I'm trying to deal a problem using this formula, but I'm unclear as to what dl represents (Or if it is the same as dx in most integrals, then in that case I don't know what the lowercase L is) [tex]{\cal{E}} = \int_{a}^{b} \vec{E} \cdot \vec{dl}[/tex]
Along a general curve l(t)=(x(t),y(t)), dl=(x'(t)*dt,y'(t)*dt). So if E=(Ex,Ey) that becomes Ex*x'(t)*dt+Ey*y'(t)*dt.