Question

The unit of absolute permittivity is
A. Fm (Farad-meter)
B. $F{m}^{-1}$ (Farad/meter)
C. $F{m}^{-2}$ $\left(Farad/mete{r}^2\right)$
D. F (Farad)
E. None of these

Answer 1

Hint: As a first step, you could recall Coulomb’s law for electrostatic force. Now, as the options are in terms of SI unit of capacitance (Farad), we could recall its expression in terms of charge Q and potential difference V. Then, you could substitute V in terms of work done and Q and thus we will get an expression for $Q^2$ which can be substituted in ohm’s law and then by substituting the respective units for each of them you will get the unit of absolute permittivity.

Formula used:
Coulomb’s law,
$F={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q^2}{r^2}}$
Expression for capacitance,
$C={\displaystyle \frac{Q}{V}}$

Complete answer:
In the question, we are asked to find the unit of absolute permittivity from the given options.
In order to answer this, let us recall the expression for force given by Coulomb’s law. So, by coulomb’s law we have,
$F={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q^2}{r^2}}$
Where, F is the force between two charges Q kept at a distance of r between them and ${\epsilon }_0$ is the absolute permittivity or the permittivity of the free space or vacuum. Rearranging the above expression we get,
${\epsilon }_0={\displaystyle \frac{Q^2}{4\pi r^{2}F}}$ ……………………………………. (1)
Since the given options contain Farad in it, let us recall the expression for capacitance of which Farad is the SI unit. So, capacitance by definition is given by,
$C={\displaystyle \frac{Q}{V}}$ ……………………………………. (2)
Where, V is the potential difference across the capacitance and we know that V by definition is given by,
$V={\displaystyle \frac{W}{Q}}$
Now (2) could be given by,
$C={\displaystyle \frac{Q^2}{W}}$
$\Rightarrow Q^2=CW$ ………………………………………….. (3)
Substituting (3) in (1), we get,
${\epsilon }_0={\displaystyle \frac{CW}{4\pi r^{2}F}}$……………………………………. (4)
We know that work done by definition is the product of force and displacement and hence Newton-metre(Nm) will be its SI unit.
Now, we could substitute the respective SI units for all the quantities in (4) to get,
${\epsilon }_0={\displaystyle \frac{F\times Nm}{m^2\times N}}$
$\therefore {\epsilon }_0=F{m}^{-1}$
Therefore, the SI unit of absolute permittivity is found to be Farad per meter$\left(F{m}^{-1}\right)$.

Hence, option B is found to be the right answer.

Note:
Absolute permittivity could be defined as the measure of the electric polarizability of a dielectric. When we use the term ‘permittivity’, we mean the relative permittivity which is a dimensionless quantity. It is defined as the ratio of permittivity of a medium to that of the permittivity of vacuum.