Question

A 7.25 μC and a -2.00 μC charge are placed 20 cm apart. Where can a third charge be placed so that it experiences no net force. Express your answer in cm.

The answer is 22.12 from the online answer key, but how do you set this up?

Answer 1

one charge is attracting and the other is repelling

the 3rd charge must be farther from the larger charge and not between the two charges

remember inverse square law

7.25 / (d + 20)² = 2.00 / d²

7.25 d² = 2 d² + 80 d + 800

5.25 d² - 80 d - 800 = 0

use quadratic formula for solution

Answer 2

Thank you!