# 5.3 Coulomb's law  (Page 4/11)

 Page 4 / 11

We can’t add these forces directly because they don’t point in the same direction: ${\stackrel{\to }{\text{F}}}_{12}$ points only in the − x -direction, while ${\stackrel{\to }{\text{F}}}_{13}$ points only in the + y -direction. The net force is obtained from applying the Pythagorean theorem to its x - and y -components:

$F=\sqrt{{F}_{x}^{2}+{F}_{y}^{2}}$

where

$\begin{array}{cc}\hfill {F}_{x}& =\text{−}{F}_{23}=-\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{{q}_{2}{q}_{3}}{{r}_{23}^{2}}\hfill \\ & =\text{−}\left(8.99\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\frac{\text{N}·{\text{m}}^{2}}{{\text{C}}^{2}}\right)\frac{\left(4.806\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(8.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{{\left(4.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{2}}\hfill \\ & =\text{−}2.16\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−14}}\phantom{\rule{0.2em}{0ex}}\text{N}\hfill \end{array}$

and

$\begin{array}{cc}\hfill {F}_{y}& ={F}_{21}=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{{q}_{2}{q}_{1}}{{r}_{21}^{2}}\hfill \\ & =\left(8.99\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\frac{\text{N}·{\text{m}}^{2}}{{\text{C}}^{2}}\right)\frac{\left(4.806\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(3.204\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\phantom{\rule{0.2em}{0ex}}\text{C}\right)}{{\left(2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{2}}\hfill \\ & =3.46\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\phantom{\rule{0.2em}{0ex}}\text{N}.\hfill \end{array}$

We find that

$F=\sqrt{{F}_{x}^{2}+{F}_{y}^{2}}=4.08\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\phantom{\rule{0.2em}{0ex}}\text{N}$

at an angle of

$\varphi ={\text{tan}}^{-1}\left(\frac{{F}_{y}}{{F}_{x}}\right)={\text{tan}}^{-1}\left(\frac{3.46\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\phantom{\rule{0.2em}{0ex}}\text{N}}{-2.16\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\phantom{\rule{0.2em}{0ex}}\text{N}}\right)=-58\text{°},$

that is, $58\text{°}$ above the − x -axis, as shown in the diagram.

## Significance

Notice that when we substituted the numerical values of the charges, we did not include the negative sign of either ${q}_{2}$ or ${q}_{3}$ . Recall that negative signs on vector quantities indicate a reversal of direction of the vector in question. But for electric forces, the direction of the force is determined by the types (signs) of both interacting charges; we determine the force directions by considering whether the signs of the two charges are the same or are opposite. If you also include negative signs from negative charges when you substitute numbers, you run the risk of mathematically reversing the direction of the force you are calculating. Thus, the safest thing to do is to calculate just the magnitude of the force, using the absolute values of the charges, and determine the directions physically.

It’s also worth noting that the only new concept in this example is how to calculate the electric forces; everything else (getting the net force from its components, breaking the forces into their components, finding the direction of the net force) is the same as force problems you have done earlier.

Check Your Understanding What would be different if ${q}_{1}$ were negative?

The net force would point $58\text{°}$ below the − x -axis.

## Summary

• Coulomb’s law gives the magnitude of the force between point charges. It is
${\stackrel{\to }{\text{F}}}_{12}\left(r\right)=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{{q}_{1}{q}_{2}}{{r}_{12}^{2}}{\stackrel{^}{\text{r}}}_{12}$

where ${q}_{2}$ and ${q}_{2}$ are two point charges separated by a distance r . This Coulomb force is extremely basic, since most charges are due to point-like particles. It is responsible for all electrostatic effects and underlies most macroscopic forces.

## Conceptual questions

Would defining the charge on an electron to be positive have any effect on Coulomb’s law?

An atomic nucleus contains positively charged protons and uncharged neutrons. Since nuclei do stay together, what must we conclude about the forces between these nuclear particles?

The force holding the nucleus together must be greater than the electrostatic repulsive force on the protons.

Is the force between two fixed charges influenced by the presence of other charges?

## Problems

Two point particles with charges $\text{+3}\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ and $\text{+5}\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ are held in place by 3-N forces on each charge in appropriate directions. (a) Draw a free-body diagram for each particle. (b) Find the distance between the charges.

Two charges $\text{+3}\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ and $\text{+12}\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the net force on a −2-nC charge when placed at the following locations: (a) halfway between the two (b) half a meter to the left of the $\text{+3}\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ charge (c) half a meter above the $\text{+12}\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ charge in a direction perpendicular to the line joining the two fixed charges

a. charge 1 is $3\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ ; charge 2 is $12\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ , ${\text{F}}_{31}=2.16\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{N}$ to the left,
${\text{F}}_{32}=8.63\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{N}$ to the right,
${\text{F}}_{\text{net}}=6.47\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{N}$ to the right;
b. ${\text{F}}_{31}=2.16\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{N}$ to the right,
${\text{F}}_{32}=9.59\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}\text{N}$ to the right,
${\text{F}}_{\text{net}}=3.12\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{N}$ to the right,
;
c. ${\stackrel{\to }{\text{F}}}_{31x}=-2.76\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}\text{N}\phantom{\rule{0.2em}{0ex}}\stackrel{^}{\text{i}}$ ,
${\stackrel{\to }{\text{F}}}_{31y}=-1.38\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}\text{N}\phantom{\rule{0.2em}{0ex}}\stackrel{^}{\text{j}}$ ,
${\stackrel{\to }{\text{F}}}_{32y}=-8.63\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{N}\phantom{\rule{0.2em}{0ex}}\stackrel{^}{\text{j}}$
$\begin{array}{cc}{\stackrel{\to }{\text{F}}}_{\text{net}}\hfill & =-2.76\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}\text{N}\phantom{\rule{0.2em}{0ex}}\stackrel{^}{\text{i}}-8.77\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{N}\phantom{\rule{0.2em}{0ex}}\stackrel{^}{\text{j}}\hfill \end{array}$

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