Electrostatics Question 2
Question: A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in [NCERT 1980; MP PET 1995; BHU 1997]
Options:
A) Reduction of charge on the plates and increase of potential difference across the plates
B) Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates
C) Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates
D) None of the above
Show Answer
Answer:
Correct Answer: C
Solution:
Battery in disconnected so Q will be constant as $ C\propto K $ .
So with introduction of dielectric slab capacitance will increase using Q = CV, V will decrease and using $ U=\frac{Q^{2}}{2C} $ , energy will decrease.
BETA
