Question

In Coulomb's law, the value of the constant $k$ is $\frac{1}{4{\mathrm{\pi \epsilon }}_0}$ . Here ${\epsilon }_0$ is the vacuum permittivity. But what actually is permittivity here?

Answer 1

Permittivity in Coulomb's Law:

1. According to Coulomb's law, the force of interaction between two stationary point charges is proportional to the product of the charges, inversely proportional to the square of the distance between them and it acts along the line joining the charges.
2. Let $\overrightarrow{r_1}$ and $\overrightarrow{r_2}$ be the position vectors of two point charges $q_1$ and $q_2$ with respect to some coordinate system.
3. Let $\overrightarrow{r_{21}}=\overrightarrow{r_2}-\overrightarrow{r_1}$ be the vector from $q_1$ to $q_2$, $r_{21}=\left|\overrightarrow{r_{21}}\right|$ be the distance between $q_1$ and $q_2$ and ${\hat{r}}_{21}=\frac{\overrightarrow{r_{21}}}{\left|\overrightarrow{r_{21}}\right|}$ be the unit vector in the direction from $q_1$ to $q_2$.
4. Then from Coulomb's law, the force on $q_2$ due to $q_1$ is $\overrightarrow{F}=k.\frac{q_1{q}_2}{{r_{21}}^2}{\hat{r}}_{21}$, where $k$ is a constant of proportionality.
5. The value of $k$ depends on the nature of the surrounding medium and the system of units used.
6. In vacuum in SI units $F_{21}$ is in newtons $\left(N\right)$, $q_1$, $q_2$ in coulombs $\left(C\right)$, $r_{21}$ in meters and $k=\frac{1}{4{\mathrm{\pi \epsilon }}_0}$.
7. Here ${\epsilon }_0$ is called the permittivity of free space which is defined as the distributed capacitance in vacuum.
8. The measured value of ${\epsilon }_0$ is $8.85\times {10}^{-12}{C}^2/N-m^2$.
9. For simplified numerical calculations one can approximately use $\frac{1}{4{\mathrm{\pi \epsilon }}_0}=9\times {10}^{9}N-m^2/C^2$.