# Capacitors, potential difference, electric displacement

1. Nov 14, 2013

### izzy93

Question: Considering a parallel-plate capacitor that has a slab of dielectric with relative permittivity epsilomd = 5 and thickness d = 2 mm occupies half of the gap. The half is air with relative permittivity epsilom2=1

Denoting the electric field, and electric displacement in the dielectric by Ed, Dd, and in the air gap by E2,D2. There is no free charge on the dielectric.

If the electric displacement in the dielectric has magnitude Dd = 2.0 x 10^8 Cm^2 and is perpendicular to the plates, what is the magnitude V of the potential difference between the two plates of the capacitor?

Using V=Ed and D=epsilom0*epsilomr*E

So V = (D/epsilom0*epsilomr)*d

Then taking the D-field to be the same in the dielectric and air gap as there is no free charge

so D2=Dd

Leads to V2/Vd = epsilomd/epsilom2 = 5

but how do I get the potential difference?

Any help would be much appreciated,

Thanks

2. Nov 14, 2013

### nasu

Find Vd and V2 and add them.