Metal sphere on a thread in a horizontal electric field

[steroidjunkie]

Homework Statement

Charged metal sphere hanging on an isolated thread of negligible mass is put in a homogeneous horizontal electric field so that the thread makes a 45 degree angle with the el. field. What angle does the thread with the sphere close with the el. field after we remove 40% of the charge from the sphere?

Homework Equations

1. $F_g=mg$
2. $F_g=T \cdot \sin(45)$
3. $F_{el}=T \cdot \cos (45)$

I have provided image of force diagram.

The Attempt at a Solution

I can substitute $F_g$ from equation 1 into equation 2 and get:
$mg=T \cdot \sin(45)$

Then I find expression for T in equation 3:
$T=\frac{F_{el}}{\cos (45)}$

and substitute it into second equation:
$mg=\frac{F_{el}}{\cos (45)} \cdot \sin(45)$
$mg=F_{el} \cdot \tan(45)$

I can express electric force as a product of electric field and charge of sphere:
$mg=q \cdot E \cdot \tan(45)$

I haven't gotten further than this. How do I show what happens with an angle when the charge is reduced 40%?

 

[TSny]

I can express electric force as a product of electric field and charge of sphere:
$mg=q \cdot E \cdot \tan(45)$

This equation looks good. Note that your derivation would work for any angle, not just for 45 degrees. If $\theta_2$ is the angle you are looking for, how would you write your equation with 45 degrees replaced by $\theta_2$?

 

[steroidjunkie]

Well, I could substitute $\Theta_2$:
$mg=q \cdot E \cdot \tan(\Theta_2)$
$\tan(\Theta_{2})=\frac{mg}{q \cdot E}$
$\Theta_2=\arctan \frac{mg}{q \cdot E}$

What do I do now? Do I just say that $\Theta_2$ is 60% of 45 degrees which is 27 degrees?

 

[TSny]

Well, I could substitute $\Theta_2$:
$mg=q \cdot E \cdot \tan(\Theta_2)$
$\tan(\Theta_{2})=\frac{mg}{q \cdot E}$
$\Theta_2=\arctan \frac{mg}{q \cdot E}$

OK. Here $q$ is the charge corresponding to $\theta_2$.

What do I do now? Do I just say that $\Theta_2$ is 60% of 45 degrees which is 27 degrees?

No. You have two equations, one for the initial charge $q_1$ and one for the final charge $q_2$. Combine them to find $\theta_2$.

 

[steroidjunkie]

OK, than:

$\tan 45=\frac{m \cdot g}{q_1 \cdot E}$
$\tan \Theta_2=\frac{m \cdot g}{q_2 \cdot E}$

$q_1=q, q_2=q_1-40\%=q-0.4q=0.6q$$\tan 45=\frac{m \cdot g}{q \cdot E}$
$\tan \Theta_2=\frac{m \cdot g}{0.6 \cdot q \cdot E}$

$q \cdot \tan 45=\frac{m \cdot g}{E}$
$0.6 q \cdot \tan \Theta_2=\frac{m \cdot g}{E}$

I equate the two equations:
$0.6 q \cdot \tan \Theta_2=q \cdot \tan 45$

$\tan \Theta_2=\frac{\tan 45}{0.6}$

$\tan \Theta_2=\frac{1}{0.6}$

$\tan \Theta_2=\frac{5}{3}$

$\Theta_2=59^{\circ} 04'$

I'm sorry for writing minutes the way I did. I couldn't find a proper code.

 

[TSny]

OK. Good work.

My calculator gives the answer as a decimal number of 59.04^o.
.04 degree is about 2 minutes.

Considering that the 45 degrees is given to 2 significant figures, it would probably be best to express the answer as 59^o or maybe 59.0^o.

 

[steroidjunkie]

You're right. It's 2 minutes, and not 4 minutes. Thank you very much for helping me.