How Do You Calculate Capacitance in a Half-Filled Spherical Capacitor?

[93157]

another for the EE pros out there

an isolated spherical capacitor has charge +Q on its inner conductor of radius r-sub-a and charge -Q on its outer conductor of radius r-sub-b. half of the volume between the two conductors is then filled with a liquid dielectric of constant K. a) find the capacitance of the half filled capacitor. b) find the magnitude of (electric field) E in th evolume between the two conductors as a function of the distance r from the center of the capacitor. give answers for both the upper and lower halves of this volume. c) find the surface density of free charge on the upper and lower halves of the inner and outer conductors. d) find the surface density of bound charge on the flat surface of the dielectric? e) what is the surface density of bound charge on the inner and outer surfaces of the dielectric.

it's mainly part a) that i can't figure out how to set up. like does Q vary across the surface of the sphere or is it uniformly spread?

thanks

 

[siddharth]

To answer the other parts, you need to solve b) first.
That is, you need to find the Electric field. There has to be spherical symmerty, and E should be a function of r only. Have you tried applying Gauss's law? You can get the electric field straight away from Gauss's law.

$$\oint \vec{D}.\vec{da} = Q_{free}$$

Hint: What's D in the upper and lower halves?

 

[Meir Achuz]

Use the fact that D_tangential is constant across the interface of the dielectric.

 

[93157]

D is E/epsilon-oh?

i know that the field generated by a conducting sphere is q/(4*pi*epsilon-oh*r^2).

and that the field in between the two spheres is only due to the inner sphere.

okay, so... (ma mind's a churnin')

E (of upper hemisphere sphere r-sub-a) = Q/(4*pi*epsilon-oh*r-sub-b^2)

and

E (of lower hemisphere sphere r-sub-a) = E (upper hemisphere r-sub-b) / K
for K is the dielectric constant


*working the rest out*


oh okay

i can calculate the capacitance based on C = Q/V which can be turned into (Q/[4*pi*epsilon-oh])*(1/r-sub-a - 1/r-sub-b), and the dielectric capacitance from C = C-sub-oh*K

and the surface density i can work out directly from the known charge and the hemisphere's surface area 2*pi*r^2. and also that induced-surface-density = surface-density * (1 - 1/K)


cool

thanks a bunch