# Where Electric Potential is Zero

1. Jul 27, 2009

### exitwound

1. The problem statement, all variables and given/known data

Charge Q1 and charge -3Q are 1 meter apart. Consider only points on the axis between the charges. A.) Where is V=0? B.) Where is E=0?

2. Relevant equations

$$V=\frac{kQ}{r}$$

3. The attempt at a solution

A.) The positions where the electric potential would be zero would be to add the potentials due to the individual charges and set equal to Zero, correct?

$$V=\frac{kQ_1}{r_1}-\frac{k3Q}{r_2}=0$$
$$r_1+r_2=1$$
$$r_1=1-r_2$$

$$V=\frac{kQ_1}{1-r_2}-\frac{k3Q}{r_2}=0$$

After this, the algebra gets very messy and I can't figure out how to approach it to isolate r_2.

B.) Haven't attempted this yet. Hints on how to from part A to B?

2. Jul 27, 2009

### cepheid

Staff Emeritus
Have you considered putting the terms on opposite sides of the equation and cross multiplying?

3. Jul 27, 2009

### exitwound

Yeah. I think I'm having problems defining where the positions are on the x-axis.

Q1---------V=0---------Q2

X1---------X0----------X2

Therefore:

$$V=\frac{kQ_1}{r_1}-\frac{k3Q}{r_2}=0$$

$$V=\frac{kQ_1}{x_0-x_1}-\frac{k3Q}{x_2-x_0}=0$$

$$\frac{kQ_1}{x_0-x_1}=\frac{k3Q}{x_2-x_0}$$

$$(kQ_1)(x_2-x_0)=(k3Q)(x_0-x_1)$$

If this is right, I should be able to solve for x_0.

But how can I be sure there are no other points of V=0 on the axis? This equation doesn't deliver multiple solutions.

4. Jul 27, 2009

### cepheid

Staff Emeritus
You do realize that for convenience, you can decide that one of the charges is sitting at the origin (x = 0)? If so, then the other one will be at x = +1 m or x = -1 m depending upon which one you pick.

Here is the answer to that question:

5. Jul 27, 2009

### exitwound

The answer in the back of the book gives two solutions, ones 25cm to the right of the Q charge, and one 50cm to the left of the Q charge.

6. Jul 27, 2009

### cepheid

Staff Emeritus
How is that consistent with these instructions?

Also, it's pretty clear that your equation for V will never lead to a quadratic. Are you sure these answers aren't for E?

7. Jul 27, 2009

### exitwound

I'm sorry. I copied the problem incorrectly. It should say, "Consider points only on the axis".

There are 2 solutions for V=0 and one solution for E=0 (40cm to the left of Q).

8. Jul 27, 2009

### Cyosis

Just pick a point a distance r to the left of charge Q and solve for r.

9. Jul 27, 2009

### exitwound

Yes but my question is how can I be sure there's a V=0 point there? Do I have to check all three regions every time a problem like this comes up? Is that the easiest way to do it? Is there a better way to determine where I should check prior to doing the calculations?

10. Jul 27, 2009

### cepheid

Staff Emeritus
I don't know which charge is where (i.e. which one is on the left and which one is on the right), but it doesn't matter. Consider the leftmost charge to be at x = 0. The distance of a point from this charge is given by:

r1 = |x|

The rightmost charge is at x = 1. The distance of a point from this charge is given by:

r2 = |x - 1|

Do you understand why I have put in the absolute value signs? The potential (or electric field) depends only on the distance from the charge, and not on in which direction along the axis that distance lies. Considering that different cases that the absolute value signs give rise to will lead you to your multiple solutions.

11. Jul 27, 2009

### cepheid

Staff Emeritus
For the electric field, yes there is. Let's say that the + charge is on the right and the - charge is on the left. Then you know that in between the two charges, their effects will NEVER cancel, but will in fact augment each other. (A positive test charge placed in between the two charges will be attracted towards the - charge, which corresponds to motion to the left, and will be repulsed from the + charge, which ALSO corresponds to motion to the left.). Therefore, you know that looking in between the charges is useless. The effects are going to cancel somewhere to the left of the - charge, where its attraction is balanced out by the repulsion from the + charge.

I guess this is what Cyosis was saying. For the potentials, I'm not sure if there is a better way than what I outlined in my previous post.

12. Jul 27, 2009

### exitwound

Okay, so checking the three regions is the only viable way to attack the potential problem. The Electric field I understood where to look as explained in the last post. But the potential I'm still trying to understand.