Consider a flat closed box with bottom and top of conducting material and side walls of isolation material. The box is half filled with oil; the relative permettivity of oil is . The other half of the box is in vacuum. The distance between bottom and top is d, and their surface is A. Bottom and top are connected to a power source with a voltage of V. Calculate E and D in the oil and the vacuum if the box is placed horizontal and on a side. Also calculate the surface charge density at the bottom and top. Calculate the capacity too.

Vertical box: vacuum and oil next to each other

Method 1: Using dielectrics

Let's draw a picture first. Note that the picture has been rotated 90 degrees.

D is independent of the medium. With Gauss' Box G we see that it is equal to the free charge density.

> D1:=sigma[free]:
D2:=sigma[free]:

The definition of the charge density is as follows.

> Q[free]:=sigma[free]*A:

Define the relation between E and D for both media.

> E1:= D1 /(1*epsilon[0]):
E2:= D2 /(epsilon[r]*epsilon[0]):

The potential V is defined as the integral of E dl . The electrical field is uniform, so we can just multiply.

> eq1:= V=E1*d/2 + E2*d/2:

Now get an expression for .

> assign(solve({eq1,Q[free]},{sigma[free],Q[free]})[1]); sigma[free];

Get expressions for E and D.

> 'E1' = E1;
'E2' = E2;
'D1' = D1;
'D2' = D2;

Test if the fields are equal when . Then the relative permittivity of oil is equal to the relative permittivity of vacuum (This is not in the real world, but it is a good test case to check if the formulas are consistent).

> subs(epsilon[r]=1,E1)=subs(epsilon[r]=1,E2);
subs(epsilon[r]=1,D1)=subs(epsilon[r]=1,D2);

Well, the above seems to be correct.

Now compute the capacity.

> C:=Q[free]/V;

Method 2: Using serial capacitors

To compute the capacity of the box, we can make use of the fact that it can be seen as two capacitors in series; one filled with vacuum and the other filled with oil. We use the standard capacitance formula for a capacitor.

> C1 := epsilon[0]*A/(d/2):
C2 := epsilon[0]*epsilon[r]*A/(d/2):

Now we just use the formula for capacitors in series and voila, we get the same result as in method 1.

> C := simplify(1/(1/C1+1/C2));

Horizontal box: vacuum and oil on top of each other

Method 1: using dielectrics.

Let's first draw a picture of this situation.

We know that the integral of E dl is equal to the potential. This is correct for the vacuum as well as the oil, so

> E1:= V/d:
E2:= V/d:

Let maple know the relation between D and E.

> D1:= epsilon[0]*E1;
D2:= epsilon[0]*epsilon[r]*E2;

Check if those are the same when

> subs(epsilon[r]=1,D2)=D1;

And this seems to be correct.

Charge density is equal to D.

> sigma1[free]:= D1;
sigma2[free]:= D2;

The charge is the charge density devided by the surface.

> Q1[free]:= D1*A/2;
Q2[free]:= D2*A/2;

The total free charge is the sum of all free charges.

> Q[free]:= simplify(Q1[free]+Q2[free]);

Now calculate the capacity

> C:=Q[free]/V;

Method 2: using parallel capacitors

The box can in this case be considered as two capacitors in parallel.

> C1:= epsilon[0]*1*A/2/d:
C2:= epsilon[0]*epsilon[r]*A/2/d:

> C := simplify(C1+C2);

This seems to correspond with method 1.

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