# Net force on Electric Charges

## Homework Statement

Consider a charge of +2.0 µC placed at the origin of an X-Y co-ordinate system and a charge of -4.0 µC placed 40.0 cm to the right. Where must a third charge be placed – between the charges, to the left of the origin, or beyond the second charge – to experience a net force of zero? Argue your case qualitatively without working out a solution. Consider both positive and negative charges.

## Homework Equations

$$F=k((q1*q2)/r^{2})$$

## The Attempt at a Solution

Wouldn't you have to place the third one directly in the middle of the two charges? because you will then have the left side of the third one being positive and the right side of the third one being negative, therefore canceling out to zero, and the opposite would apply for a negative third charge...

But i'm not sure that i am correct with my reasoning...

## Answers and Replies

Delphi51
Homework Helper
Say the 3rd charge is positive. If you put the 3rd charge between the first two, the first charge (A) will push it to the right and the second charge (B) will attract it to the right. The two forces to the right do not cancel - they add.

Think about the directions of the two forces in the other two possible positions and you'll find the correct answer.

then have the left side of the third one being positive and the right side of the third one being negative
It sounds like you are thinking of induced charge separation in the 3rd charge. It actually will be the other way round with some positive charge being attracted to the negative charge (B) on the right and repelled by (A) on the left. So the right side will be more positive due to induction. This will partly cancel out because the left side will be attracted to A more strongly than the right side will be repelled by A while the opposite is true for its interaction with B. But the bottom line on induction is that we don't know how it compares in strength with the basic force due to the charge on C. It will be easily overwhelmed by the net charge on C unless that is very small (C is almost neutral). If it is neutral, the force due to induced charge separation will still be to the right in your proposed solution because the positive charge on the right side will be more strongly attracted to the larger B charge than the left side is to the smaller A charge.

Say the 3rd charge is positive. If you put the 3rd charge between the first two, the first charge (A) will push it to the right and the second charge (B) will attract it to the right. The two forces to the right do not cancel - they add.

Think about the directions of the two forces in the other two possible positions and you'll find the correct answer.

It sounds like you are thinking of induced charge separation in the 3rd charge. It actually will be the other way round with some positive charge being attracted to the negative charge (B) on the right and repelled by (A) on the left. So the right side will be more positive due to induction. This will partly cancel out because the left side will be attracted to A more strongly than the right side will be repelled by A while the opposite is true for its interaction with B. But the bottom line on induction is that we don't know how it compares in strength with the basic force due to the charge on C. It will be easily overwhelmed by the net charge on C unless that is very small (C is almost neutral). If it is neutral, the force due to induced charge separation will still be to the right in your proposed solution because the positive charge on the right side will be more strongly attracted to the larger B charge than the left side is to the smaller A charge.
Hey, i'm somewhat confused now... Don't you just add the forces to find the net acting on the third?

Delphi51
Homework Helper
Yes. Forget about the induced charge separation.

I'm still not sure about this question.. because it says to not work the problem out.. would I just use repel/attract theory...