# Charged Ring again.

1. Aug 30, 2009

### exitwound

1. The problem statement, all variables and given/known data

2. Relevant equations

$$E=\frac{kqz}{(z^2+r^2)^{3/2}}$$

3. The attempt at a solution

(c) is asking where the maximum value of the electric field would be in terms of R. In order to do this, I have to take the derivative of this function, set it equal to zero, correct?

$$E=kq\frac{z}{(z^2+r^2)^{3/2}}$$

Is this how I do this?

$$\frac{dE}{dz} = \frac{uv\prime - u\prime v}{v^2}$$

2. Aug 30, 2009

### rock.freak667

Yes that is how to differentiate it.

3. Aug 30, 2009

### exitwound

I end up with

[edited]

First of all, is this right?

Last edited: Aug 30, 2009
4. Aug 30, 2009

### exitwound

Okay, I'm completely lost on this one. I need help.

5. Aug 31, 2009

### rock.freak667

6. Aug 31, 2009

### exitwound

$$kq(\frac{(z)(3/2)(z^2+r^2)^{1/2}(2z)-(1)(z^2+r^2)^{3/2}}{(z^2+r^2)^3})$$

$$kq\frac{(3z^2)(z^2+r^2)^{1/2}-(z^2+r^2)^{3/2}}{(z^2+r^2)^3}$$

$$(3z^2)(z^2+r^2)^{1/2}=(z^2+r^2)^{3/2}$$

$$(3z^2)=(z^2+r^2)$$

$$2z^2-r^2=0$$

7. Aug 31, 2009

### rock.freak667

so z= ± r/√2

8. Aug 31, 2009

### exitwound

But what does that tell me in terms of the question asked?

9. Aug 31, 2009

### rock.freak667

so if Emax occurs for z=R/√2

To find Emax, put z=r/√2 into your equation for E

10. Aug 31, 2009

### exitwound

I inadvertently dropped the kq from the post above.

$$kq(3z^2)(z^2+r^2)^{1/2}=(z^2+r^2)^{3/2}$$

$$kq(3z^2)=(z^2+r^2)$$

$$kq(2z^2-r^2)=0$$

$$z=\frac{r}{\sqrt{2kq}}$$

If I put it back into the original equation, I still have an unknown r then. This equation is a mess.

11. Aug 31, 2009

### rock.freak667

well you could just compute the value of z and then put that number into the equation with E

12. Aug 31, 2009

### exitwound

Okay I can't solve this. The algebra is way too messy and I can't follow what I'm doing.