Finding the Maximum Electric Field of a Charged Ring Using Derivatives

[exitwound]

Homework Statement

Homework Equations

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

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?

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

 

[rock.freak667]

Yes that is how to differentiate it.

 

[exitwound]

I end up with

[edited]

First of all, is this right?

 

[exitwound]

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

 

[rock.freak667]

Where did your differentiation go?

 

[exitwound]

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

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

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

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

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

 

[rock.freak667]

so z= ± r/√2

 

[exitwound]

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

 

[rock.freak667]

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

so if E_max occurs for z=R/√2


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

 

[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.

 

[rock.freak667]

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

 

[exitwound]

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