# Deriving units using equation

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 $$QV^{-1}m^{-1}$$.

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".
• Coulombs are the SI units of charge $Q$. Oct 19, 2016 at 0:32
• $~[Q]=\mathrm C~,~ [V]=\mathrm V~,~ [r]=\mathrm m~$ so $[Q/(Vr)]=\mathrm{C V^{-1}m^{-1}}$. Notice: Bobbie uses distinct fonts for units and variables . Oct 19, 2016 at 0:41