Question

An air-insulated parallel-plate capacitor is connected to a battery that imposes a potential difference V across the capacitor. If a dielectric slab is inserted between the capacitor plates, what happens to (a) the potential difference, (b) the capacitance, (c) the capacitor charge?



2. Which capacitor can store more energy: a 1-μF capacitor rated at 250 V or a 470-pF capacitor rated at 3 kV? Show the calculations for your answer.

Answer 1

potential difference will remain constant but the electric field within the dielectric will decrease
now if k=relative permittivity of the dielectric =E/E0
=(permittivity of dielctric medium/permittivity of free space)
then capacitence C will be K times C, C'=KC
The capacitor charge increases from Q to KQ because the capacitance increased K times but the voltage is still constant

for ur second problem use energy stored= E = CV²/2 and comapre them

Answer 2

thanks