# Homework Help: Help with partial derivative electric field

1. Oct 27, 2008

### krugertown

We found on-axis potential of a ring of radius R and Charge Q to be:
V=(1/4pi*Epsilon naught) * (Q/sqrt(z^2 + R^2))

Find on axis electric field of ring of charge

I know i just derive that equation, but am getting stuck.

i got d/dz((1/4pi*Epsilon naught) * (Q/sqrt(z^2 + R^2))

Any ideas how i do that partial derivative?

2. Oct 27, 2008

### Dick

Use the chain rule. d/dz(f(z)^(-1/2))=(-1/2)*(f(z)^(-3/2))*d/dz(f(z)). Same as always.

3. Oct 27, 2008

### Staff: Mentor

If I understand your function, it's a function of z alone, so you don't need to take a partial derivative.

d/dz((1/4pi*Epsilon naught) * (Q/sqrt(z^2 + R^2))
= Q*(1/4pi*Epsilon naught) *d/dz[ z^2 + R^2]^(-1/2)

Now, just use the chain rule to differential (z^2 + R^2)^(-1/2)

I'm assuming that Q, R, and $$\epsilon$$0 are constants.

Also, it's not clear whether the expression with $$\epsilon_{0}$$ is $$\frac{1}{4pi * \epsilon_{0}}$$ or $$\frac{1}{4pi} \epsilon_{0}$$. The use of parentheses will make it clear.

4. Oct 27, 2008

### krugertown

so z becomes (-1/2)*(f(z)^(-3/2))*d/dz(f(z))? i understand that but how does that then become:

E=(Q/(4pi*Epsilon naught)) (zQ/(z^2 + R^2)^3/2?

thanks!

5. Oct 28, 2008

### Dick

In your best interests, I'm not going to tell you. Work it out. What do you get? Show us how you did it if you still have questions. BTW I think E is the negative of the gradient of the potential.

6. Oct 28, 2008

### Staff: Mentor

No, z doesn't become what you have above; that's dV/dz. Also, you started with V as a function of z, so you should end up with dV/dz, not E.

7. Oct 28, 2008

### Dick

Hi Mark44, It's actually a physics problem. V is voltage and E is electric field. The gradient of V is E (up to a sign). And it is a partial derivative. The d/dx and d/dy parts are zero because V is cylindrically symmetric and krugertown is working on the z-axis.

8. Oct 28, 2008

### Staff: Mentor

My mistake. I was thinking that E was EMF.