# Electric charge and the distance

The strength of an electric field is: $E = 200\ \mathrm{N/C}$

The potential (of the test charge) is: $V = 600\ \mathrm{V}$

$\epsilon_r=1$

I need to calculate the distance between this point and the charge, the size of the charge. I think that the distance between this point and the charge is $E/V\implies E=V/m$, so it's $3$. To calculate the size of the charge, I used:

$$E=\frac{kQ}{r^2},$$ So,

$$Q=\frac{E r^2}{k},$$

but I'm not sure what $k$ and $\epsilon_r$ are.

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Did you try to look it up in a textbook?! –  Michiel Mar 16 '13 at 13:42
yes, I tried... –  TomDavies92 Mar 16 '13 at 14:09

$k=\frac1{4\pi\epsilon_0}=98.9\times 10^9 \:\mathrm{N m^2 C^{-2}}$

This is Coulomb's constant, part of Coulomb's law: $$\vec F=\frac{kq_1q_2\hat{r}}{r^2}=\frac{q_1q_2\hat{r}}{4\pi\epsilon_0r^2}$$

($\epsilon_0$ is the permittivity of free space, and is another constant)

In a dielectric medium, you replace $\epsilon_0$ with $\epsilon_r\epsilon_0$(the permittivity in that medium) in this equation, since the induced charges decrease the force. In case of a vacuum (and to an approximation, air), $\epsilon_r=1$

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Thank you very much for the explanation! :) –  TomDavies92 Mar 16 '13 at 14:13