Electric potential due to two point charges

[aftershock]

Homework Statement

Two point charges (opposite signs, equal in magnitude) are a distance d apart. Point P is a distance z above both charges and horizontally equidistant. Find the electric potential at point P.

Homework Equations

Kq/r

The Attempt at a Solution

It's my understanding that direction does not matter with potential so r is the same value for both. We can add the potentials together to get total potential. Plugging in q for the first charge and -q for the second gives 0.

I know that's not right, that would imply the electric field is zero which is obviously incorrect.

What am I doing wrong?

 

[TSny]

What's your reason for saying that zero potential implies zero electric field?

 

[aftershock]

What's your reason for saying that zero potential implies zero electric field?

E = -∇V

gradient of zero is zero

 

[TSny]

Not necessarily. For example, sin(x) = 0 at x = 0. But the derivative of sin(x) is not zero at x = 0. Likewise, if the potential happens to be zero at some point, it doesn't mean that E has to be zero at that point.

But, if V = 0 (or any other constant) throughout some region of space, then E would be zero in that region.

 

[aftershock]

Not necessarily. For example, sin(x) = 0 at x = 0. But the derivative of sin(x) is not zero at x = 0. Likewise, if the potential happens to be zero at some point, it doesn't mean that E has to be zero at that point.

But, if V = 0 (or any other constant) throughout some region of space, then E would be zero in that region.

Yeah, what I mean was that it would imply zero electric field only at point P. Is that incorrect?

EDIT: Never mind I just read the "if the potential happens to be zero at some point, it doesn't mean that E has to be zero at that point."So does that mean that zero potential is correct?

 

[TSny]

Yeah, what I mean was that it would imply zero electric field only at point P. Is that incorrect?

No. Suppose V as a function of position (x, y, z) happens to be V = -x - y² + 2z. Then note that V = 0 at (x,y,z) = (1, 1, 1). What would E be at that point?

 

[TSny]

So does that mean that zero potential is correct?

Yes.

 

[aftershock]

Yes.

I really appreciate your help but I'm a little confused.

The problem states "Compute E = -∇V , and compare your answer with Prob. 2.2a"

2.2a is the same problem but instead asks to calculate the electric field.I understand now (thanks to you) why I can't simply differentiate the value of potential at some specific point, but then how do I go about this problem?

 

[TSny]

Good question. You can see that V = 0 as long as you stay on the z axis. So, if you are at point P and move up or down along the z axis, V remains constant. So you should be able to deduce what the derivative of V is with respect to z at point P. That will give you one of the components of the electric field.

To get the other components you need the derivative of V with respect to x and y at point P. This would require knowing how V varies as a function of x and y as you move parallel to the x and y axes from point P. Your calculation of V at point P does not give you that information. So, the question seems to me to be a bad question unless the question is just to get you to see that the one component of E that you can calculate is consistent with what you found for that component in the other problem.

 

[frogjg2003]

You should already know the electric field of a point charge. Just like you can add potentials, you can add fields. The total field is the sum of all the fields of the point charges. Remember, the electric field is a vector, with components in the x, y, and z directions. TSny has already explained that the z component of the electric field will be 0, but you don't necessarily need that when using the superposition method.