# What happens to the negative charge

1. May 11, 2014

### negation

A negative charge -q is placed at the origin.
Two positive charge of +4q are on both sides of the negative charge at distance d.
what happens?

2. May 11, 2014

If they are placed simultaneously,then the negative charge will stay still,because the net force is 0 and the initial state of the negative charge was "rest".

3. May 11, 2014

### negation

Why is the net force = zero?

4. May 11, 2014

### Staff: Mentor

What do you think the net force is?

5. May 11, 2014

### negation

I have no idea. Can you shed some light?

6. May 11, 2014

### ZapperZ

Staff Emeritus
Are you not able to calculate the force between two point charges using the straightforward Coulomb's law? Shouldn't this be something you already know BEFORE dealing with a problem such as "...radial electric field outside coaxial cable..."?

Zz.

7. May 11, 2014

### Staff: Mentor

Use Coulomb's law to find the force that each positive charge exerts on the negative charge. But you should be able to use symmetry to figure out the net force on that center charge. No calculations needed.

Then you can consider the forces on the positive charges to see what happens to them.

8. May 11, 2014

### negation

Spot on. I'm looking for an interpretation without calculation.
I do know the calculation and the formula but in many cases, the question wants me to know what is happening without any empirical calculation.

9. May 11, 2014

### Staff: Mentor

Well, what's stopping you?

How do the forces exerted by each positive charge compare?

10. May 11, 2014

### negation

Both positive charge exerts an outward electric field in a radial direction. The electric field from the positive charge to the left of the origin points in the positive x-direction. The electric field from the positive charge to the right of the origin point to the negative x-direction.

Since the negative charge is at an equal distance from both positive charge, the electric force experienced by the negative charge from the left is +F and -F from the right.
Fnet = -F+F=0?

11. May 11, 2014

Correct!

12. May 11, 2014

### Staff: Mentor

Yes. Simple as that!

13. May 11, 2014

### negation

Thanks guys!

14. May 11, 2014

### jbriggs444

There is another argument that may be appealing:

The layout is symmetric about the plane midway between the positive charges. If you were able to predict motion in one direction then you would also be able to predict motion in the symmetric direction. The only case in which this does not result in a contradiction is when the two predicted motions are identical -- i.e. when there is no motion at all.

Last edited: May 11, 2014