# Homework Help: Potential Difference between Two Points affected by a Dipole

1. Feb 7, 2016

### Oribe Yasuna

1. The problem statement, all variables and given/known data
A dipole is centered at the origin, with its axis along the y axis, so that at locations on the y axis, the electric field due to the dipole is given by
E vector = 0, 1/4πε0 * 2qs/y^3, 0 V/m
The charges making up the dipole are q1 = +6 nC and q2 = -6 nC and the dipole separation is s = 4 mm (see figure below). What is the potential difference along a path starting at location P1 = 0, 0.02, 0 m and ending at location P2 = 0, 0.06, 0 m?

2. Relevant equations
delta V = - integral of i -> f (E vector * delta L vector)

3. The attempt at a solution
delta V = 2qs/4πε0 ( 1/P2^3 - 1/P1^3 ) * 0.04
delta V = 9e+9 * 2(6e-9)(0.004) * ( 1/(0.06)^3 - 1/(0.02)^3 ) * 0.04
delta V = -2.08e+3

Last edited by a moderator: Apr 17, 2017
2. Feb 7, 2016

### Staff: Mentor

In your relevant equations you indicated that finding the potential difference involves integrating the electric field along the path between the points of interest. I don't see you doing that in your solution attempt. It would appear that you're just taking the difference in the field strength at each location and multiplying by the path length.

3. Feb 13, 2016

### Oribe Yasuna

I wanted to avoid doing an integral and instead get an estimate. I suppose that's an unreasonable expectation, though.

Then, this is the integral I set up:
delta V = integral of 0.02 -> 0.06 [(1 / 4*pi*E0) * (2*6e-9*0.004 / y^3) * (<dx, dy, dz>)]

I know:
delta L = <dx, dy, dz>
delta L = 0.04 m

Does <dx, dy, dz> mean I have to derive delta L? If so, how do I derive a constant without getting 0?

4. Feb 13, 2016

### Staff: Mentor

The path indicated is along one direction, and happens to be along one particular axis, so you can ignore the other components. So your "delta L" is just dy.

I suggest that you perform the integration symbolically first, using variables as the limits of the integration. Plug in numbers at the end.

5. Feb 13, 2016

### Oribe Yasuna

Alright.

Since this is on the y-coordinate I'll exclude the x & z coordinates of the vectors:
delta V = integral of P1 -> P2 (1 / 4 pi E0 * 2 q s / y^3) dy
delta V = 1 / 4 pi E0 * 2 q s * integral of P1 -> P2 (1 / y^3) dy

In this form, the problem seems to become easier than I expected it to be.
If I recall correctly, the integral of 1 / y^3 is:
-1 / 2 y^2

So the integral of 1 / y^3 from P1 to P2 would be:
[ - 1 / 2 (P1)^2 + 1 / 2 (P2)^2 ]

Plugging this back in:
delta V = 1 / 4 pi E0 * 2 q s * (- 1 / 2 P1^2 + 1 / 2 P2^2)

is this correct so far?

Last edited: Feb 13, 2016
6. Feb 13, 2016

### Oribe Yasuna

I plugged in the numbers and I got -4.8e+2 (-4.8 * 10^2).

Thanks for the help, I probably would have over-complicated things without it.

7. Feb 13, 2016

### Staff: Mentor

Yes, looking good. Carry on!

Edit: Oops! too late!

Well done.