A) The charge on the capacitor
B) The stored energy in the capacitor
C) The potential difference between the plates
D) The electric field in the capacitor
Correct Answer: A
Solution :
If a dielectric slab is inserted between the plates of a charged capacitor, the intensity of electric field potential difference of capacitor and the energy stored all reduce to\[\frac{1}{K}\]times and K. capacity of the capacitor increases K times. But the charge on the capacitor remains unchanged. Here K is the dielectric constant of dielectric.
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