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We can’t add these forces directly because they don’t point in the same direction: ${\overrightarrow{\text{F}}}_{12}$ points only in the − x -direction, while ${\overrightarrow{\text{F}}}_{13}$ points only in the + y -direction. The net force is obtained from applying the Pythagorean theorem to its x - and y -components:
where
and
We find that
at an angle of
that is, $58\text{\xb0}$ above the − x -axis, as shown in the diagram.
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 $58\text{\xb0}$ below the − x -axis.
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?
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
$12\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ ,
${\text{F}}_{31}=2.16\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\phantom{\rule{0.2em}{0ex}}\text{N}$ to the left,
${\text{F}}_{32}=8.63\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\phantom{\rule{0.2em}{0ex}}\text{N}$ to the right,
${\text{F}}_{\text{net}}=6.47\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\phantom{\rule{0.2em}{0ex}}\text{N}$ to the right;
b.
${\text{F}}_{31}=2.16\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\phantom{\rule{0.2em}{0ex}}\text{N}$ to the right,
${\text{F}}_{32}=9.59\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}\phantom{\rule{0.2em}{0ex}}\text{N}$ to the right,
${\text{F}}_{\text{net}}=3.12\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\phantom{\rule{0.2em}{0ex}}\text{N}$ to the right,
;
c.
${\overrightarrow{\text{F}}}_{31x}=\mathrm{-2.76}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}\phantom{\rule{0.2em}{0ex}}\text{N}\phantom{\rule{0.2em}{0ex}}\widehat{\text{i}}$ ,
${\overrightarrow{\text{F}}}_{31y}=\mathrm{-1.38}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}\phantom{\rule{0.2em}{0ex}}\text{N}\phantom{\rule{0.2em}{0ex}}\widehat{\text{j}}$ ,
${\overrightarrow{\text{F}}}_{32y}=\mathrm{-8.63}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\phantom{\rule{0.2em}{0ex}}\text{N}\phantom{\rule{0.2em}{0ex}}\widehat{\text{j}}$
$\begin{array}{cc}{\overrightarrow{\text{F}}}_{\text{net}}\hfill & =\mathrm{-2.76}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}\phantom{\rule{0.2em}{0ex}}\text{N}\phantom{\rule{0.2em}{0ex}}\widehat{\text{i}}-8.77\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\phantom{\rule{0.2em}{0ex}}\text{N}\phantom{\rule{0.2em}{0ex}}\widehat{\text{j}}\hfill \end{array}$
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