Wo parallel plates of area 100cm2are given charges of equal magnitudes 8.9 ×10−7c but opposite signs. the electric field within the dielectric material filling the space between the plates is 1.4 ×106v/m. (a) calculate the dielectric constant of the material. (b) determine the magnitude of the charge induced on each dielectric surface.

(a) 7.18

The electric field within a parallel plate capacitor with dielectric is given by:

(1)

where

is the surface charge density

k is the dielectric constant

is the vacuum permittivity

The area of the plates in this capacitor is

while the charge is

So the surface charge density is

The electric field is

So we can re-arrange eq.(1) to find k:

(b)

The surface charge density induced on each dielectric surface is given by

where

is the initial charge density

k = 7.18 is the dielectric constant

Substituting,

And by multiplying by the area, we find the charge induced on each surface:

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explanation: