# Electrostatics: sign of the potential

• torito_verdejo
In summary: 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
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.

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
torito_verdejo said:
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
vanhees71 said:
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. :)

PeroK said:
##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

## 1. What is the sign of the potential in electrostatics?

The sign of the potential in electrostatics depends on the type of charge present. If the charge is positive, the potential will be positive. If the charge is negative, the potential will be negative.

## 2. How is the sign of the potential determined?

The sign of the potential is determined by the direction of the electric field. If the electric field is pointing towards a positive charge, the potential will be positive. If the electric field is pointing away from a positive charge, the potential will be negative.

## 3. Can the sign of the potential change?

Yes, the sign of the potential can change if the type of charge present changes. For example, if a negative charge is brought near a positive charge, the potential at that point will change from positive to negative.

## 4. What is the significance of the sign of the potential in electrostatics?

The sign of the potential is important because it determines the direction of the electric field. A positive potential indicates that the electric field is pointing towards the positive charge, while a negative potential indicates that the electric field is pointing away from the positive charge.

## 5. How does the sign of the potential affect the movement of charges?

The sign of the potential affects the movement of charges by determining the direction of the electric field. Charges will always move from areas of higher potential to areas of lower potential, following the direction of the electric field. This movement of charges is what creates electric current.

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