Converting the Electric Field of a Dipole from Cartesian to Spherical Coords

[albertoid]

Homework Statement

Show that $$E_z = \frac{p}{4 \pi \epsilon_0} \left( \frac{3z^2}{r^5} - \frac{1}{r^3} \right)$$ is equivalent to the electric field on the positive z-axis from $$E_r = \frac{2 p \cos \theta}{4 \pi \epsilon_0 r^3}$$

Homework Equations

The unit normal for a sphere, sin0cos%, sin0sin%, cos0 (0 is theta; % is phi)
z = rcos0

The Attempt at a Solution

I multiplied E_r by cos0 (the z component of the unit vector in spherical coordinates)
and got 2cos^2 theta /r^3 (times p/4piEo)However, simplifying E_z I get [2cos^2 theta - sin^2 theta]/r^3 (times p/4piEo)

I'm not sure what I'm doing wrong, any guidance would be appreciated!Edit: solved. Forgot to consider a value for theta

 

[mark726]

.

Hello,

Thank you for sharing your attempt at a solution. It looks like you are on the right track, but there are a few things that need to be corrected in your approach.

First, when you multiplied E_r by cos0, you should have also multiplied the p/4piEo term. This would give you [2p*cos^2(theta)] / [4piEo*r^3].

Next, in order to simplify E_z, you need to substitute in the expression for z in terms of r and theta. This would give you [3z^2/r^5 - 1/r^3] * (p/4piEo).

Now, using the equation z = rcos(theta), we can substitute in for z to get [3(r*cos(theta))^2/r^5 - 1/r^3] * (p/4piEo).

Simplifying this expression gives you [3cos^2(theta) - 1] / r^3 * (p/4piEo).

Finally, we can use the trigonometric identity cos^2(theta) = 1 - sin^2(theta) to substitute in for cos^2(theta) and get [2cos^2(theta) - sin^2(theta)] / r^3 * (p/4piEo).

As you can see, this is the same expression that you got for E_r when you multiplied by cos0, but now we have properly substituted in for z in terms of r and theta.

I hope this helps clarify your approach and leads you to the correct solution. Good luck!