Answered

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A +1.4 nC charge exerts a repulsive force of 20.0 mN on a second charge
which is located a distance of 2.2 m away from it. What is the magnitude
and sign of the second charge?

Sagot :

Answer:

+ .00769 C

Explanation:

Whoever 'katie' is , removed my previous, innocuous answer for some reason....

The force is repulsive , so the charge is also POSITIVE

f = k q1 q2 / r^2  

   k = Coulombs constant = 8.98755 x 10^9  kg m^3 . (s^2 C^2)

   q 1 = 1.4 x 10^-9 C     f = 20 x 10^-3  N

    r = 2.2 m       q2 = ?

Sub in the values to get q 2 = + .00769 C

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