Spherical capicator w/ dielectric

[genxhis]

I am not sure if I have done the following problem correctly.

An isolated spherical capacitor has charge +Q on its inner conductor (radius a) and charge -Q on its outer conductor (radius b). The lower half of the capacitor is filled with a liquid dielectric of constant K. a) Find the capacitance of the half-filled capacitor. b) Find the magnitude of E in the volume between the conductors as a function of the distance r from the center of the capacitor. Give answers for the both the upper and lower halves of this volume. c) Find the suface 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 inner and outer surfaces of the dielectric. e) What is the surface density of bound charge on the flat surface of the dielectric? Explain.

To start, I do not know how to prove the electric field must be directed radially. Assuming this is true, then the magnitude of the field must be uniform along any spherical boundary (otherwise we could form a nonconservative loop that straddles this boundary). If we then apply Gauss's Law for dielectrics on a spherical surface of radius r, we have: 2 pi r²(K E + E) = Q/e₀ or E = 1/(2 pi e₀) Q/(1+K)r². From this it easy follows V_ab = 1/(2 pi e₀) Q/(1+K) (1/a - 1/b) and then C = 2 pi e₀ (1 + K) (a b)/(b - a). On reflection this does at least make some sense since the capacitance rises with stronger dielectrics and reduces to the standard expression when K = 1.

But the wording in part B implies the field strength is not the same in the lower and upper portions of the capacitor. And how one shows the field is directed radially throughout troubles me (though I can't imagine it being much else since it must at least be radial near the conductors).

 

[s_a]

OK suppose there is a (non zero) circumferential component of the E-field. Then there is a circumferential component of the D-Field (electric displacement vector) too.

Because there's no free charge in the dielectric nor in the air, the normal component of the D-field at the air-dielectric boundary (which also happens to be the circumferential component) is continuous (i.e. the circumferential component of the D-field is the same in the air and dielectric). Therefore if the circumferential component of the E-field in the dielectric is E_d, then the circumferential component of the E-field in the air is k*E_d, where k is the relative permittivity of the dielectric (provided these two points are the same radial distance from the centre sphere).

So the line integral of the E-field along any closed circular loop of circumference L circling the charged inner sphere (sharing the same centre), is E_d(1 + k)*L/2. This must be zero (a condition of electrostatics). Since k != -1, L != 0, then E_d = 0. So the circumferential component of the E-field is zero. The E-field is purely radial.

 

[wais]

Overall, I think you have made a good attempt at solving this problem. However, there are a few things that could be clarified or corrected.

First, in part a) you have found the capacitance for the entire capacitor, not just the lower half. To find the capacitance of the lower half, you would need to divide your answer by 2.

In part b), you are correct that the electric field must be directed radially. This can be shown by considering the symmetry of the problem and using Gauss's Law. The electric field must also be continuous at the boundary between the two dielectrics, so it will be the same in both the upper and lower halves of the volume between the conductors.

In part c), you have not specified which surface you are referring to. The surface density of free charge will be different on the inner and outer surfaces of both conductors.

In part d), the surface density of bound charge will be the same on both the inner and outer surfaces of the dielectric.

In part e), the surface density of bound charge on the flat surface of the dielectric will be zero. This is because bound charge only exists at the interfaces between different materials, and there is no interface on the flat surface of the dielectric.

Overall, your approach is correct, but be sure to clarify and specify which surfaces you are referring to in your calculations. Also, make sure to divide your answer in part a) by 2 to find the capacitance of the lower half of the capacitor. Keep up the good work!