Two point charges $q_1 = 4\ \mu$C and $q_2 = -2\ \mu$C are placed 0.3 m apart. Find the force between them.
Two point charges $q_1 = 4\ \mu$C and $q_2 = -2\ \mu$C are placed 0.3 m apart. Find the force between them.
1 Answer
DMDivya Mehta
✓ Accepted
· 2mo ago
▲ 11
$$F = \frac{kq_1q_2}{r^2} = \frac{9\times10^9 \times 4\times10^{-6} \times 2\times10^{-6}}{(0.3)^2}$$
$$F = \frac{9\times10^9 \times 8\times10^{-12}}{0.09} = \frac{72\times10^{-3}}{0.09} = 0.8 N (attractive)$$
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Discussion (2)
SP
I solved it a slightly different way and got the same answer, good sign.
IM
Can someone explain why we ignore the other root here?