- #1

kc8ukw

- 1

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## Homework Statement

Hi. I'm wondering how one determines the direction of

*dl*(vector) when calculating

*V*, the electric potential. So, suppose we want to calculate the electric potential caused by a point charge

*q*at a distance

*a*from the point charge, if we know the electric field of a point charge.

## Homework Equations

The textbook I have says to calculate the potential by

*V*

_{a}-

*V*

_{b}= integral from

*a*to

*b*of

*E**

*dl*.

## The Attempt at a Solution

So, I can let

*b*= infinity, to get

*V*

_{a}-

*V*

_{infinity}= integral from

*a*to

*infinity*of

*k**

*q*/

*r*

^{2}

*dr*, where I have said that

*dl*=

*dr**

*rhat*, since I am integrating from

*r*=

*a*to

*r*=

*infinity*. Doing the integral, and assuming

*V*

_{infinity}= 0, I get the correct answer,

*V*=

*k*

*q*/

*a*.

But suppose instead that I had integrated from

*infinity*to

*a*instead of the other way. So,

*V*

_{infinity}-

*V*

_{a}= integral from

*infinity*to

*a*of

*E**

*dl*.

Now, since I am coming from

*infinity*toward

*a*, I make

*dl*= -

*dr**

*rhat*. That introduces one negative sign. But the order of integration is also flipped, which introduces another. BUT, since I have

*V*

_{infinity}-

*V*

_{a}, that introduces a third negative, so I end up with

*V*= -

*k*

*q*/

*a*

the opposite of what I got before, and the wrong answer. Where did I go wrong? I'm guessing it must have been when I said that

*dl*= -

*dr**

*rhat*- but why isn't that correct? What sets the proper direction of

*dl*?