Thread: 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,



Answer from book: 9.5 J

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,

= -19/2 = -9.5

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

Consider more simpler task:
if E=i
P1(0,0,0) to P2(1,0,0)
According to your formular
W/q=- 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.

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

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,



Answer from book: 9.5 J

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,

= -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 ....?