Physics 230 March 20 2002
Find the electric energy density at point P as shown in the diagram.

The electric energy density is

*u = *(epsilon0)*E*^{2}/2
First we find the electric field at P. This is the sum of the two
electric fields produced by each of the two charges. Each has the
same magnitude, since each charge has the same magnitude and is the same
distance from P. The vectors also make equal angles with the perpendicular
bisector of the line between the charges, as shown in the diagram.
Thus the horizontal components add to zero, and the vertical component
is
*E*_{y }= 2*kqd*/(*l*^{2} + *d*^{2
})^{3/2}
Thus the energy density is:
*u* = 2(epsilon0)(*kqd*)^{2}/(*l*^{2}
+ *d*^{2 })^{3}

=(*k*/2pi)(*qd*)^{2}/(*l*^{2} + *d*^{2
})^{3}