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I have a uniform electric field $E$ projected from left to right and I placed an insulator or dielectric right in the middle of it. To the left of the insulator, the $E$ is the original $E$. Inside the insulator, there is an induced field and thus the net field is less than the original. What happen to the $E$ to the right of the insulator? Will it be the same as the original? If it does, then the $E$ will have to increase back? Why?

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Hi lowmow. Welcome to Physics.SE. Here, we use an unique TeX markup called MathJax, same as Math.SE. The markup is very much helpful in understanding equations, etc. Please have a look here for an introductory, or atleast have a look at our FAQ for an overview. For now, I'll help revising your post. – Waffle's Crazy Peanut Jan 20 '13 at 10:39

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I have a thought that you're mentioning about the polarization of dielectric in the presence of external field (say $E_0$). When you place some dielectric slab in an electric field, it gets polarized (i.e) surface charges are induced on both sides of the dielectric.

This constitutes an induced electric field (say $E_i$) in the dielectric in the exactly opposite direction to that of the applied one. You're right that the resultant electric field (effective one, say $E$) is less than the applied field inside the dielectric ($E=E_0-E_i$). This $E$ is nothing but the electric field acting through the dielectric of permittivity $\epsilon$. Roughly, $$E=\frac{q}{4\pi\epsilon r^2}, E_0=\frac{q}{4\pi\epsilon_0 r^2}$$ $$\frac{E_0}{E}=\frac{\epsilon}{\epsilon_0}=\epsilon_r$$

It should be noted that the electric field depends only upon the permittivity of the medium. So, its magnitude is low, when in dielectric medium (since $\epsilon_r>1$) and relatively high when in air medium ($\epsilon_r=1$). So, Yes. $E$ has to increase.

Have a look at this table in wiki. The electric field increases when it exits a material in the table and enters a material above it somewhere in the table.

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