the slab
• B. energy stored in the capacitor reduces
• C. potential difference across the capacitor is decreased
• D. potential difference across the capacitor is increased

Answer 1

Answer: (D)

potential difference across the capacitor is increased

When a capacitor is charged and then the supply is disconnected charge present on it becomes constant (law of conservation of charge)
Initial capacitance $C=\frac{k{\epsilon }_{0}A}{d}$ (with slab)
Final capacitance $C=\frac{{\epsilon }_{0}A}{d}$ (slab removed)
We know $Q=CV$
As C decreases and Q remains constant V has to increase.