# Homework Help: Potential Difference between two positive point charges

1. Apr 18, 2013

### flannabhra

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

Two equal positive charges Q are fixed on the x-axis, one at +a and the other at -a.
(a) The electric field E at the Origin O
(b) The electric potential V at the origin O

2. Relevant equations

E=-dV/dr --> V=kQ/r

3. The attempt at a solution

VNet = V0 + V1

I got V0 = -kQ/r (because the vector r is negative)
..and V1 = kQ/r
V = 0

Is this correct?

2. Apr 18, 2013

### Dick

No, the potential of a charge q located at x=a is kq/|x-a|, notice the absolute value.

3. Apr 18, 2013

### JC Palmer

Hi Flannabhra!

a.) What kind of quantity is the electric potential, V?
b.) Describe to me the significance of 'r'.

4. Apr 18, 2013

### SammyS

Staff Emeritus
That's not correct.

First of all, what was your answer for part (a) and how did you get it.

For you attempt at part (b):
What does the variable, r, represent?

The operation of division is not defined for a vector --- that is to say, division by a vector is not defined.

Electric potential is a scalar quantity.​

5. Apr 18, 2013

### flannabhra

I got E = 0 for part (a) but i now feel that it is wrong based on your responses. The variable r represents the distance from the point charge to the Origin.

6. Apr 18, 2013

### SammyS

Staff Emeritus
That is the correct answer for part (a), but do you know why it is?

Distance is a non-negative quantity, as pointed out by Dick. So r can't change the sign of the potential.

7. Apr 18, 2013

### flannabhra

Well a logical explanation sufficed for me(the Electric Fields would be of equal magnitude in opposite directions), but I would like to understand the explanation of it based on formulas.

On the matter of r, I was under the impression that you could take the direction that r is pointing from the reference point and use that as the direction for r. I.e. the equation for E would be:

E = kQ/r3 * r

- in which r is the vector r and the r in the denominator is raised to the third power so as to keep the equation true.

8. Apr 18, 2013

### SammyS

Staff Emeritus
Yes. That answers the question regarding part (a) very nicely.

9. Apr 18, 2013

### flannabhra

I don't understand this.. What is x in this situation if r is the distance from the point Q to the origin?

10. Apr 18, 2013

### flannabhra

Hi!

a) I believe that in this situation, V is scalar.
b) 'r' is the distance from the point Q to the origin

11. Apr 18, 2013

### SammyS

Staff Emeritus
For the charge at x = -a, r = ai , where i is the unit vector in the positive x direction.

For the charge at x = a, r = -ai .

In each case the distance is a, and the vector, r, points away from the charge.

12. Apr 18, 2013

### flannabhra

So since the Electric Potential is scalar, is V = 2kQ/a correct?

Thank you so much for your help SammyS!!!

13. Apr 18, 2013

### SammyS

Staff Emeritus
That's correct.

14. Aug 21, 2017

### amirgul2007

My qs is what is the potential difference between two similar charges?

15. Aug 22, 2017

### haruspex

You'll need to be more precise. The potential at a point charge is theoretically infinite (+ or -), point charges being purely theoretical things.