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Three point charges are arranged along the x-axis. charge q1 = +3 µc is at the origin, and charge q2 = "5 µc is at x = 0.200 m. charge q3 = "8 µc. where is q3 located if the net force on q1 is 7.00 n in the -x-direction?

Respuesta :

Force on charge q1 due to charge q2 is given as

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

[tex]F = \frac{9*10^9 * 3*10^{-6}*5 * 10^{-6}}{0.2^2}[/tex]

[tex]F = 3.375 N[/tex]

Now the net force on q1 is 7 N so the force due to q3 must be

[tex]F = 7 - 3.375 = 3.625 N[/tex]

[tex]F = \frac{kq_1q_3}{r^2}[/tex]

[tex]3.625 = \frac{9*10^9 * 3*10^{-6}*8*10^{-6}}{r^2}[/tex]

by solving above

[tex]r = 0.24 m[/tex]

so q3 is at distance 0.24 m from q1

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