Electric Potential of Point Charge (sign problem)

Question

This question has been asked before, but the answers didn't clarify the problem for me, so I thought I might ask again.

It's really a simple question. Let's say we're calculating the electric potential of a point charge, so we have

$$V(r) - V(\infty) = -\int^r_{\infty} \vec{E}\cdot d\vec{\ell}$$

And the electric field is $$\vec{E} = \frac{q}{4\pi\epsilon r^2} \hat{r}.$$ What I'm not understanding is the choice of $d \vec{\ell}$. It seems to me that since we're coming from infinity to $r$, the path should be in the $-\hat{r}$ direction, so $d\vec{\ell} = -\hat{r} dr$. But this gives $$V(r) = -\int^r_{\infty} - \frac{q}{4\pi\epsilon_0 r'^2} dr' = -\frac{q}{4\pi\epsilon_0 r}$$ which is obviously the wrong answer. The problem is that apparently $d\ell = \hat{r} dr$. Can someone tell me why this is the case?

Answer 1

The use of the formula $\displaystyle V(r) - V(\infty) = -\int^r_{\infty} \vec{E}\cdot d\vec{\ell}$ requires a little care.

If there is an electric field $\vec E$ then the force on unit positive charge due to this electric field is $\vec E$.

The work done by the electric field in moving this charge $d \vec r$ is $\vec E \cdot d\vec r$ and the work done by an external force is $\left(- \vec E \right) \cdot d\vec r$.

There are two equivalent ways to define the potential at a point.

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The first definition is that the potential at a point is minus the work done by the electric field in taking unit positive charge from infinity to the point.

The work done by the electric field in taking unit positive charge from infinity to the point is $\displaystyle \int_\infty ^r \vec E \cdot d\vec r$

The potential at point $r$ is $V_r - V_\infty = V_r = \displaystyle \int_\infty ^r \vec E \cdot d\vec r$

Your uncertainty seems to come from what to do with $\vec E \cdot d\vec r$.

Since the path is radial there will be no angular dependence and so the dot product will be $E\; dr$ where $E$ is positive if the electric field is radially outwards or negative if the electric field is radially inwards.

The $dr$ is just a step length and its direction is taken care of by the limits of integration.

On working out $\displaystyle \int_{\infty}^{r} \dfrac {kq}{r^2}\;dr$ you get $- \dfrac {kq}{r}$ as the work done by the electric field.

If the path had been in the opposite direction ie from $r$ to $\infty$ the work done by the electric field would have been $+\dfrac {kq}{r}$.

This is exactly what you should expect.

With the first integral the force is outwards (positive direction) and the path is inwards (negative direction) and so the dot product will be negative.

With the second integral the force is outwards (positive direction) and the path is outwards (positive direction) and so the dot product will be positive.

To finish off remembering that the potential at a point is minus the work done by the electric field in taking unit positive charge from infinity to the point we get

$V_r – V_\infty = V_r = - \left ( - \dfrac {kq}{r } \right ) = + \dfrac {kq}{r } $

So in your original equation $\displaystyle V(r) - V(\infty) = -\int^r_{\infty} \vec{E}\cdot d\vec{\ell}$ is minus the work done by electric field.

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The second definition is that the potential at a point is the work done by an external force in taking unit positive charge from infinity to the point.

For the external force to move the unit positive charge this force must equal $-\vec E$ and so the work done by the external force in taking unit positive charge from infinity to the point is

$\displaystyle \int_\infty ^r \left(- \vec E \right)\cdot d\vec r$

which is the same as

$\displaystyle - \int_\infty ^r \vec E \cdot d\vec r = $ minus the work done by the electric field.

Doing the integration remembering that the external force is $-\vec E$ gives

$\displaystyle \int_\infty ^r \left(- \dfrac {kq}{r^2}\right) dr = + \dfrac {kq}{r} = V_r$ as before.

So rewriting you original equation slightly $\displaystyle V(r) - V(\infty) = \int^r_{\infty} \left(-\vec{E}\right)\cdot d\vec{\ell}$ is the work done by the external force.

Answer 2

While computing the final integral, we must ensure that the $r$ within the function $\vec E$ and the differential element $\mathrm dr$ must increase in the same direction, that is $\mathrm d \vec \ell$ and $\hat r$ should be aligned along the same direction, which is the reason why we should take $ \displaystyle \mathrm d\vec{\ell} = \hat{r} \, \mathrm dr $.

When we put the limits of the integral, we integrate from a higher quantity $\infty$ to a lower quantity $r$, giving the required negative sign that should appear because of "moving" in the opposite direction of the field (by "moving" i mean carrying out the process of integration). We do not have to take this into consideration and change the element of integration to account for something that is going to happen during integration itself!

As a final hint, try integrating the integral from $r$ to $\infty$. Then keeping $\mathrm d \vec \ell$ aligned with $\hat r$, the potential difference would come out to be negative as expected. While interchanging the limits back, there is no reason to change the differential element $\mathrm dr$!