The charge q3 must be placed to the right of q2
let the charge q3 be x m to the right of q2
force on q3 by q1 = force on q3 by q2
kq1q3 / r1^2 = kq2q3 / r2^2
47*23 / (3.5 +x)^2 = 21*23 / x^2
47x^2 = 21(3.5 + x)^2
47x^2 = 21x^2 + 257.25 + 147x
26x^2 - 147x - 257.25 = 0
x = 147 + sqrt(147^2 + 4*26*257.25) / 52
= 7.05 m
so the charge q3 should be placed at (3.5 + 7.05) = 10.55 m
I need help on this, I already have 2 attempts left! Point charges q, 7ad 2-21...
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I need the answers to both of these please because I think I’m
wrong ?
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Please explain how you get x. Thanks!
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21-80
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