Question

Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a factor of three?

Answer 1

Answer:

force becomes one - ninth

Explanation:

According to Coulomb's law in electrostatics, two charges can exert a force of attraction or repulsion on each other which is directly proportional to the product of two charges and inversely proportional to the square of distance between them.

Here both the charges remains same but the distance is variable.

So, we can say that

$F \alpha \frac{1}{d^{2}}$    .... (1)

Where d be the distance between the tow charges

As the distance between two charges increases by factor of three, let the new force be F'.

$F' \alpha \frac{1}{9d^{2}}$   .... (2)

Divide equation (2) by equation (1), we get

$\frac{F'}{F}=\frac{d^{2}}{9d^{2}}$

$F'=\frac{F}{9}$

Thus, the force becomes one - ninth times the initial force.