Question

The force of repulsion between to like charge table tennis balls is 8.2 X 10 ^-7 N if the charge on the two objects a 6.7 X10^-9 coulombs each what is the distance between the two charges?

Answer 1

Answer:

0.702 m

Explanation:

The magnitude of the electrostatic force between two charged objects is given by Coulomb's Law:

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

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the objects

And the force is:

- Attractive if the two charges have opposite signs (+-)

- Repulsive if the two charges have same sign (++ or --)

In this problem we have:

$F=8.2\cdot 10^{-7}N$ is the force between the two balls

$q_1 = q_2 = 6.7\cdot 10^{-9}C$ is the charge on each ball

Solving for r, we find the separation between the balls:

$r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(9\cdot 10^9)(6.7\cdot 10^{-9})^2}{8.2\cdot 10^{-7}}}=0.702 m$

Answer 2

Answer: 0.70

Explanation: