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I've got: $Q=\frac{Er^2}{k}$ how to check the units? I start with $\left[\frac{\text V}{\text m} \, \text m^2\right]$, tried replacing $[ \text V ]$ with $\left[ \frac{\text J}{\text C} \right]$, but it's not leading me to $[\text C]$.

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please note that you can use LateX on this website –  Michiel Mar 16 '13 at 15:55

2 Answers 2

up vote 5 down vote accepted

First of all note that $k$ is not dimensionless, it is $k = \frac{1}{4 \pi \varepsilon_0}$, and $\varepsilon_0$ has dimensions of $\frac{ \text C^2}{ \text {N m}^2}$. So you have already $\frac{ \text{V C}^2 \text m^2}{ \text {N m m} ^2}$. Also, volt can be expanded as $ \text V = \frac{ \text {N m}}{ \text C}$, so one gets $$ \frac{ \text C^2}{ \text {N m}^2} = \frac{ \text{V C}^2\text { m}^2}{ \text {N m m}^2} = \frac{ \text C^2}{ \text {N m}} \frac{ \text {N m}}{ \text C} = \text C$$ - exactly what you were looking for.

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When in doubt, break everything down into a set of indivisible units: kilograms, meters, seconds, and coulombs. So ask yourself what is a joule in terms of those units, and also consider that the constant $k$ has units. You can figure out what those units are by writing out Coulomb's law.

Edit: the units of $\epsilon_0$ can be found as follows. You know Coulomb's law.

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

Rearrange this to solve for $\epsilon_0$:

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

You know that $E$ is measured as force per unit charge. The other quantities are charges and lengths (or dimensionless constants). The result is

$$\epsilon_0 = \frac{\text{[charge]}^2}{\text{[force]} \times \text{[length]}^2}$$

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it seems k is just a number (4, pi and epsilons). Am I right? –  TomDavies92 Mar 16 '13 at 16:09
No, $k$ has units because $\epsilon_0$ has units. Do a dimensional analysis of Coulomb's law and you will see what these units are. –  Muphrid Mar 16 '13 at 16:14
k is F/m, right? –  TomDavies92 Mar 16 '13 at 16:27
That's $\epsilon_0$. –  Muphrid Mar 16 '13 at 17:26
ok, but the rest is just numbers (I mean 4, pi, epsilon r)? –  TomDavies92 Mar 16 '13 at 18:03

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