Questions based on derivation of electrical potential energy

1. Nov 13, 2015

gracy

Consider a system of two charges $q_1$ and $q_2$ separated by distance $r_1$.This configuration is associated with a potential energy $U_1$.When the separation is increased to $r_2$.Potential energy becomes $U_2$

$dW_E$=$\vec{F}$.$\vec{dr}$

$dW_E$=$Fdrcos0$=$\frac{1}{4πε0}\frac{q_1q_2}{r^2}$dr

⇒$W_E$=$\int_{r_1}^{r_2}$$\frac{1}{4πε0}$$\frac{q_1q_2}{r^2}$dr

$W_E$=$\frac{-q_1q_2}{4πε0}$$[\frac{1}{r_2}$-$\frac{1}{r_1}]$

By definition of potential energy ,

$U_2$-$U_1$=$-W_E$

⇒$U_2$-$U_1$=$\frac{q_1q_2}{4πε0}$$[\frac{1}{r_2}$-$\frac{1}{r_1}]$

Taking infinity as reference i.e $r_1$=∞ and $U_1$=0

⇒$U_2$-0=$\frac{q_1q_2}{4πε0}$$[\frac{1}{r_2}$-$\frac{1}{∞}]$

⇒$U_2$=$\frac{1}{4πε0}$$\frac{q_1q_2}{r^2}$

Taking $U_2$=$U$ and $r_2$=$r_1$

$U$=$\frac{1}{4πε0}$$\frac{q_1q_2}{r}$

I want to ask as we can see $r_2$is >$r_1$

and then problem assumes $r_1$ to be ∞ my question is what is $r2$ then?how can $r_2$ be greater than $r_1$?How can any number be greater than infinity?

2. Nov 13, 2015

jbriggs444

The only place where I see an assumption that r2 > r1 is in the limits of integration in $W_E =\int_{r_1}^{r_2} \frac{1}{4πε0}\frac{q_1q_2}{r^2}dr$

But that is not an assumption that r2 > r1. You can evaluate an integral from a larger endpoint to a smaller. The result is the negative of the same integral evaluated from smaller to larger.

3. Nov 13, 2015

Mister T

This is what we see looking at the figure, yes.

Yes, now they want you to imagine increasing $r_1$ so that it's not only bigger than $r_2$ but bigger than any value of $r$ you can imagine.

They replaced $r_2$ with $r$, meaning that instead of thinking of it as a fixed value for the separation distance, it's now thought of as a variable.

It can't. Very poorly authored example, if you ask me.

4. Nov 14, 2015

gracy

But they don't want to increase $r_1$ such that distance between them becomes greater than $r_2$ instead they want to increase $r_1$ such that distance between them becomes $r_2$

5. Nov 14, 2015

haruspex

Unless you have mistyped the final equation, it contains r, not r1. So it's a misprint: they mean replacing r2 with r.

6. Nov 14, 2015

gracy

Yes.

7. Nov 14, 2015

haruspex

It isn't. When you write an integral from a to b, there is no requirement for b to be greater than a:
$\int_a^bf(x)dx = -\int_b^af(x)dx$

8. Nov 14, 2015

gracy

Yes,you have told me once.Upper and lower limit just indicate final and initial positions but this line confused me

9. Nov 14, 2015

haruspex

Ok, but the equation is valid either way. With hindsight, the author might have preferred to write "changed to r2".

10. Nov 14, 2015

gracy

I can also refer to this http://aakashtestguru.com/document/askExpert/08-10-2015-18:26:171608102015DP.pdf [Broken]
for derivation of potential energy between the two charges ,right?

Last edited by a moderator: May 7, 2017
11. Nov 14, 2015

Mister T

Is this another typo then?

I suggest you present the entire example, as it was written by the author, along with a reference to the text you are quoting from.