Question

Constants Part A wo charged dust particles exert a force of 7.5x102 N n each other What will be the force if they are moved so they are only one-eighth as far apart? Express your answer using two significant figures.

Answer 1

Answer:

New force, $F'=48\times 10^3\ N$

Explanation:

It is given that,

Force acting between two charged particles, $F=7.5\times 10^2\ N$

We need to find the force if they are moved so they are only one-eighth as far apart.

The force between two charged particles separated at a distance of r is given by :

$F=k\dfrac{q_1q_2}{r^2}$............(1)

If the charges are one-eighth as far apart then, r' =(1/8)r and new force is given by :

$F'=k\dfrac{q_1q_2}{(\dfrac{r}{8})^2}$..........(2)

Dividing equation (1) and (2) :

$\dfrac{F}{F'}=\dfrac{1}{64}$

$F'=7.5\times 10^2\ N\times 64$

F' = 48000 N

or

$F'=48\times 10^3\ N$

Hence, this is the required solution.