Find the point where the eletrostatic force is maximum. (With drawing)

[tsuwal]

Homework Statement

https://www.physicsforums.com/attachments/56019

Homework Equations

The Attempt at a Solution

The net force will be:

F=\frac{2*Sin(\alpha)*q1*q}{(d/cos(\alpha))^{2}}=cos(\alpha)^{2}*Sin(\alpha)*qq1/d^{2}

Taking the rerivative to find the maximum we get:

\frac{d cos(\alpha)^{2}*Sin(\alpha)}{d\alpha}=0 \Leftrightarrow cos(\alpha)^{2}=2*Sin(\alpha)

How do I solve this?

 

[voko]

The attachment is broken.

 

[tsuwal]

Sorry, here it is:

 

[voko]

I do not think the final equation is correct. Show how you differentiated the previous equation.

 

[tsuwal]

\frac{d cos(\alpha)^{2}*Sin(\alpha)}{d\alpha}=0 \Leftrightarrow cos(\alpha)^{2}´*sin(\alpha)+cos(\alpha)^{2}*sin(\alpha)´=0 \Leftrightarrow <br /> -2*sin(\alpha)*cos(\alpha)+cos(\alpha)^{3}=0 \Leftrightarrow cos(\alpha)^{2}=2*Sin(\alpha)

 

[voko]

\frac{d cos(\alpha)^{2}*Sin(\alpha)}{d\alpha}=0 \Leftrightarrow cos(\alpha)^{2}´*sin(\alpha)+cos(\alpha)^{2}*sin(\alpha)´=0 \Leftrightarrow <br /> -2*sin(\alpha)*cos(\alpha)+cos(\alpha)^{3}

This should be $-2*sin(\alpha)*cos(\alpha)*sin(\alpha)+cos(\alpha)^{3}$

 

[tsuwal]

Thanks, don't know how I missed that!

 

[voko]

I assume you can continue from here.