Question

how the electrostatic force between two charges changes when the distance between those two charges is tripled.

A. electrostatic force remains same
B. electrostatic force increases by factor of 9
C. electrostatic force decreases by

Answer 1

Answer:

answer is option B

hope it helps u!!

Answer 2

Answer: None. The force decreases by a factor of 9.

Explanation:

Force between any two charges q1, and q2 separated at a distance r is given by:

$\bf F = k \dfrac{q_1 q_2}{r^2}$

where, k = 9 × 10^9 Nm²/C² (in vacuum)

Let the initial distance be R here and the charges are qa and qb.

So,

$F = k \dfrac{q_a q_b}{R^2}$ __(1)

Now,

distance = 3R

Therefore,

$F^{\prime} = k \dfrac{q_a q_b}{(3R)^2}$

= $F^{\prime} = \dfrac{1}{9} (k \dfrac{q_a q_b}{R^2})$

= $F^{\prime} = \dfrac{1}{9} F$.

So, the force will decrease by a factor of 9.

More:

• In the equation, $\bf k = \dfrac{1}{4 \pi {\epsilon}_{0} }$. Where $\bf \epsilon_{0}$ is the permittivity in free space.
• Electric potential at a distance r on a charge Q is given by kQ/r.