Question

A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.55 is placed between the plates, completely filling the space between them. A) How much energy is stored in the capacitor before the dielectric is inserted?
B) How much energy is stored in the capacitor after the dielectric is inserted?
C) By how much did the energy change during the insertion? Did it increase or decrease?
D) Explain why inserting the dielectric (or equivalently exchanging air with the material) causes a change in the stored energy of the capacitor?

Answer 1

Answer:

Explanation:

Capacitance of the capacitor = 13.5μF

Voltage across plate is 24V

Dielectric constant k=3.55.

a. Energy in capacitor is given by

E=1/2CV^2

We want to calculate energy without the dielectric substance

Given that C=13.5 μF and V=24V

The capacitance give is with dielectric so we need to remove it

C=kCo

Co=C/k

Then the Co=13.5μF/3.55

Co=3.803μF

Then

E=(1/2)×3.803×10^-6×24^2

E=1.1×10^-3J

E=1.1mJ

b. Energy in capacitor is given by

E=1/2CV^2

The capacitance given is with a dielectric, so we are going to apply it direct.

Given that C=13.5 μF and V=24V

Then

E=(1/2)×13.5×10^-6×24^2

E=3.89×10^-3J

E=3.9mJ

c. The energy without dielectric is 1.1mJ and the energy with dielectric is 3.9mJ

The energy increase when the dielectric material is added

d. Dielectrics in capacitors serve three purposes: to keep the conducting plates from coming in contact, allowing for smaller plate separations and therefore higher capacitances;

Therefore, Since dielectric allow higher capacitance, and energy of a capacitor is directly proportional to the capacitance, then the higher the capacitance the higher the energy.

Answer 2

(A) The energy stored in the capacitor before inserting dielectric is

1.1 × 10⁻³ J

(B) The energy stored in the capacitor after inserting dielectric is

3.89 × 10⁻³ J

(C) There is an increase in energy equal to 2.88 × 10⁻³ J after inserting the dielectric.

(D) The use of dielectric increases the capacitance and hence increases the energy stored.

Given Information:

The capacitance of the capacitor C = 13.5μF

The voltage across the plate is V = 24V

Dielectric constant k =3.55

Energy stored in the capacitor:

(A) Energy stored in the capacitor is given by:

$E=\frac{1}{2} CV^2$

Given that

C = 13.5 μF and V = 24V

The capacitance  with dielectric :

C = kC₀

where C₀ is capacitance without the dielectric

C₀= C/k

C₀ = 13.5 / 3.55

C₀ = 3.803μF

So the energy stored is:

$E=(1/2)\times3.803\times10^{-6}\times24^2\\\\E=1.1\times10^{-3}J$

(B) Energy in the capacitor is given by

E = ¹/₂CV²

The capacitance given is already with a dielectric.

We have C=13.5 μF and V=24V

Then

$E=(1/2)\times13.5\times10^{-6}\times24^2$

E = 3.89 × 10⁻³ J

(C) From the above results we can conclude that:

The energy increase when the dielectric material is added.

The amount of change in energy is = (3.89 - 1.1) × 10⁻³ J = 2.88 × 10⁻³ J

(D) Dielectrics in capacitors allow smaller plate separations and therefore the capacitance increases since capacitance is inversely proportional to plate separation.

Also, the energy stored in a capacitor is directly proportional to the capacitance. So, the higher the capacitance the higher the energy.

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