# Electrostatics: sign of the potential

• torito_verdejo
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.

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2022 Award
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.

• torito_verdejo
Homework Helper
Gold Member
2022 Award
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
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.

• PeroK