Coulomb's Law

The law states that "Unlike charges attract, like charges repel."

Charges attract or repel due to a force they exerted on each other. These forces attract or repel each other. To solve for the force of attraction or repulsion between the charges, the amount of charge and the distance between them must be measured. The equation for force is expressed as

$$F=\frac {kq_1q_2}{r^2}$$

where $$F$$ is the force between the charges, $$q_1$$ and $$q_2$$ are the charges attracting or repelling each other, $$r$$  is the distance between the charges, and $$k$$ is the electrostatic constant with a value of $$k=9\times 10^9\;N\cdot m^2/C^2$$.

Example 1.

An electron and a proton are separated by by a distance of $$2.0\times 10^{-5}\;m$$. solve for the force between the proton and electron.

Given:

$$r=2.0\times 10^{-5}\;m\\ k=9.0\times 10^9\;N\cdot m^2/C^2\\ \text{charge of proton}=1.6\times 10^{-19}\\ \text{charge of electron}=-1.6\times 10^{-19}$$

Solution:

To solve for the force between the charges, we use the formula $$F=\frac {kq_1q_2}{r^2}$$.

$$F=\frac {(9.0\times10^9\;N\cdot m^2/C^2)(1.6\times 10^{-19}\;C)(-1.6\times 10^{-19}\;C)}{(2.0\times 10^{-5}\;m)^2}\\ \;\;\,=\frac{-2.304 \times 10^{-28}}{4.0 \times 10^{-10}\;m^2}\\ \;\;\,=-5.76 \times 10^{-19}\;N$$

Example 2.

Three protons and 2 electrons are separated by a force of $$-3.60 \times 10^{-3}\;N$$.

Given:

$$F=-3.60 \times 10^{-3}\;N\\ k=9.0 \times 10^9\;N\cdot m^2/C^2\\ \text{charge of proton}=1.6\times 10^{-19}\\ \text{charge of electron}=-1.6\times 10^{-19}$$

Solution 1.

Calculate the total charge of proton and electron.

$$q_p=3(1.6 \times 10^{19}\;C)=4.8 \times 10^{-19}\;C$$

$$q_{\text{e}}=2(-1.6 \times 10^{19}\;C)=-3.2 \times 10^{-19}\;C$$

Solution 2.

Derive the formula for the distance from the equation $$F=\frac {kq_1q_2}{r^2}$$.

$$F=\frac {kq_1q_2}{r^2} \implies r^2=\frac{kq_1q_2}{F} \implies r=\sqrt {\frac{kq_1q_2}{F}}$$

Therefore, the equation $$r=\sqrt {\frac{kq_1q_2}{F}}$$ will be used to solve the distance between the charges.

$$r=\sqrt{\frac {(9.0\times10^9\;N\cdot m^2/C^2)(4.8\times 10^{-19}\;C)(-3.2\times 10^{-19}\;C)}{-3.6\times 10^{-3}\;N}}\\ \;\;=\sqrt{\frac{-1.38 \times 10^{-27}\;N \cdot m^2}{-3.6\times 10^{-3}\;N}}\\ \;\;=\sqrt{3.84 \times 10^{-11}\;m^2}\\ \;\;=6.2 \times 10^{-6}\;m$$

The distance between the charges is $$6.2 \times 10^{-6}\;m$$.