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Kisachek [45]
3 years ago
12

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 ?
Mathematics
1 answer:
marin [14]3 years ago
3 0
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, <span>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</span>
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