Question

Point charges q1 = 2.0 µc and q2 = 4.0 µc are located at →r 1 = (4.0 i ^ − 2.0 j ^ + 5.0k ^ )m and →r 2 = (8.0 i ^ + 5.0 j ^ − 9.0k ^ )m . what is the force of q2 on q1 ?

Answer 1

Given two point charges $q_1$ and $q_2$ which are a distance $r_{12}$ apart.

The force $\overrightarrow{F}$ on one of the charges is proportional to the magnitude of its own charge and inversely proportional to the square of the distance between them.
i.e.
$\overrightarrow{F}= \frac{k_e|q_1q_2|}{r_{12}^2}$

where $k_e$ is the constant of proportionality $8.99\times10^9 \ Nm^2C^{-2}$

Given that $q_1=2.0\mu C$ and $q_2=4.0\mu C$

Also, given that $r_ 1 = (4.0i-2.0 j+5.0k)$ m and $r_2 = (8.0 i+ 5.0 j-9.0)$ m, then

$|r_{12}|=|r_2-r_1| \\ \\ =|(8.0 i+ 5.0 j-9.0)-(4.0i-2.0 j+5.0k)| \\ \\ =|4.0i+7.0j-14k|= \sqrt{4^2+7^2+(-14)^2} = \sqrt{16+49+196} \\ \\ = \sqrt{261} =16.2 \ m$

Therefore, the force of $q_2$ on $q_1$ is given by

$\overrightarrow{F}= \frac{8.99\times10^9\cdot2\times10^{-6}\cdot4\times10^{-6}}{(16.2)^2} \\ \\ = \frac{7.192\times10^{-5}}{261} =2.76\times10^{-7} \ N$