Work required to remove a metal sheet from a capacitor

[hitemup]

Homework Statement

How much work would be required to remove a metal sheet from between the plates of a capacitor, assuming

a) the battery remains connected so the voltage remains constant
b) the battery is disconnected so the charge remains constant

Homework Equations

When a metal sheet with a thickness "l" is put between the plates

C = \frac{\epsilon _0A}{d-l}

Without the metal

C = \frac{\epsilon _0A}{d}

The Attempt at a Solution

There is an example named "moving parallel capacitor plates" in my textbook, It asks the work needed to move the capacitor plates until the separation between them is 3x from x.

Then an explanation is given to the solution.

"Unlike Example 9(charge was constant in ex. 9), here the capacitor remains connected to the battery. Hence charge and energy can flow to or from the battery, and we cannot set the work W = \Delta U. Instead the work can be calculated from the equation \int_{a}^{b}Fdl.
As you might expect, the work required to pull these oppositely charged plates apart is positive."I tried to use this approach for the problem in the topic because it was like moving the plates. (from "d-l" to "d"). I ended up with a positive answer but the correct answer is the negative of what I have found. In the solutions manual, it just calculates the work from W = \Delta U

So was it wrong to see this problem as a moving plates question?

 

[Hesch]

Hint: In a) and b), you can find C₁, U₁ and C₂, U₂.

The energy in a capacitor, E_cap = ½*U²*C.

W = E₂ - E₁

 

[hitemup]

Hint: In a) and b), you can find C₁, U₁ and C₂, U₂.

The energy in a capacitor, E_cap = ½*U²*C.

W = E₂ - E₁

Sorry but that doesn't answer my question.