Deriving expression for potential of a point charge

[jimmyoung]

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?

 

[TSny]

Hello, Welcome to PF!

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.

 

[jimmyoung]

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.

 

[TSny]

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.

 

[jimmyoung]

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.

 

[TSny]

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, $ds = -dr$.

 

[kuruman]

You may wish to check an alternate way of handling line integrals here
https://www.physicsforums.com/threads/direction-of-tension.907922/#post-5719074
Using unit vectors avoids confusion and helps a great deal when doing line integrals in 2 and 3 dimensions.