Find the electric potential at point P due to the dipole shown.
link to picture: http://i1360.photobucket.com/albums/r646/dressXXI/ScreenShot2013-07-21at52313PM_
a) Give your answer in terms of q, k, a, and b. (d, r, and θ donít matter here)
b) Prove that if r >> d, (is much greater than), the electric potential at P can be accurately approximated by: V = kq(dcosθ/r^2).
Hint: in the second situation a ≈ b ≈ r ; therefore a and b can also be considered parallel.
a) The 2 charges are point charges and therefore the potential at P due to either charge is calculated with the formula
V = k*q/r
In this case, q is positive for the upper charge and negative for the lower charge and r=a for the upper charge and r=b for the lower charge. The net potential at P is the sum of the 2 results.
Va = k*q/a
Vb = -k*q/b
The potential at P = Va + Vb = k*q/a - k*q/b
The potential at P = k*q(1/a + 1/b) = k*q*((b-a)/ab)
b) For this part we don't want a or b in the result. We need to find an expression for a and b in terms of r and theta. Since a and b and r are approximately the same, we can say that
The potential at P = k*q*((b-a)/r^2)
We can't reduce b-a to zero though. Approximately is not the same as exactly. And here we have to figure out what a-b actually is.
Using the figure I attached, since a, b, and r are considered parallel, the angle between b and the vertical line is also theta. Consider the right triangle formed by d, the line I drew from +q that is perpendicular to b, and the segment of b whose length is how much longer b is than a, in other words > b-a. We can use trigonometry to find an expression for the length b-a.
cos(theta) = (b-a)/d
b-a = d*cos(theta)
Substituting that in for b-a
The potential at P = k*q*(d*cos(theta)/r^2)
Sorry that I forgot to include the figure. It's attached now.
I hope this helps,