Electrostatics: sign of the potential

[torito_verdejo]

The final result will only differ in its sign, but this is crucial. Having a positively, radially oriented electric field $\textbf{E}$, I understand that the sign of the integral should be positive ($- (- A) = A$), but it is not! How and why is this the case? A line integral where the vector field is antiparallel to the displacement vector should be negative; is this negativity already taken into account by the conventional minus sign in $V(\textbf{r})=-\int_\Lambda \textbf{E}\cdot d\textbf{l}$?

Thank you very much.

 

[vanhees71]

It's a convention that conservative fields' potentials are defined with the - sign. It comes from classical mechanics, where you like to have the total (conserved) energy being the sum of the kinetic and potential energy.

As for your example. I guess you think about the Coulomb field. You know that
$$\vec{E}(\vec{x})=\frac{q}{4 \pi r^3} \vec{x} \quad \text{with} \quad r=|\vec{x}|.$$
To get the potential, simply use the definition and think a bit about symmetries. Since the charge $q$ is sitting at the origin everything is symmetric under rotations around the origin. Thus it makes sense to try the ansatz that the potential depends only on $r=|\vec{x}|$. Then you have
$$\vec{E}=-\vec{\nabla} \Phi(r)=-\frac{\vec{x}}{r} \Phi'(r).$$
Comparing the Coulomb law you see that
$$\Phi'(r)=-\frac{q}{4 \pi r^2} \; \Rightarrow \; \Phi(r)=\frac{q}{4 \pi r}.$$
I've chosen the arbitrary additive constant such that $\Phi(r) \rightarrow 0$ for $r \rightarrow \infty$ which is usually the most convenient choice.

Physicswise the sign is also clear. For a test charge $q'$ the potential energy is $U=q' \Phi(r)$, which is the work to be done to bring the test charge at the position $\vec{x}$ from infinity. If $q'>0$ and $q>0$ the Coulomb force is repulsive and you need to do work against this repulsive force to bring the test charge from infinity to a place with distance $r$ from the charge $q$, i.e., you put energy into the system, and that's why this energy must be positive, which is indeed the case $U=q q'/(4 \pi r)$. If $q q'<0$ the Coulomb force is attractive and you have to take energy out of the system, which is why in this case $U$ is negative.

 

[PeroK]

Summary:: Given the potential $V(\textbf{r})=-\int_\infty^r \textbf{E}\cdot d\textbf{l}$, and given that $\textbf{E}$ is antiparallel to $d\textbf{l}$, should I evaluate the positive integral $V(\textbf{r})=\int_\infty^r E\ dr$?

The final result will only differ in its sign, but this is crucial. Having a positively, radially oriented electric field $\textbf{E}$, I understand that the sign of the integral should be positive ($- (- A) = A$), but it is not! How and why is this the case? A line integral where the vector field is antiparallel to the displacement vector should be negative; is this negativity already taken into account by the conventional minus sign in $V(\textbf{r})=-\int_\Lambda \textbf{E}\cdot d\textbf{l}$?

Thank you very much.

$d\vec l$ is defined by the direction you are moving. Let's forget about coordinates for a moment:

If $\vec E$ is radially outward and $d \vec l$ is radially inward, then the inner product is negative. The integral itself must be negative (and the potential difference must be postive).

If we use $r$ as a coordinate and integrate from $0$ to $\infty$, then we get a positive integral, which is not right. But, if we integrate from $\infty$ to $0$, then by the properties of the integral this gives us a negative integral, as we expect.

In summary:

$\int_{\infty}^{P} \vec E \cdot \vec dl = \int_{\infty}^{r_0} E dr = - \int_{r_0}^{\infty} E dr$

Where $P$ is a point at radius $r_0$.

In summary, havng the bounds on $r$ go from $\infty$ to $0$ already factors in the negative sign. Alternatively, you can have $r$ go in its "normal" direction, from $0$ to $\infty$, then then you do need to have a negative factor because that is antiparallel to the direction of your line element.

 

[torito_verdejo]

It's a convention that conservative fields' potentials are defined with the - sign. It comes from classical mechanics, where you like to have the total (conserved) energy being the sum of the kinetic and potential energy.

As for your example. I guess you think about the Coulomb field. You know that
$$\vec{E}(\vec{x})=\frac{q}{4 \pi r^3} \vec{x} \quad \text{with} \quad r=|\vec{x}|.$$
To get the potential, simply use the definition and think a bit about symmetries. Since the charge $q$ is sitting at the origin everything is symmetric under rotations around the origin. Thus it makes sense to try the ansatz that the potential depends only on $r=|\vec{x}|$. Then you have
$$\vec{E}=-\vec{\nabla} \Phi(r)=-\frac{\vec{x}}{r} \Phi'(r).$$
Comparing the Coulomb law you see that
$$\Phi'(r)=-\frac{q}{4 \pi r^2} \; \Rightarrow \; \Phi(r)=\frac{q}{4 \pi r}.$$
I've chosen the arbitrary additive constant such that $\Phi(r) \rightarrow 0$ for $r \rightarrow \infty$ which is usually the most convenient choice.

Physicswise the sign is also clear. For a test charge $q'$ the potential energy is $U=q' \Phi(r)$, which is the work to be done to bring the test charge at the position $\vec{x}$ from infinity. If $q'>0$ and $q>0$ the Coulomb force is repulsive and you need to do work against this repulsive force to bring the test charge from infinity to a place with distance $r$ from the charge $q$, i.e., you put energy into the system, and that's why this energy must be positive, which is indeed the case $U=q q'/(4 \pi r)$. If $q q'<0$ the Coulomb force is attractive and you have to take energy out of the system, which is why in this case $U$ is negative.

Thank you for the detailed explanation. :)

 

[torito_verdejo]

$d\vec l$ is defined by the direction you are moving. Let's forget about coordinates for a moment:

If $\vec E$ is radially outward and $d \vec l$ is radially inward, then the inner product is negative. The integral itself must be negative (and the potential difference must be postive).

If we use $r$ as a coordinate and integrate from $0$ to $\infty$, then we get a positive integral, which is not right. But, if we integrate from $\infty$ to $0$, then by the properties of the integral this gives us a negative integral, as we expect.

In summary:

$\int_{\infty}^{P} \vec E \cdot \vec dl = \int_{\infty}^{r_0} E dr = - \int_{r_0}^{\infty} E dr$

Where $P$ is a point at radius $r_0$.

In summary, havng the bounds on $r$ go from $\infty$ to $0$ already factors in the negative sign. Alternatively, you can have $r$ go in its "normal" direction, from $0$ to $\infty$, then then you do need to have a negative factor because that is antiparallel to the direction of your line element.

Thank you, this solved my doubt.