The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separation reaches 3d3d. Find the new energy U1U1U_1 of the capacitor after this process.

U₁ = (ϵAV²)/6d

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

Explanation:

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

Energy in the capacitor initially

U = 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

A = Cross sectional Area of the capacitor

d = separation between the capacitor

So,

U = CV²/2

Substituting for C

U = ϵAV²/2d

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

U₁ = ϵAV²/2d₁

U₁ = ϵAV²/(2(3d))

U₁ = (ϵAV²)/6d

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

the area between the graph and the t axis

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right