1. ## Work

If $\displaystyle \bold{F}(x,y) = \frac{k(x \bold{i} + y \bold{j})}{x^{2}+y^{2}}$ find the work done by $\displaystyle \bold{F}$ in moving a unit charge along a straight line segment from $\displaystyle (1,0)$ to $\displaystyle (1,1)$.

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

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

$\displaystyle u = 1+y^{2}$

$\displaystyle du = 2y \ dy$

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

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

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

Is this correct?

2. Essentially it the work is just in the y-direction right?

3. 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