IMAT 2019 Question 60

Practice Question Solving
1

Q60) Two charged particles P and Q are 0.10 m apart. The charge on P is 1.50 × 10–7 C and the
charge on Q is 1.50 × 10–7 C. Particle P experiences an electrostatic force of magnitude F
because it is near to the charge on particle Q.
The distance between the two particles is increased to 0.20 m. The charge on P is increased to
4.50 × 10–7 C and the charge on Q is increased to 6.00 × 10–7 C.
What is the magnitude of the force that particle P experiences now?

A)F/4
B)6F
C)3F/4
D)3F

2

Hi!
Here we need to use the Coulomb’s Law:
F=kQ1xQ2/r^2

In the first scenario
F1=k x (1.5x10^-7)^2/(0.10)^2=2.25k x 10^-12.

In the second scenario
F2=k X (4.50x10^-7) x (6.00x10^-7)/(0.20)^2=6.75k x 10^-12.

Divide F2 by F1 and we get 3. We are told the force has a magnitude of F so the answer is 3F. The answer is thus D.
Hope it helps:)

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