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I have a simple doubt. The textbook I'm working with says that if $E_0$ is the electric field in some region without diletric, then if we introduce a dieletric of constant $k$ the new field will simply be $E=k^{-1}E_0$. Is it always true? I mean, suppose we have concentric spheres of radii $a$ and $b$ being $a$ the radius of the inner sphere and $b$ the radius of the outer sphere and charge $Q_a$ in the inner and $Q_b$ in the outer.

If there's no dieletric between them the field in the space between at a distance $r$ from the center will be given by Gauss' Law as:

$$E_0=\frac{1}{4\pi\epsilon_0} \frac{Q_a}{r^2}$$

Now if we insert a dieletric on the space between is it true that the field on the dieletric will simply be:

$$E=\frac{1}{4\pi k \epsilon_0}\frac{Q_a}{r^2}$$

Thanks very much in advance.

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1 Answer 1

up vote 3 down vote accepted

Not always. All of your Gaussian surface should be in a linear dielectric with constant electric permittivity $\epsilon$ to be able to use gauss law and derive that formula. With this conditions it's true most of the times.

Here you can use again the gauss law:

$\vec D = {Q_a \over 4 \pi r^2} \hat r$

But we know that for linear dielectrics: $\vec D = \epsilon \vec E = k \epsilon_0 \vec E$

So: $\vec E = {Q_a \over 4 \pi k \epsilon r^2} \hat r$

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