Deriving expression for potential of a point charge

AI Thread Summary
The discussion focuses on deriving the expression for the electric potential of a positive point charge as a test charge is brought from infinity to a distance R. The initial calculation yields a negative potential, which raises questions about the interpretation of the negative sign. Clarifications are provided regarding the integration limits and the nature of the displacement vector, emphasizing that while the radial distance decreases, the magnitude of displacement remains positive. The distinction between the change in radial coordinate (dr) and the magnitude of displacement (ds) is crucial, with ds always being positive. Ultimately, understanding these concepts resolves the confusion regarding the potential's sign.
jimmyoung
Messages
3
Reaction score
0

Homework Statement


I am trying to derive an expression for the potential of a positive point charge by bringing in another positive test charge in from infinity to a point at a distance R from the point charge.

Homework Equations


$$V_f - V_i = - \int \vec E \cdot d \vec r \, dr$$

The Attempt at a Solution


Being that the test charge is coming in from infinity ##\vec E \cdot d \vec r = Edrcos180 = -Edr##
So then $$V_f - V_i = \int E \, dr$$
Then taking ##V_i = 0## at infinity
$$ V= \frac q {4\pi\epsilon_o}
\int_\infty^R \ {r^{-2}} \, dr =
\left. - \frac q {4\pi\epsilon_o r} \right|_\infty^R$$

This gives me ##V = - \frac q {4\pi\epsilon_o R}##

Which is almost right except I have that negative sign that says the potential is decreasing...
I thought since the potential is taken to be 0 at infinity and that the test charge is getting closer to the charge I would get a positive result saying the potential is increasing.

Could someone help explain why I got this negative result?
 
Physics news on Phys.org
Hello, Welcome to PF!
jimmyoung said:
Being that the test charge is coming in from infinity ##\vec E \cdot d \vec r = Edrcos180 = -Edr##
This should be written ##\vec E \cdot d \vec r = E|dr|cos180##. What happens if you now remove the absolute value sign on dr? Keep in mind that when integrating from ∞ to R, dr will be a negative quantity.

Alternately, before doing the dot product, reverse the limits on your integral so that you integrate from R to ∞. Then dr will be a positive quantity.
 
TSny said:
Hello, Welcome to PF!

This should be written ##\vec E \cdot d \vec r = E|dr|cos180##. What happens if you now remove the absolute value sign on dr? Keep in mind that when integrating from ∞ to R, dr will be a negative quantity.

Alternately, before doing the dot product, reverse the limits on your integral so that you integrate from R to ∞. Then dr will be a positive quantity.

Thank you for replying.
However, I am still not getting it.

Wouldn't dr be a negative quantity since it is pointing radially inward? So then its magnitude would still be positive? Why would I need to remove the absolute value sign?

Thanks.
 
I think part of the confusion is using ##d\vec{r}## for the displacement along a path and also using ##dr## for the change in the radial coordinate ##r## (which increases in the radially outward direction). Suppose we let ##d\vec{s}## be the displacement along an arbitrary path and imagine doing ##\int_a^b \vec{E}\cdot d\vec{s}## along some arbitrary path connecting points ##a## and ##b##. If we agree to let ##E## stand for the magnitude of ##\vec{E}## and ##ds## stand for the magnitude of ##d\vec{s}##, then the path integral may be written as ##\int_a^b E ds \cos\theta## .

Now suppose we apply this to a radial path that starts at infinity and goes to a radial distance ##R## from the origin. Let ##r## be the radial distance from the origin as used in the expression ##E = \frac{kq}{r^2}##. Then if we choose ##r## as the integration variable along the radial path from infinity to ##R##, ##dr## will be negative for each infinitesimal step along the path. So, the magnitude of the step, ##ds##, would be ##ds = -dr##. So, you would have ##\int_\infty^R E ds \cos\theta = -\int_\infty^R E dr \cos\theta##. In this integral, ##dr## does not represent the magnitude of the displacement (i.e., a positive quantity); rather, ##dr## represents the actual change in the radial distance ##r## when making a step along the path. So, ##dr## is a negative quantity while ##ds## is a positive quantity.
 
Last edited:
TSny said:
I think part of the confusion is using ##d\vec{r}## for the displacement along a path and also using ##dr## for the change in the radial coordinate ##r## (which increases in the radially outward direction). Suppose we let ##d\vec{s}## be the displacement along an arbitrary path and imagine doing ##\int_a^b \vec{E}\cdot d\vec{s}## along some arbitrary path connecting points ##a## and ##b##. If we agree to let ##E## stand for the magnitude of ##\vec{E}## and ##ds## stand for the magnitude of ##d\vec{s}##, then the path integral may be written as ##\int_a^b E ds \cos\theta## .

Now suppose we apply this to a radial path that starts at infinity and goes to a radial distance ##R## from the origin. Let ##r## be the radial distance from the origin as used in the expression ##E = \frac{kq}{r^2}##. Then if we choose ##r## as the integration variable along the radial path from infinity to ##R##, ##dr## will be negative for each infinitesimal step along the path. So, the magnitude of the step, ##ds##, would be ##ds = -dr##. So, you would have ##\int_\infty^R E ds \cos\theta = -\int_\infty^R E dr \cos\theta##. In this integral, ##dr## does not represent the magnitude of the displacement (i.e., a positive quantity); rather, ##dr## represents the actual change in the radial distance ##r## when making a step along the path. So, ##dr## is a negative quantity while ##ds## is a positive quantity.

ahh ok, yeah, that clears up a lot of my confusion. Thank you.
So with ##d\vec s,## being the displacement vector along the path could it be represented by ##dr\hat r## ?
In this case since it is in the decreasing radial direction ##d\vec s = -dr\hat r##
I may just be confusing these two again, but I am trying to find how to mathematically arrive at
ds = -dr.
 
jimmyoung said:
So with ##d\vec s,## being the displacement vector along the path could it be represented by ##dr\hat r## ?
Yes, this is correct for a radial path
In this case since it is in the decreasing radial direction ##d\vec s = -dr\hat r##
No. Whether you are going outward along a radial path or inward along a radial path, you would always have ##d\vec s = dr\hat r## (assuming that ##\hat r## is always defined in the radially outward direction). When going outward, ##dr## is positive. When going inward, ##dr## is negative.

I may just be confusing these two again, but I am trying to find how to mathematically arrive at ds = -dr.
##ds## represents the magnitude of the displacement. So, ##ds## is always positive. ##dr## represents the change in the radial coordinate ##r##. So, ##dr## is positive when moving radially outward, it is negative when moving radially inward. For moving outward or inward along a radial path, you could always write ##ds = |dr|##. When moving inward, ##|dr| = -dr##. So, when moving inward, ##ds = -dr##.
 
Back
Top