SPHERICAL CAPACITOR - with two insulating shells

[Grommit]

PLEASE HELP ASAP - Spherical capacitor

I could really use some help with this one, and I need to submit very soon...
Problem:
A spherical capacitor is made of two insulating spherical shells with different dielectric
constants K1 and K2 situated between two (inner and outer, shown by thick lines)
spherical metallic shells and separated by a vacuum gap. Geometrical dimensions of
the cross-section are as shown in figure 2. Outer metallic shell has charge +Q and the
inner metallic shell has charge −Q. What is the potential difference V between these
metallic shells?
Image: http://tinyurl.com/yazcvq2

If I am understanding this correctly, this can be interpreted as three capacitors connected in series. To that end, I thought I could utilize the equation for capacitors in series: 1/Ceq = 1/C1 + 1/C2 +1/C3. Then, Ceq=Q/V would indicate the potential difference. (Im thinking this may be flawed logic though).

Using K*(4*pi*Epsilon(0))*(a*b)/(b-a) for the spherical capacitances seems to be making a cumbersome mess of numbers more than anything.
Im starting to think now that maybe I should be setting up an integral along the lines of V=Q/(4*pi*Epsilon(0)) INTEGRAL (from a to b) db/b^2... hope that makes sense.
Thanks in advance for any help.

 

[Grommit]

FWIW, I found the problem partially worked out in this manner:
Basically you have 3 capacitors in series, the first has a dielectric K2 with a thickness b -a, the second has a vacuum dielectric with a thickness c-b and the third capacitor with a dielectric K3 with a thickness d -c.
C1 = K2/(b-a), C2 = 1/(c-b), and C3 = K1/(d-c)
1/Ct = 1/C1 + 1/C2 + 1/C3
1/Ct = (b-a)/K2 + (c-b) + (d-c)/K1
1/Ct = (K1(b-a) +K1(K2)(c-b) + K2(d-c))/K1(K2)
Ct = K1(K2)/(K1(b-a) +K1(K2)(c-b) + K2(d-c))
http://answers.yahoo.com/question/i...F112UmVhoyDA.YirA--&paid=add_comment#openions
However, this solution doesn't seem to have the variables for spherical capacitance included ==> K*(4*pi*Epsilon(0))*(a*b)/(b-a)

 

[kerimek]

I understand your frustration and urgency in solving this problem. Let's break down the problem and see if we can come up with a solution together.

Firstly, we need to understand the concept of a capacitor. A capacitor is a device that stores electrical energy by accumulating opposite charges on two conductors separated by an insulating material, also known as a dielectric. In this case, our capacitor is made up of two insulating spherical shells with different dielectric constants (K1 and K2) between two metallic shells.

Now, let's look at the geometry of the problem. We have an outer metallic shell with charge +Q and an inner metallic shell with charge -Q. The potential difference V is the difference in electric potential between these two shells. In other words, it is the work required to move a unit of electric charge from the outer shell to the inner shell.

To solve for V, we can use the equation V = Q/C, where C is the capacitance of the system. In this case, we have three capacitors connected in series, so we can use the equation you mentioned: 1/Ceq = 1/C1 + 1/C2 + 1/C3. However, we need to find the capacitance of each individual capacitor first.

To find the capacitance of a spherical capacitor, we can use the equation C = 4*pi*Epsilon(0)*K*a*b/(b-a), where a and b are the radii of the inner and outer shells, respectively. Since we have two different dielectric constants, we will have two different capacitance values. Let's call them C1 and C2.

Now, we need to find the equivalent capacitance, Ceq, of the three capacitors in series. This can be done by substituting the values of C1 and C2 into the equation 1/Ceq = 1/C1 + 1/C2 + 1/C3. We can also substitute the value of C3, which is the capacitance of the vacuum gap between the shells. Since the vacuum has a dielectric constant of 1, the capacitance of the vacuum gap will simply be C3 = 4*pi*Epsilon(0)*a*b/(b-a).

Once we have the value of Ceq, we can use the equation V = Q/Ceq to solve for the potential difference V between the metallic shells.

Alternatively, as you