Question

Two charges q1 and q2 are separated by a distance d and exert a force F on each other. What is the new force F ′ , if charge 1 is increased to q′1 = 5q1, charge 2 is decreased to q′2 = q2/2 , and the distance is decreased to d′ = d 2 ?

Answer 1

Answer:

The new force is 10 times the force F

Explanation:

Electric force between charged particles q1 and q2 at distance d is:

$F=k\frac{\mid q_{1}q_{2}\mid}{d^{2}}$ (1)

A new force between two different particles at a different distance is:

$F'=k\frac{\mid q_{1}'q_{2}'\mid}{d'^{2}}=k\frac{\mid 5q_{1}\frac{q_{2}}{2}\mid}{(\frac{d}{2})^{2}}=\frac{5}{\frac{2}{4}}k\frac{\mid q_{1}q_{2}\mid}{d^{2}}$

$F'=10k\frac{\mid q_{1}q_{2}\mid}{d^{2}}$

Note that on the right side of the equation the term $k\frac{\mid q_{1}q_{2}\mid}{d^{2}}=F$ on (1), so:

$F'=10F$