# Uniformly charged ring

1. Jun 1, 2013

### dido28

hi every one on an exercise on book we ask us to find the electrical field for a uniformly charged ring where we gonna find : E=kqz/(z2+R2)3/2
then we have to derivate it wth respect to z and we find : dE/dz=kq*(R2+2z2)/(z2+R2)5/2
so my question is how do we get this derivation ?

2. Jun 1, 2013

### WannabeNewton

You will have show us what you have *tried* thus far. Where did you get stuck in the derivation when you tried it yourself? Also note that the given expression is only the electric field produced by the uniformly charged static ring at points along its symmetry axis (in case you didn't know already).

3. Jun 1, 2013

### dido28

well me i found : dE/dz=qk[ (z2+R2)3/2-3z2(z2+R2)-1/2 ]/(z2+R2)3
i just applied the rule that i know but in the book it seems that they did some simplification so my question is how they did this

4. Jun 1, 2013

### WannabeNewton

Oh are you only interested in finding the derivative? Do you already know how to get the electric field itself?

5. Jun 1, 2013

### dido28

yes that's it

6. Jun 1, 2013

### WannabeNewton

Are you sure it wasn't $R^2 - 2z^2$ as opposed to $R^2 + 2z^2$ in the numerator of the final answer?

7. Jun 1, 2013

### dido28

oh yes sorry it's R2-2z2 :shy:

8. Jun 1, 2013

### WannabeNewton

Ok. Would you agree that, before doing any simplifications, the derivative comes out to $\frac{\mathrm{d} E}{\mathrm{d} z} = \frac{kq}{(z^2 + R^2)^{3/2}} - \frac{3kqz^2}{(z^2 + R^2)^{5/2}}$? All I have done is use the product and chain rule; I haven't done any simplifications at all. Now, can you find a way to combine these two expressions?

9. Jun 1, 2013

### dido28

yes it's done i get the result . thanks for your help WannabeNewton

10. Jun 1, 2013

### WannabeNewton

I didn't do anything mate! It was all you :) Good luck with your studies.