A parallel plate capacitor is charged to a potential difference of 100 V and disconnected from the source. A slab of dielectric is then inserted between the plates. (a) Compared to the energy before the slab was inserted, the energy stored in the capacitor with the dielectric is introduced to the energy stored in the empty capacitor is U/U0 = k.

(b) Give a physical explanation for this increase in stored energy.

(c) What happens to the charge on the capacitor? Note: This situation is not the same as when the battery is removed from the circuit before the dielectric is introduced.

Question

Julio Summers

Physics

8 January, 05:29

A parallel plate capacitor is charged to a potential difference of 100 V and disconnected from the source. A slab of dielectric is then inserted between the plates. (a) Compared to the energy before the slab was inserted, the energy stored in the capacitor with the dielectric is introduced to the energy stored in the empty capacitor is U/U0 = k.

(b) Give a physical explanation for this increase in stored energy.

(c) What happens to the charge on the capacitor? Note: This situation is not the same as when the battery is removed from the circuit before the dielectric is introduced.

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Answers (1)

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Jimena Lester · Answer 1

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Explanation:

a) The energy stored in a capacitor is given by (1/2) (CV²)

Energy in the capacitor initially

U₀ = CV²/2

V = voltage across the plates of the capacitor

C = capacitance of the capacitor

But the capacitance of a capacitor depends on the geometry of the capacitor is given by

C = ϵA/d

ϵ = Absolute permissivity of the dielectric material

ϵ = kϵ₀

where k = dielectric constant

ϵ₀ = permissivity of free space/air/vacuum

A = Cross sectional Area of the capacitor

d = separation between the capacitor

If air/vacuum/free space are the dielectric constants,

So, k = 1 and ϵ = ϵ₀

U₀ = CV²/2

Substituting for C

U₀ = ϵ₀AV²/2d

The slab of dielectric is now inserted in the space between the plates of the capacitor,

The dielectric material has a dielectric constant of k

ϵ = kϵ₀

U = (kϵ₀AV²) / 2d

Compared to U₀

U = (kϵ₀AV²) / 2d

U₀ = (ϵ₀AV²) / 2d

(U/U₀) = k (Proved)

b) The dielectric constant of a dielectric material is an expression that shows how much the material concentrates the electric flux between the plates of the capacitor. As it is a ratio, it compares this ability with the ability of air/vacuum/free space to concentrate the required electric flux.

So, any material with a dielectric constant greater than 1 has the ability to enable the capacitor to store more charges, thereby leading to more energy stored in that capacitor.

c) As shown in (a), the capacitance, C is directly proportional to the absolute permissivity of the dielectric material between the plates of the capacitor. So, as the dielectric constant of the dielectric material increases, the capacitance of the the capacitor increases as long as the O the parameters stay constant.