# Thread: work done in a vector field

1. ## work done in a vector field

Determine the work done by the electric field E=i*x-j*2y in moving a unit positive charge from position P1(-2,0,0) to position P2(5,-1,3).

The work is defined as the integral from p1 to p2 of E "dot" dl,

$\frac{W}{q}=-\int _{\text{P1}}^{\text{P2}}E\circ dl$

My attempted solution

Break dl into constituent vectors: dl = i*dx+j*dy

The dot product of E dot dl = x*dx - 2y*dy

And the integration becomes,

$\frac{W}{q}=-\left[\int _{-2}^5x \text{dx}-\int _0^{-1}2y \text{dy}\right]$= -19/2 = -9.5

So the negative sign was removed b/c work cannot be negative?

if E=i
P1(0,0,0) to P2(1,0,0)
W/q=- $\int^1_0$ dx=-1.
To me must be +1 (the field on the particle).
Please check the definition of the work.
The field on the particle
or particle on the field.

3. This is what my book says about work (see attachment)

4. Originally Posted by jut
Determine the work done by the electric field E=i*x-j*2y in moving a unit positive charge from position P1(-2,0,0) to position P2(5,-1,3).

The work is defined as the integral from p1 to p2 of E "dot" dl,

$\frac{W}{q}=-\int _{\text{P1}}^{\text{P2}}E\circ dl$

My attempted solution

Break dl into constituent vectors: dl = i*dx+j*dy

The dot product of E dot dl = x*dx - 2y*dy

And the integration becomes,

$\frac{W}{q}=-\left[\int _{-2}^5x \text{dx}-\int _0^{-1}2y \text{dy}\right]$= -19/2 = -9.5

So the negative sign was removed b/c work cannot be negative?
Here is something for you to think about: What would be the work done if the charge was negative ....?