P19-080 – Electric Force between Two Point Charges: Coulomb’s Law

Electric Force Between Two Point Charges – Coulomb’s law

Coulomb’s Law:

Two point charges $q_1$ and $q_2$ separated by a distance $r$ exert an electric force on each other because of the electric field each one of the charges creates.

This force can be attractive or repulsive and its magnitude is given by Coulomb’s law

\boxed{F=\frac{k|q_1q_2|}{r^2}\ \ \ \ \ \ \ \ \ \ \ \ where\ \ \ \ k=\frac{1}{4\pi {\varepsilon }_0}}

Attractive force: point charges with opposite signs

Force on $q_1$:

The point charge $-q_2$ creates an electric field $\displaystyle{{\overrightarrow{E}}_2=\frac{kq_2}{r^2}\hat{r}}$ at the location of $q_1$. As a consequence, the point charge $q_1$ experiences an electric force equal to

\boxed{{\overrightarrow{F}}_{2\ on\ 1}=q_1{\overrightarrow{E}}_2=\frac{kq_1q_2}{r^2}\hat{r}}

Force on $-q_2$:

The point charge $q_1$ creates an electric field $\displaystyle{{\overrightarrow{E}}_1=\frac{kq_1}{r^2}\hat{r}}$ at the location of $-q_2$. As a consequence, the point charge $-q_2$ experiences an electric force equal to

{\overrightarrow{F}}_{1\ on\ 2}=-q_2{\overrightarrow{E}}_1=-\frac{kq_1q_2}{r^2}\hat{r}

Overall, the forces ${\overrightarrow{F}}_{1\ on\ 2}$ and ${\overrightarrow{F}}_{2\ on\ 1}$ are equal in magnitude and opposite in direction and we have

\boxed{{\overrightarrow{F}}_{2\ on\ 1}=-{\overrightarrow{F}}_{1\ on\ 2}\ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ F_{1\ on\ 2}=F_{2\ on\ 1}=\frac{k|q_1q_2|}{r^2}}

Both forces are said to form a Third Law Pair (opposite in direction and equal in magnitude).

Repulsive force: point charges with the same sign

Force on $q_1$:

The point charge $q_2$ creates an electric field $\displaystyle{{\overrightarrow{E}}_2=-\frac{kq_2}{r^2}\hat{r}}$ at the location of $q_1$. As a consequence, the point charge $q_1$ experiences an electric force equal to

\boxed{{\overrightarrow{F}}_{2\ on\ 1}=q_1{\overrightarrow{E}}_1=-\frac{kq_1q_2}{r^2}\hat{r}}

Force on $q_2$:

The point charge $q_1$ creates an electric field $\displaystyle{{\overrightarrow{E}}_1=\frac{kq_1}{r^2}\hat{r}}$ at the location of $q_2$. As a consequence, the point charge $q_2$ experiences an electric force equal to

{\overrightarrow{F}}_{1\ on\ 2}=q_2{\overrightarrow{E}}_1=\frac{kq_1q_2}{r^2}\hat{r}

Overall, the forces ${\overrightarrow{F}}_{1\ on\ 2}$ and ${\overrightarrow{F}}_{2\ on\ 1}$ are equal in magnitude and opposite in direction and we have

{\overrightarrow{F}}_{2\ on\ 1}=-{\overrightarrow{F}}_{1\ on\ 2}\ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ F_{1\ on\ 2}=F_{2\ on\ 1}=\frac{kq_1q_2}{r^2}

Both forces are said to form a Third Law Pair (opposite in direction and equal in magnitude).

Important note: Coulomb’s law only applies to point charges. You may not use it if you have any charge distribution other than point charges.