Potenial difference from electric field and line integral

Question

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.

Answer 1

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

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

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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.

Answer 2

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.

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