Combined Electric field for two charges using electric potential

[gralla55]

Homework Statement

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.

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 appreciate any help on this one!

 

[SammyS]

Homework Statement

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.

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 appreciate any help on this one!

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

 

[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?

 

[SammyS]

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?

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

 

[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?

 

[SammyS]

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?

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

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

 

[gralla55]

You are correct! Thanks alot!

 

[SammyS]

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