Question

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.

Answer 1

(a) 7.18

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

$E=\frac{\sigma}{k \epsilon_0}$ (1)

where

$\sigma$ is the surface charge density

k is the dielectric constant

$\epsilon_0$ is the vacuum permittivity

The area of the plates in this capacitor is

$A=100 cm^2 = 100\cdot 10^{-4} m^2$

while the charge is

$Q=8.9\cdot 10^{-7}C$

So the surface charge density is

$\sigma = \frac{Q}{A}=\frac{8.9\cdot 10^{-7} C}{100\cdot 10^{-4} m^2}=8.9\cdot 10^{-5} C/m^2$

The electric field is

$E=1.4\cdot 10^6 V/m$

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

$k=\frac{\sigma}{E \epsilon_0}=\frac{8.9\cdot 10^{-5} C/m^2}{(1.4\cdot 10^6 V/m)(8.85\cdot 10^{-12} F/m)}=7.18$

(b) $7.66\cdot 10^{-7}C$

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

$\sigma' = \sigma (1-\frac{1}{k})$

where

$\sigma=8.9\cdot 10^{-5} C/m^2$ is the initial charge density

k = 7.18 is the dielectric constant

Substituting,

$\sigma' = (8.9\cdot 10^{-5} C/m^2) (1-\frac{1}{7.18})=7.66\cdot 10^{5} C/m^2$

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

$Q' = \sigma' A = (7.66\cdot 10^{-5} C/m^2)(100 \cdot 10^{-4}m^2)=7.66\cdot 10^{-7}C$