Question

How do you find r in Coulombs Law?

Answer 1

Coulombs Law:

According to Coulomb's Law, the electrostatic force between two stationary electrically charged point particles is inversely proportional to the square of the distance between them and directly proportional to the sum of the charges.

Calculate $r$ in coulombs law

$\begin{array}{rcl}& \Rightarrow & \mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1\times {\mathrm{q}}_2}{{\mathrm{r}}^2}\\ & \Rightarrow & {\mathrm{r}}^2=\frac{\mathrm{F}\times 4{\mathrm{\pi \epsilon }}_0}{{\mathrm{q}}_1\times {\mathrm{q}}_2}\\ & \Rightarrow & \mathrm{r}=\sqrt{\frac{\mathrm{F}\times 4{\mathrm{\pi \epsilon }}_0}{{\mathrm{q}}_1\times {\mathrm{q}}_2}}\end{array}$

Where,$F=$ Electrostatic force in $newton$, $q1andq2=$ Magnitude of charged in$coulomb$, and $r=$ distance between the charges in $meters$.

Note:

$\Rightarrow \frac{1}{4{\mathrm{\pi \epsilon }}_0}=k=9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$

1. Along the line connecting the point, charges are where the force is applied. The force is appealing if the signs of the two charges are different. The force is repulsive if the charges have the same sign (positive or negative).
2. Here, ${\epsilon }_0$ is the open space's permittivity. It is a physical constant that expresses a vacuum's ability to support electric fields. If there is a medium other than air between the charges, its permittivity is often indicated by $\epsilon$. Since force and permittivity have an inverse relationship, adding a medium between the charges will alter the force's strength.