# Find electric field inside a material

Istiak
Homework Statement:
An infinite slab of insulating material with
dielectric constant K and permittivity ##\epsilon = K \epsilon_0## is placed in a uniform electric field of magnitude ##E_0## . The field is perpendicular to the surface of the material. Find the magnitude of the electric field inside the material.]
Relevant Equations:
##\vec D=\epsilon\vec E##

##\oint \vec D\cdot d\vec a=q_{f_{enc}}##
From the second equation I get that,
##\vec D =\frac{q}{4\pi \vec r^2}\hat r##
From first equation I get that

##\vec E = \frac{q}{4\pi \vec r^2 \epsilon}=\frac{q}{4\pi \vec r^2 K \epsilon_0}##
But I saw that the answer is ##\vec E=\frac{\vec E_0}{K}##
While writing the comment my mind said, ##\vec E_0=\frac{q}{4\pi \vec r^2 \epsilon_0}##

So easily, ##\vec E= \frac{\vec E_0}{K}##

Or should I do the process some other way?

Gold Member
Homework Statement:: An infinite slab of insulating material with
dielectric constant K and permittivity ##\epsilon = K \epsilon_0 is placed in a uniform electric field of magnitude ##E_0## . The field is perpendicular to the surface of the material. Find the magnitude of the electric field inside the material.]
Relevant Equations:: ##\vec D=\epsilon\vec E##

##\oint \vec D\cdot d\vec a=q_{f_{enc}}##

From the second equation I get that,
##\vec D =\frac{q}{4\pi \vec r^2}\hat r##
From first equation I get that

##\vec E = \frac{q}{4\pi \vec r^2 \epsilon}=\frac{q}{4\pi \vec r^2 K \epsilon_0}##
But I saw that the answer is ##\vec E=\frac{\vec E_0}{K}##
While writing the comment my mind said, ##\vec E_0=\frac{q}{4\pi \vec r^2 \epsilon_0}##

So easily, ##\vec E= \frac{\vec E_0}{K}##

Or should I do the process some other way?
This may help:

### Electric field inside a material​

Homework Helper
Gold Member
2022 Award
While writing the comment my mind said, ##\vec E_0=\frac{q}{4\pi \vec r^2 \epsilon_0}##
That's the field a distance r from an isolated point charge (or outside a spherically symmetric charge-distribution where r is the distance to the centre). So the equation is not applicable here.

(Also the left side of the equation is a vector but the right side is a scalar.)