Given two positively charged particles, of equal magnitude, separated by a distance “d”. What will happen to the force field bet

Question

4 years ago

15

ween the two particles when d is halved

1 answer:

Ulleksa [173] · Answer 1 · 2021-12-11T14:24:29+03:00

The force between the two particles will quadruple

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{d^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

d is the separation between the two charges

In this problem, let's call F the initial force between the two charges, when they are at a distance of d.

Later, the distance between the two particles is halved, so the new distance is:

$d'=\frac{d}{2}$

This means that the new electrostatic force will be:

$F'=k\frac{q_1 q_2}{(d/2)^2}=4(k\frac{q_1 q_2}{d^2})=4F$

Therefore, the force between the two particles will quadruple.

Learn more about electrostatic force:

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