Part A

Find the energy U0 stored in the capacitor. Express your answer in terms of A, d, V, and ϵ0. Remember to enter ϵ0 as epsilon_0.

Part B

The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separation reaches 3d. Find the new energy U1 of the capacitor after this process.

Express your answer in terms of A, d, V, and ϵ0.

Part C

The capacitor is now reconnected to the battery, and the plate separation is restored to d. A dielectric plate is slowly moved into the capacitor until the entire space between the plates is filled. Find the energy U2 of the dielectric-filled capacitor. The capacitor remains connected to the battery. The dielectric constant is K.

Express your answer in terms of A, d, V, K, and ϵ0.

A) U₀ = ϵ₀AV²/2d

B) U₁ = (ϵ₀AV²)/6d

This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.

C) U₂ = (kϵ₀AV²)/2d

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

B) Now, for U₁, the new distance between plates, d₁ = 3d

U₁ = ϵ₀AV²/2d₁

U₁ = ϵ₀AV²/(2(3d))

U₁ = (ϵ₀AV²)/6d

This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.

C) U₂ = CV²/2

Substituting for C

U₂ = ϵAV²/2d

The dielectric material has a dielectric constant of k

ϵ = kϵ₀

U₂ = (kϵ₀AV²)/2d

inverse v=i*r

nice nice