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The doctor in university was deriving a formula and I can't understand how it works

A sphere with charge Q The Sphere's radius is R, and we are trying to derive a formula for potential difference at a point B at r distance from the center of the $$ K_e = \frac{1}{4\pi\epsilon_0}$$ He supposed that there's a point at infinity as $V_\infty$ is zero, so we can write

$$V_b - V_\infty = V_2 = -\int_i^f E•dl$$ $$ V_b = -\int_\infty^B \frac{K_eQ}{r^2}•dl $$ $$ V_b=-\int_\infty^B \frac{K_eQ}{r^2} dl \cos\theta$$ $$ V_b = \int_\infty^B \frac{K_eQ}{r^2}dl$$

this is clear for me, but when he said

our variable is r and our integral variable is dl, substitute dl with - dr

he said, imagine a distance coming infinity to point B, that's our definate integral limits right? and we don't know infinity so we must change the limits to something we know,

So if the vector Infinity B- is dl , and the line from the center of the sphere to point B is r, we can change the limits such that instead of saying we will walk 15 meters from a 20 meters line, we say we have 5 meters left and he explained it but tbh I didn't understand that, then he changed the integral to

$$ V_b = -\int_\infty^r \frac{K_eQ}{r^2}dr $$

he also said that if dl decreases, dr increases so we can say that dl = - dr now, I have two questions

  • how did he replace a vector dl with a length -dr ?
  • why is dl = - dr ?
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In this case forget about any angles and change of variable.

$$V_{\rm final} - V_{\rm initial} = V_{\rm r} - V_\infty =V_{\rm r} = -\int_\infty^r \vec E\cdot d\vec r$$

where $\vec E = \dfrac{K_eQ}{r^2}\, \hat r$ and $d\vec r = dr \,\hat r\Rightarrow \vec E\cdot d\vec r = \dfrac{K_eQ}{r^2}\, \hat r \cdot dr \,\hat r=\dfrac{K_eQ}{r^2}\, dr $

$$ V_{\rm r} = -\int_\infty^r\frac{K_eQ}{r^2}\cdot dr = \frac{K_eQ}{r}$$


The sign of $dr$ is dictated by the limits of integration and should not be assumed by you.

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  • $\begingroup$ it's not like that, the limits were from inf to to B and that's the path we will use but at the end, it was inf to r and that's the remaining path $\endgroup$ – Ahmed I. Elsayed Apr 20 at 9:13
  • $\begingroup$ @AhmedI.Elsayed My integration was from infinity to $r$ which is the point that you have labelled $B$?. $\endgroup$ – Farcher Apr 20 at 12:22
  • $\begingroup$ $B$ is a point $r$ is a distance from center of sphere $C$ to point $B$ $\endgroup$ – Ahmed I. Elsayed Apr 20 at 12:27
  • $\begingroup$ just read my answer and see the picture if you want to understand what I was talking about $\endgroup$ – Ahmed I. Elsayed Apr 20 at 12:30
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$dr$ is radially outward, but the integral is radially inward hence the difference in sign. As to why the signs are opposite, I think it comes down to convention.

We often associate a potential energy of zero when interacting particles are infinitely far apart. Since only changed in potential energy matter we can put the zero wherever we want. Zero at infinity makes some integrals easier. It intuitively implies non-interaction. We need to add energy to a bound system to separate the particles from their current position to infinity, so we consider the bound system to have a negative energy.

Changes in potential energy are the result of force doing work, the application of a Force through a distance, $dE=\vec{F}\cdot d\hat{l}$. The change in energy is the same in magnitude whether we put a negative sign or not. We slap on the negative sign to respect the convention that 0 is at infinity and a bound system has negative energy.

Because we do this with potential energy and the electric potential is PE per charge, the sign convention carries over to considerations of potential.

So to keep your signs straight when doing the integral, ask yourself what it must be to separate a bound system from their starting distance to infinity.

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  • $\begingroup$ Nah, he kept saying that if L increases, r decreases thus we can say L is -r and I lost him at the beginning $\endgroup$ – Ahmed I. Elsayed Apr 19 at 15:02
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In my opinion, what your prof did was to swap the direction.

You know that too.

I have no idea why he did that, but you get a negative potential in this scenario. But that does not really mean much because your professor has changed the coordinate system (keeping the origin at infinity).

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  • $\begingroup$ he didn't swap direction, he wanted to swap dl with dr to integrate so he got a relation between L & r that I don't understand $\endgroup$ – Ahmed I. Elsayed Apr 19 at 19:46
  • $\begingroup$ He took the sign of $cos \theta $ to be -1 and then "saved" the day by changing the sign again. He did not have to do that, as the dot product under the integral is positive. See Farcher's answer. There is nothing deep to understand. $\endgroup$ – nasu Apr 20 at 2:18
  • $\begingroup$ Yes. By multiplying something by negative, you are changing the direciton of one of the vectors. Remember that you were using a "dot product" $\vec E\cdot \vec dl$? If you multiply minus one to it, you are changing one of the vectors in this case. $\endgroup$ – KV18 Apr 20 at 7:28
  • $\begingroup$ I got it and answered it here, thanks for your time! $\endgroup$ – Ahmed I. Elsayed Apr 20 at 11:07
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This question was in my mind for 2 days and I posted it a day ago, I kept thinking till I got it 12 hours ago, so $\int$dl is the distance between $\infty$ and $B$ right? and $r$ is the distance between $B$ and the center of the sphere let's call it $C$ so the when distance $\infty$B increases, BC must decrease, as he said and I quoted above, when you walk a 20 meters line you can say you walked 15 meters or you've 5 meters left to walk and from this... it's like moving the point $B$ to the left or the right.

let's imagine the picture, what do we want? we want to make this integral solvable

a nice design that you don't wanna miss

$$V_b = -\int_\infty^B\frac{K_eQ}{r^2}∙dl$$ and we know that any length dx is equal to any length dy from the fact that we take the smallest possible part to integrate, $dx, dy \;$ are the smallest possible parts of a distance so we can replace them with each other right? so we can actually replace dl with dr, and from the fact that when $\int dl$ increases, $r$ decreases, we can deduce that the rate of change of both of them is equal but in negative signs right?

so

$$V_b = -\int_\infty^B\frac{K_eQ}{r^2}∙dl$$ $$V_b = -\int_\infty^r\frac{K_eQ}{r^2}dr$$

$cos(180)$ is $-1$, I substituted and cancelled the negative. also substituted $dl = -dr$

In the first one, we were going from $\infty$ to $B$ and that's our path

in the second one, he changed the reference, instead of saying we are going to the point $B$ from $\infty$ a distance $\infty$B, we say we have the distance $r$ left to reach $C$ coming from $\infty$ and it's actually the same distance but $B$ was a point which is $r$ distance from the center while the second one says we have $r$ distance left to reach $C$

in other words, instead of coming from $\infty$ to $B$, we come from $\infty$ to a point which is $r$ distance from $C$. just alert me if einstein is offering any jobs.

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