2
$\begingroup$

I have the question "Using the equation and information below, derive the units for the permittivity of free space , $\epsilon_{_0}$.

$V = \dfrac 1 {4 \pi \epsilon_{_0}} ~ \dfrac Q r$

$V$ is voltage in volts, $Q$ is charge in coulombs, $r$ is distance in metres, $4$ and $\pi$ are numbers and have no units."

From this information do I need to rearrange the formula to make e0 the subject and then workout the si units of each value?

When I checked the solutions the answer to this is $Q/Vm$ or $QV^{-1}m^{-1}$.

$\endgroup$
0

1 Answer 1

Reset to default
3
$\begingroup$

Yes, solving for $\epsilon_0$ first is a good idea:

$$V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r} \\ \epsilon_0 = \frac{1}{4\pi V}\frac{Q}{r} \\ [\epsilon_0] = \frac{1}{\textrm{V}}\frac{\textrm{C}}{\textrm{m}} = \textrm{C} \textrm{V}^{-1}\textrm{m}^{-1}$$

Read out, $\epsilon_0$ has units of "Coulombs per volt per meter".

$\endgroup$
3
  • $\begingroup$ I don't quite understand where C came from could you please elaborate ? Thanks. $\endgroup$
    – Dan
    Oct 19, 2016 at 0:31
  • 1
    $\begingroup$ Coulombs are the SI units of charge $Q$. $\endgroup$
    – Bobbie D
    Oct 19, 2016 at 0:32
  • 1
    $\begingroup$ $~[Q]=\mathrm C~,~ [V]=\mathrm V~,~ [r]=\mathrm m~$ so $[Q/(Vr)]=\mathrm{C V^{-1}m^{-1}}$. Notice: Bobbie uses distinct fonts for units and variables . $\endgroup$
    – Graham Kemp
    Oct 19, 2016 at 0:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .