Capacitance of spherical capacitor with mixed dielectrics:

Let us consider A and B are the two concentric spherical shells of radii a and b respectively (b >a). Shell A is charged by Q and the outer surface of shell B is earthed.

Let a shell filled with dielectric medium of dielectric constant K_{2} of inner and outer radii r_{1} and r_{2} respectively as shown in figure. The remaining part of capacitor is filled with another dielectric medium of dielectric constant K_{1}.

The electric field intensity in medium 1 of dielectric constant K_{1 }at a distance r from the centre is Â and that in medium 2 of dielectric constant K_{2 }is .

As dV = – Edr, then for medium 1 dV = – dr and for medium 2 dV = – dr.

Then the potential difference between shell A and B is V_{B} – V_{A} = – dr + – dr + – dr

V_{A} â€“ V_{B} = – drÂ + – dr + – dr

Or, V_{A} â€“ V_{B} = [Â + [ Â + [

Or, V_{A} â€“ V_{B} = [ Â – ] + [Â – ] + [ – ]

Or, V_{A} â€“ V_{B} = [ – ] – [ – ] + [ – ]

Or, V_{A} â€“ V_{B} =

Or, V_{A} â€“ V_{B} = Â

Or, V_{A} â€“ V_{B} = []

Capacitance C = = Â .

If K_{1} = 1 and K_{2} = K then, C = Â .

If K_{1} = K_{2} = 1 then, C = .

Force acting between the plates of the capacitor: Let us consider a parallel plate capacitor consists of two parallel plates each of plate area A are separated by a distance d. If F is force acting between two plates then

F = [see theory]

If V is the potential difference between the plates, then, V = Ed and F = .

If K is the dielectric constant between the plates, then, F = .

Click the button to go to the previous part of this chapter.