Question

A parallel plate capacitor is connected to a battery. The
quantities charge, voltage, electric field and energy
associated with this capacitor are given by Q₀, V₀, E₀, and
U₀ respectively. A dielectric slab is now introduced to fill
the space between the plates with the battery still in
connection. The corresponding quantities now given by
Q, V, E and U are related to the previous ones as
(a) Q > Q₀ (b) V > V₀ (c) E > E₀ (d) U < U₀

Answer 1

Answer:

(b) V > V

follow me......

Answer 2

Explanation:

C = k∈A /d

as  capacitor is connected to a battery, V is constant

1) Q =CV  increases

       Q > Q₀

2) V is constant

3)E decreases

4) U = 1/2 CV²

   As c increases , U increases

     U > U₀

 ------------Hope you understand -------