Calculate the total force on Q1

[JustAStudent]

Homework Statement

Q1<------>Q2<------>Q3

1. 
   In the above figure, the distance between Q1 and Q2 is equal to the distance between Q2 and Q3. That distance is R=1.5 m.

1. 
   Q1= 2.24x10-6 C, Q2=+Q1 and Q3=-Q1.

1. 
   Calculate the total force on Q1. Give your answer with a positive number for a force directed to the right, or a negative number for a force directed to the left.

Homework Equations

|F|=((k)(Q1)(Q2))/r^2

The Attempt at a Solution

|F13|=((k)(2.24E-6)(2.24E-6))/(1.5+1.5)^2
=0.0050176

Because Q1=positive and Q3=negative there is an attraction and therefore the value should be negative (to the left)

SO, =-0.0050176

|F12| =((k)(2.24E-6)(2.24E-6))/(1.5)^2
=0.0200704

Because Q1=positive and Q2=positive, there is a repulsion and therefor the value should be positive (to the right)

SO: 0.0200704

Add them together to get the net force: -0.0050176 +0.0200704=0.0150528

I have no idea what I am doing wrong

 

[Doc Al]

Because Q1=positive and Q3=negative there is an attraction and therefore the value should be negative (to the left)

If they attract then the force on Q1 from Q3 should be toward Q3 and thus to the right.

 

[JustAStudent]

If they attract then the force on Q1 from Q3 should be toward Q3 and thus to the right.

Okay, so I'm thinking of it incorrectly then. Force of Q3 ON Q1 means which way Q1 is going (attraction v repulsion) and not which way Q3 is going?

 

[Doc Al]

Okay, so I'm thinking of it incorrectly then. Force of Q3 ON Q1 means which way Q1 is going (attraction v repulsion) and not which way Q3 is going?

Correct. The attraction between Q1 and Q3 creates a force on Q1 to the right and a force on Q3 to the left. But all you care about here are the forces on Q1.