# Spherical capicator w/ dielectric

1. Feb 9, 2005

### genxhis

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

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 Vab = 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).

2. Feb 9, 2005

### 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.

Last edited: Feb 9, 2005