I am really confused about the relation of potential difference and the electric field.

The relation between potential and electric field is $$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} \, .$$

Now let's look at a solid sphere uniformly charged with $q$ and let's find the potential difference $V_{ab}$ while $b\to\infty$.

The electric field of a solid sphere is in the $\vec{r}$ direction hence if we are going from $b$ to $a$ we are going against the electric field (because $b>a$).

With that we get $$V_{ab} = -\int_{b}^a{\vec{E}\cdot d\vec{r}} = -\int_{b}^a E \, dr \, \cos(\pi) = \int_{b\to\infty}^a \frac{kq}{r^2} \, dr = -\frac{kq}{a} < 0 \, .$$

The result makes no sense because it means that $V(a) < V(b)$ which means that the electric field of the solid sphere is in the $-\vec{r}$ direction, which is wrong.

What I'm doing wrong? I saw many solutions of exercises which are using this relation and it seems that each one of them is just solving the integral without considering the dot product.

  • $\begingroup$ The quantity $\vec{E}\cdot d\vec{r}$ is negative. You're moving from $b$ to $a$, against the electric field, so d$\vec{r}$ and $\vec{E}$ are pointing in opposite directions. $\endgroup$ Commented Oct 23, 2018 at 19:24
  • $\begingroup$ i don't understand your answer, this is exactly what i wrote. $\endgroup$
    – Guy
    Commented Oct 23, 2018 at 19:47

2 Answers 2


The relation between potential and electric field is $$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} \, .$$

is correct.

Now what are $\vec E$ and $d\vec r$ in terms of the unit vector $\hat r$?

$\vec E = E\,\hat r$ and $d\vec r = dr\,\hat r$ where $E$ and $dr$ are components of those two vectors in the $\hat r$ direction and they can be either positive or negative.

This gives you $\vec E \cdot d\vec r = E\,\hat r \cdot dr\,\hat r = E\,dr$

$$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} = -\int_{b}^aE\,dr$$

In your example the electric field is radially outwards and so $E$ will be a positive quantity.

It is the sign of $dr$ which causes the confusion, so is $dr$ positive or negative?

The sign of $dr$ is entirely determined by the limits of integration.

You do not need to assign a sign to $dr$ all you need to do is state the limits of integration which will then determine the direction of travel.

In other words if $a>b$ then whilst doing the integration $dr$ is positive but if $a<b$ then whilst doing the integration $dr$ is negative.

Going back to your example with $b>a$ you have $E$ is positive and $dr$ is negative so $\vec E \cdot d\vec r = E\,dr$ will be a negative quantity.
Thus $-E\,dr$ will be a positive quantity and it will give you that $V_{\rm a} - V_{\rm b}$ will be a positive quantity leading to the expected result that $V_{\rm a} > V_{\rm b}$

So finishing off the example

$$V_{\rm a} - 0 = V_{\rm a}=-\int_{\infty}^a{\vec{E}\cdot\vec{dr}} = -\int_{\infty}^a\frac{kq}{r^2}\,dr=+\frac{kq}{a}$$

You have stated that $\vec{E}\cdot d\vec{r} = E \, dr \, \cos(\pi)$

How did you get this relationship?

You said that $\vec E = E \,\hat i$ and that $d\vec r = dr \left( -\hat i\right)$.
In other words you have looked at the problem, noticed that the direction of travel will be in the $-\hat i$ direction and so assigned a positive value to $dr$.

What you cannot do now is use limits of integration such that the direction of travel will result in $dr$ being negative.

Doing it your way you proceed as follows:

$$V_{\rm a} - 0 = V_{\rm a}=-\int_{\infty}^a{\vec{E}\cdot\vec{dr}} = -\int^{\infty}_a\frac{kq}{r^2}\,\left(-dr\right)=+\frac{kq}{a}$$

Notice that the limits of integration reflect the fact that $dr$ is positive.

  • $\begingroup$ Excellent explanation! $\endgroup$
    – fich
    Commented May 3, 2021 at 9:01

There is actually no need for the factor $\cos(\pi)$ even if $\vec{dr}$ is directed against $\vec{E}$.

It would be needed if we wanted to write the integral in terms of $|E|$ and $|dr|$: $$ \int_b^a \vec{E}\cdot\vec{dr} = \int_b^a |\vec{E}| \, |\vec{dr}|\,\cos(\pi)~~~. $$ But it is more useful to write it in terms of magnitudes $E,dr$, as ordinary definite integral.

The correct way to express the line integral as ordinary definite integral is

$$ \int_b^a \vec{E}\cdot\vec{dr} = \int_b^a E \, dr~~~. $$

This is because although $E,dr$ are not vectors, they still have signs, and this those signs are such that the product $Edr$ has the correct sign.

$E>0,dr<0 \implies Edr<0.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.