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1 vote
1 vote
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?

User Rdxdkr
by
3.1k points

1 Answer

17 votes
17 votes

Answer:

+ .00769 C

Step-by-step 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

User Derryl Thomas
by
3.3k points