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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 don’t 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.

Hello Basila,

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,

Steve

Physics

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