# Discrepancy in potential of two opposite charges

1. Aug 22, 2014

### ELB27

1. The problem statement, all variables and given/known data
Suppose that we changed the right-hand charge in the image attached to -q (instead of +q); what then is the potential at P? What field does that suggest? Compare your answer to Prob. 2.2 [answer written below] and explain carefully any discrepancy.

2. Relevant equations
The definition of potential: $$E = -\nabla V$$
The formula for the potential at a certain point due to a collection of point charges:
$$V = \frac{1}{4\pi \epsilon_0} \sum_{i=1}^{n} \frac{q_i}{r_i}$$ where r is the magnitude of the distance between the point and the charge.
$$\vec{E} = \frac{1}{4\pi \epsilon_0} \frac{qd}{\left(z^2 + \left(\frac{d}{2}\right)^2\right)^{3/2}}\hat{x}$$

3. The attempt at a solution
Well, I substitute the relevant quantities in the above formula and get $V = 0$ which implies $\vec{E} = 0$. But from Prob. 2.2 it is obvious that the field is non-zero and points in the x-direction. I'm not sure how to explain this - the only thing that comes to mind is that in previous problems the field was always in the z-direction. However, I'm not sure why it would cause problems because in the derivation of the above formula for $V$ the general formula for $\vec{E}$ in the $\hat{r}$ direction is used and therefore should cover all axis.
Thanks in advance for any help!

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2. Aug 22, 2014

### BvU

V=0 at P does NOT imply E = 0 ! You need $\nabla V$ to get E !

It is probably easier to look at the E vectors you found earlier on, change the sign of the one due to -q and then add them up ...

 and what about the $\hat x$ in the answer to 2.2 ?

3. Aug 22, 2014

### ELB27

Thank you for the reply. However, I'm not sure I understand you. If $V=0$ then $\vec{E} = -\nabla V = -\frac{\partial V}{\partial x} \hat{x} -\frac{\partial V}{\partial y} \hat{y} -\frac{\partial V}{\partial z} \hat{z} = 0$. I'm supposed to first calculate $V$ using the formula for it and then explain why it doesn't agree rather than getting from $\vec{E}$ to $V$. And the answer to 2.2 already takes into account the -q replacement.

4. Aug 22, 2014

### ehild

The potential is zero only along the z axis. V(0,0z)=0 but V(x,y,z)≠0 if x≠0 and y≠0.

The components of the electric fields are the negative partial derivatives of the potential function. Recall how a partial derivative is defined:

$$E_x(0,0,z )=-\partial V/\partial x = -\lim _{x{\rightarrow}{0} }\frac{V(x,0,z)-V(0,0,z)}{x}\neq{0}$$

But
$$E_z(0,0,z_P )=-\partial V/\partial z = -\lim _{z{\rightarrow}{z_P} }\frac{V(0,0,z)-V(0,0,z_P)}{z-z_P}={0}$$

ehild

5. Aug 22, 2014

### ELB27

So if I understand correctly: basically I just computed $V(0,0,z)$ and in order to find the general expression for $V$ I need to find the potential at a point $P=(x,y,z)$ and then to find $\vec{E}$ take $-\nabla V(x,y,z)$ and only then substitute the coordinates of the desired point? That means that a potential at a certain point does not necessarily measures the field at that point as I thought?
EDIT: If so, how to think about the potential? (i.e. what does it measure?)

Last edited: Aug 22, 2014
6. Aug 22, 2014

### Orodruin

Staff Emeritus
Exactly, the potential is not directly proportional to the field. In fact you can add a constant to any potential and get the same resulting field. They are related in such a way that the field is the gradient of the potential.

Think of the potential as you would think about potential energy (in fact, electric potential is just the potential energy of a test particle of unit charge - or the potential energy of a test particle with charge $q$ is $q V$, where $V$ is the electric potential). In the same way as the force is related to the gradient of the potential energy, the electric field (which gives the force on a charge $q$) is related to the gradient of the electric potential.

7. Aug 22, 2014

### ehild

The electric potential is not measure of the electric field intensity. By definition, the electric potential at a point P is equal to the work done by the electric field on a unit positive charge when the charge moves from P to the point where the potential is zero.
You can get the work done between two points by taking the difference of the potential values and multiply by the charge.

If you know the potential function, you can find the electric field as E=-∇V.
Concerning the problem, what does it mean for the electric field that V=0 along the z axis? Is it necessary any work to move a charge along the z axis?

ehild

8. Aug 22, 2014

### ELB27

Thank you both very much! I just computed the general potential and took it's gradient and the answer agrees with Prob. 2.2. So in the future, if I know that $\vec{E}$ points in a certain direction, I can simplify the calculation by finding $V$ only in this direction's axis (if it points in x-direction, I calculate $V$ at $(x,0,0)$ for example and then substitute the coordinates of the point at which I want to find $\vec{E}$)?
And regarding the work - the fact that $V(0,0,z)$ is zero just means that $\vec{E}$ is perpendicular to the z-axis (so it's work is zero)?

9. Aug 22, 2014

### BvU

We have to take your word for whatever is in the mysterious 2.2

In the future ... yes, but only if E points in the same direction everywhere. So that's not a big help, I'm afraid.

And yes, V(0,0,z) = constant is already enough to make the component Ez = 0

10. Aug 22, 2014

### ehild

Yes, if the displacement of the charge is along the z axis.

ehild

11. Aug 22, 2014

### ELB27

Thanks alot! I feel that I understand this concept much better now.

12. Aug 22, 2014