# work done in a vector field

• Mar 31st 2010, 02:31 PM
jut
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?
• Mar 31st 2010, 03:30 PM
zzzoak
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.
• Mar 31st 2010, 03:35 PM
jut
This is what my book says about work (see attachment)
• Mar 31st 2010, 07:14 PM
mr fantastic
Quote:

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 ....?