Question

Two electric charges repel each other with a force of 0.0460 N when they are 1.20 a cm apart. The force between them when the charges are 2.40 cm apart is about

A) 0.1840 N
B) 0.0230 N
C) 0.0920 N
D) 0.0651 N
E) 0.0115 N​

Answer 1

Hi there!

Comparing the two distances:

2.40 cm is DOUBLE the distance of 1.20 cm.

According to the equation:

$F_E = \frac{kq_1q_2}{r^2}$

An inverse-square relationship exists between the electric force and distance.

Thus, if you DOUBLE the distance:

$F_E' = \frac{kq_1q_2}{(2r^2)} = \frac{1}{4}\frac{kq_1q_2}{r^2}$

The resulting electric force will be a QUARTER of the original, so:

0.0460 ÷ 4 = 0.0115 N