# Solving an equation involving Coulomb's law

• logearav
However, in the second solution, you incorrectly factor the equation to get the solution. In order to find the solution, you need to factor the equation.

## Homework Statement

Hi Members,
I have solved an equation involving Coulomb's law in two ways and have posted the images of the same. I get two types of equations. Please help me to find where i gone wrong?

## The Attempt at a Solution

#### Attachments

• problem.pdf
186.6 KB · Views: 199
• problem1.pdf
360.5 KB · Views: 202
I looked at your second solution where you get a quadratic equation (you almost have the solution), but you apparently don't know how to factor it to get the solution. It is an interesting problem where the instructor combined the algebra of a quadratic expression with the physics. This solution of your quadratic equation is quite simple: It factors ## (Q_1-(8.0 \cdot E-6))(Q_1+(2.0 \cdot E-6))=0 ## This gives two possible solutions for ## Q1 ##. One is positive and one is negative. When you solve for ## Q_2 ## you will see that the results are such that Q1 and Q2 are a pair of charges of two (different) positive numbers, or a pair of negative charges with the same values (with the absolute values interchanged). Please proceed to solve the equation that I factored for you, and get the results. If you have any additional questions on this one, I'd be happy to assist. ... editing...the image is a little blurred, but in your final line the first term should read ## Q_1^2 ##. (It looks like you may have written ## Q_1^3 ##, but I can't see it real clearly to tell for sure.)

Last edited:
• logearav
Thanks Mr. Charles Link. In the first solution i get the q12 -6*10-6q1 -16*10-12 = 0
and for second one i get q12 - 6*10-6q1-16*106 = 0.
My question is why i get different powers for the term involving -16.

logearav said:
Thanks Mr. Charles Link. In the first solution i get the q12 -6*10-6q1 -16*10-12 = 0
and for second one i get q12 - 6*10-6q1-16*106 = 0.
My question is why i get different powers for the term involving -16.