# Work

• November 26th 2007, 03:41 PM
shilz222
Work
If $\bold{F}(x,y) = \frac{k(x \bold{i} + y \bold{j})}{x^{2}+y^{2}}$ find the work done by $\bold{F}$ in moving a unit charge along a straight line segment from $(1,0)$ to $(1,1)$.

So $\bold{F}(1,y) = \frac{k(\bold{i} + y \bold{j})}{1 + y^{2}}$. Then $x = 1, \ y = y$.

$k \int_{0}^{1} \frac{y}{1+y^{2}} \ dy$

$u = 1+y^{2}$

$du = 2y \ dy$

$\frac{k}{2} \int \frac{du}{u}$

$= \frac{k}{2} \int_{0}^{1} \ln|1+y^{2}|$

$= \frac{\ln 2}{2}$.

Is this correct?
• November 26th 2007, 06:33 PM
shilz222
Essentially it the work is just in the y-direction right?
• November 27th 2007, 07:38 AM
topsquark
Quote:

Originally Posted by shilz222
Essentially it the work is just in the y-direction right?

Yes it is, because the displacement is also in the +y direction. (Typically we'd have to take a component of the force along the y direction and wind up with a cosine term, but the force was conveniently in coordinate form to begin with.)

-Dan