# Work done in moving a 1C positive charge from one point to another

1. Dec 23, 2016

### Abdulwahab Hajar

1. The problem statement, all variables and given/known data
Determine the Work done by the electric field E = axX - ay2y in moving a unit positive charge from position p1 (-2,0,0) to position p2 (5,-1,3) the distances are in m

2. Relevant equations

3. The attempt at a solution
I'm not really experienced with forums therefore my attempted solution is an attached image.https://scontent-lht6-1.xx.fbcdn.net/v/t34.0-12/15683157_1201819119912120_1486498649_n.jpg?oh=0aad954d8ecb161d62c98aa8d1d685e7&oe=585F5529
is my displacement vector in my solution correct ? I did it as such because I have +ve displacement in the ax direction and -ve displacement in the ay direction.

2. Dec 23, 2016

### TSny

The general way to represent an infinitesimal displacement is $\vec{dl} = dx \, \hat{a}_x + dy \, \hat{a}_y + dz \, \hat{a}_z$. The quantities $dx$, $dy$, and $dz$ could be positive, negative, or zero depending on the direction of the displacement. When you integrate with respect to $y$ from 0 to -1, $dy$ will be negative. You should not write $\vec{dl} = dx \, \hat{a}_x - dy \, \hat{a}_y + dz \, \hat{a}_z$ with a negative sign for the y component. This would imply that the y component of displacement would be in the $+\hat{a}_y$ direction when $dy$ is negative.

3. Dec 23, 2016

### TSny

Also, can you explain why you have a negative sign in front of the integral in the expression $W = -q\int{\vec{E} \cdot \vec{dl}}$? Keep in mind that you are asked to find the work done by the electric field, not the work done by an external agent.

4. Dec 23, 2016

### Abdulwahab Hajar

Wow dude thanks for the reply..
W=−q∫⃗E⋅→dl well I got the minus from the book, it's just like −∫⃗E⋅→dl for electrical potential apparently..
Could you enlighten me perhaps and so as you said dl=dx^ax+dy^ay+dz^az always stands right the integration limits determine the rest .
Thank you sir

5. Dec 23, 2016

### TSny

The definition of the work done by a force along a path from point $a$ to point $b$ is $\int_a^b \vec{F} \cdot \vec{dl}$. There is no negative sign in the definition. Definitions of electric potential and electric potential energy will have a negative sign.

6. Dec 24, 2016

Awesome bro
Thanks a lot