# Combined Electric field for two charges using electric potential

1. Mar 14, 2013

### gralla55

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

Two equal and positive point charges Q1 and Q2 are a distance d from each other on the x-axis. Find the electric potential at the point P which lies a distance z on the z-axis from their centre. Then find the Electric field at the point, and then find the force F on a charge q placed there.

3. The attempt at a solution

Since the charges are equal, and the distance to the point is the same for both of them, I figured I could just find the potential due to one of the charges, and multiply it by 2. When it comes to the second part, I just took the derivative of the potential and multiplied it times a unit vector. Then I tought the unit vector should point in the z-direction, because the horizontal components of the field will cancel each other out.

But thinking about it, I don't think that will give the correct answer, as some part of the field as stated will disappear... I've attached a picture of my described attempt at a solution.

Would appriciate any help on this one!

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2. Mar 14, 2013

### SammyS

Staff Emeritus
Along the z axis, the potential due to the two charges depends only upon z, so $\displaystyle \ E_z=-\,\frac{dV}{dz}\ .$ Also, the E field along the z-axis has only a z-component.

Last edited: Mar 14, 2013
3. Mar 14, 2013

### gralla55

Thank you for your reply! Was the potential I found at the point P the correct one? And if so, to find E at that point I just differentiate that expression with regards to z?

4. Mar 14, 2013

### SammyS

Staff Emeritus
There was a typo, leaving out a negative sign. I have since fixed it.

5. Mar 15, 2013

### gralla55

So I took the negative partial derivative of my potential with respect to z, and got the result:

E = -Qz / (2pi*epsilon(z^2 + (d/2)^2)^(3/2)) times the unit vector k.

But this does not look right... Both charges are positive, so the electric field should point upwards. Was there something wrong with my potential function?

6. Mar 15, 2013

### SammyS

Staff Emeritus
You should get an additional negative sign from taking the derivative.

$\displaystyle \frac{d}{du}\left(u^2+a\right)^{-1/2}=2u(-1/2)\left(u^2+a\right)^{-3/2}$

7. Mar 15, 2013

### gralla55

You are correct! Thanks alot!

8. Mar 16, 2013

### SammyS

Staff Emeritus
... and again, all is right with the world .