Energy stored in the electric field of a capacitor

[jersey]

Homework Statement

The figure shows cross-sectional views of two cubical capacitors, and a cross-sectional view of the same two capacitors put together so that their interiors coincide. A capacitor with the plates close together has a nearly uniform electric field between the plates, and almost zero field outside; these capacitors don’t have their plates very close together compared to the dimensions of the plates, but for the purpose of this problem, assume that they still have approximately the kind of idealized field pattern shown in the figure. Each capacitor has an interior volume of 1.00 m3, and is charged up to the point where its internal field is 1.00 V/m.

Diagram 1
---┴---
↓↓↓↓↓
↓↓↓↓↓
---┬---

Diagram 2
-→→-
-→→-
-→→-
┤→→├
-→→-
-→→-

Diagram 3 (This diagram has the same four sides and is empty on the inside)
---┴---

(a) Calculate the energy stored in the electric field of each capacitor when they are separate.

(b) Calculate the magnitude of the interior field when the two capacitors are put together in the manner shown. Ignore effects arising from the redistribution of each capacitor’s charge under the influence of the other capacitor.

(c) Calculate the energy of the put-together configuration. Does assembling them like this release energy, consume energy, or neither?

Homework Equations

Energy stored, W = ½ QV = ½ CV2 joules
Charge Q = CV where C is the capacitance in Farads

charge Q is measured in coulombs (C)

If the dielectric (the material between the plates) is a vacuum, Capacitance C = e0 (A / l) where A is the area of the capacitor plates, and l is the distance between them.

e0 is the permittivity of free space (8.85X10-12)

If the dielectric is another material, capacitance is given by:

C = e0 (A / l) where er is the relative permittivity, which varies between materials.

Putting capacitors in parallel increases the total capacitance:

C = C1 + C2 + C3 ...

The Attempt at a Solution

a) C=8.85 x 10^-12 x 1/1
=8.85 x 10^-12

W= 1/2cV^2
=1/2 (8.85 x 10^-12) x 1.0^2
=4.425 x 10^-12 J

I'm stuck, am i even on the right track here?

 

[willem2]

you answered a) correctly

 

[jersey]

What is the formula for magnitude of the inner field?