Potential Difference between Two Points affected by a Dipole

[Oribe Yasuna]

Homework Statement

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?

Homework Equations

delta V = - integral of i -> f (E vector * delta L vector)

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

 

[gneill]

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.

 

[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?

 

[gneill]

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.

 

[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?

 

[Oribe Yasuna]

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

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

 

[gneill]

Yes, looking good. Carry on!

Edit: Oops! too late!

Well done.