Question

A 12.5 µF capacitor is connected to a power supply that maintains a constant potential difference of 24.0 V across the plates. A piece of material with a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. How much energy is stored in the capacitor before and after the dielectric is inserted?

Answer 1

Answer:

The energy stored before dielectric slab insert is $3.6 \times 10^{-3}$ J and after is $13.5 \times 10^{-3}$ J

Explanation:

Given :

Capacitance $C = 12.5 \times 10^{-6}$ F

Potential difference $V = 24$ V

Dielectric constant $k = 3.75$

The energy stored in the capacitor before dielectric slab inserted is,

  $U = \frac{1}{2} C V^{2}$

  $U = \frac{1}{2} (12.5 \times 10^{-6} ) (24)^{2}$

  $U = 3.6 \times 10^{-3}$ J

When dielectric slab placed the energy stored in capacitor increase by (k) times.

 $U' = \frac{1}{2} C k V^{2}$

 $U' = \frac{1}{2} (12.5 \times 10^{-6} ) \times 3.75 \times (24)^{2}$

 $U' = 13.5 \times 10^{-3}$ J

Therefore, the energy stored before dielectric slab insert is $3.6 \times 10^{-3}$ J and after is $13.5 \times 10^{-3}$ J