Q. A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is
A) Charge $Q$
B) Potential $V$
C) Capacity $C$
D) Energy $U$
Explanation
The correct answer is Option A: Charge $Q$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.