# 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.

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