1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Metal sphere on a thread in a horizontal electric field

  1. Jun 1, 2016 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant 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.

    3. 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%?
     

    Attached Files:

    • img.jpg
      img.jpg
      File size:
      51.1 KB
      Views:
      57
  2. jcsd
  3. Jun 1, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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##?
     
  4. Jun 1, 2016 #3
    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?
     
  5. Jun 1, 2016 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

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

    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##.
     
  6. Jun 2, 2016 #5
    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.
     
  7. Jun 2, 2016 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    OK. Good work.

    My calculator gives the answer as a decimal number of 59.04o.
    .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 59o or maybe 59.0o.
     
  8. Jun 2, 2016 #7
    You're right. It's 2 minutes, and not 4 minutes. Thank you very much for helping me.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Metal sphere on a thread in a horizontal electric field
Loading...