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.
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.