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}= (k_e|q_1q_2|)/(r_(12)^2)`

where
`k_e` is the constant of proportionality
`8.99*10^9 \ Nm^2C^(-2)`

Given that
`<span>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|= √(4^2+7^2+(-14)^2) = √(16+49+196) \\ \\ = √(261) =16.2 \ m`

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


`\overrightarrow{F}= (8.99*10^9\cdot2*10^(-6)\cdot4*10^(-6))/((16.2)^2) \\ \\ = (7.192*10^(-5))/(261) =2.76*10^(-7) \ N`