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Concept of the Electric Field(picture included)

  1. Sep 11, 2011 #1
    I did search for this problem found a few but they were all different and I couldnt follow the math


    1. The problem statement, all variables and given/known data
    http://session.masteringphysics.com/problemAsset/1169191/1/jfk.Figure.20.P24.jpg

    What is the strength of the electric field at the position indicated by the dot in the figure?

    What is the direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above the horizontal line.


    2. Relevant equations

    E=K*Q/r^2
    E1+E2= total

    3. The attempt at a solution
    8.99*10^9*1*10^-9/.05^2 = 3596 =E1
    same equation for e2 get 3596 again add them together i got 7192nC

    for the second problem i took a stab at it and got 0 degrees i used some logic and got that not exactly sure for the math to back that up but 0 is correct

    Would appreciate any help or general tips, thanks

    I'm not going to lie to you guys I have no idea what I'm doing
     
  2. jcsd
  3. Sep 11, 2011 #2
    Hi zha28 and welcome to PF (you're the 3rd person I've welcomed tonight!) Anyway...,

    Your equations are correct. The main problem with your calculation for the first question is your distance is incorrect. What is the distance between one of the charges and the dot?
     
  4. Sep 11, 2011 #3
    Thanks for the warm welcome and thanks for your quick response


    The distance between the point and one of the charges is 5 cm isnt it?
     
  5. Sep 11, 2011 #4
    Not quite. If you look the point is 5cm up/down and 5cm to the right from the charges. Do you know how to use Pythagoras' theorem?
     
  6. Sep 11, 2011 #5
    A^2+B^2=c^2?


    or am i going in the wrong direction.
     
  7. Sep 11, 2011 #6
    8.99*10^9*1*10^-9/.071^2 = 1783.38 times that by 2 since the formula is the same I get 3566.76 Does that sound about right? I only have 1 more guess on this problem so I'm trying to get some positiving assurance that I'm doing this properly
     
  8. Sep 12, 2011 #7
    You're correct up until multiplying the formula by 2. The issue is that, because both charges are positive, and either side of the point, the field doesn't perfectly add up. As a result, you need to calculate the x and y components of the electric field from each charge at the point. I'm guessing you're not familiar with using vectors, so you'll find this easiest if you use trigonometry.
     
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