Electric field and potential

Question

Suppose we consider a point dipole i.e. an ideal dipole. At the origin the potential:

$$ \phi(\mathbf r) = \frac{1}{4\pi\epsilon_0} \frac{\mathbf p\cdot\hat{\mathbf r}}{r^2}$$

is undefined due to the $1/r^2$ term in the potential, but electric field:

$$ \mathbf E = \frac{1}{4\pi\epsilon_0}\frac{3(\mathbf p\cdot\hat{\mathbf r})\hat{\mathbf r} - \mathbf p}{r^3} - \frac{1}{3\epsilon_0}\mathbf p\delta^3(\mathbf r) $$

remains well defined. How can we calculate electric field at the origin by equation $\mathbf E=-\nabla\phi$?

Hi Manoj. Do you mean an ideal dipole, i.e. a point dipole, or an arrangement of two closely separated charges? — John Rennie, Mar 21 '19 at 06:53
If you're thinking of a point dipole see the derivation of the field in this question — John Rennie, Mar 21 '19 at 06:56
In that case look at the question I linked as that goes through the derivation. The electric field has a $\delta^3(r)$ term that keeps it well defined at $r=0$. — John Rennie, Mar 21 '19 at 07:10

score 3 · Accepted Answer · answered Mar 21 '19 at 12:09

Both potential and electric field are singular at the point dipole, there is not much difference in "well-definedness" - there are no real values there. Electric field above is a singular distribution, not a function. It has no real value at the dipole. Only integrals of such distribution have value.

There are some fringe cases where electric field has value but isn't calculable as gradient of potential either. For example, electric field of charged sphere at any point on the sphere has definite value (half of electric field just above the sphere), but since potential is not differentiable at that point, electric field at such points is not a gradient of electric potential.

score 2 · Answer 2 · answered Mar 21 '19 at 06:49

2

Just because a variable is zero at a certain position does not mean that the rate of change of that variable at the position is zero.

When a ball is thrown up, at its greatest height the ball’s vertical velocity is zero but its rate of change of velocity, the acceleration of free fall, is not zero.

In your case the potential at a distance $x$ from a charge $+q$ is $\phi=\frac{k(+q)}{x}+\frac{k(-q)}{d-x}$ where $d$ is the separation of the charges.

When $x=\frac d2$ the potential $\phi$ is zero but the electric field (minus the gradient of the potential) is non zero.

answered Mar 21 '19 at 06:49

Farcher

95,680

1

The OP is asking about a point dipole not two separated charges. – John Rennie Mar 21 '19 at 07:11
@JohnRennie When I read the question it did not occur to me that the OP was asking about an ideal dipole. – Farcher Mar 21 '19 at 07:15

Electric field and potential

2 Answers2