Two charges, Q1 and Q2, experience an attractive force between them, F1. If the distance between the charges is tripled, the sign of Q2 is changed, and the magnitude of the charge on Q1 increases by a factor of nine, what will the new force be between the charges?
A) F1
B) -F1
C) -3F1
D) 9F1
Click for Explanation
The force between the charges is given by Coulomb’s law:
F = kQ1q2 / r2
We may start by noting that changing one of the charges from positive to negative (or vice versa), as the question indicates when it says the valence on one of the charges is flipped, would immediately change the force from attractive to repulsive, or vice versa. Thus, we may immediately narrow our choices down to (B) and (C).
Next, if the distance between the charges in tripled and one charge in increased by a factor of 9, we have the following:
F = k(9Q1)q2 / (3r)2
F = 9kQ1q2 / (9)(r2)
F = kQ1q2 / r2
Thus we are left with a magnitue of force that is unchanged. Therefore, the answer is (B). Same magnitude, opposite direction.
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