E field magnitude

  • #1
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Homework Statement


What is the magnitude of the collective contribution to the E field at the origin from both the + and - charges located 5 cm above the horizontal axis?

Homework Equations


kQ/d^2



The Attempt at a Solution


Im just having a problem setting up this problem. Could someone help me on getting started
 

Attachments

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Answers and Replies

  • #2
hi sasuke07! :smile:

(try using the X2 button just above the Reply box :wink:)

first, find the magnitude of each contribution,

from that, find the x and y components …

what do you get? :smile:
 
  • #3
I already figured out the magnitued of the the positive charge 5 cm above the axis and got 5.4X10^4. What do you mean by the x and y components. WOuldn't the negative charge have the same magnitude of the positive charge.
 
  • #4
WOuldn't the negative charge have the same magnitude of the positive charge.

yes :smile:
What do you mean by the x and y components.

draw an arrow at the origin showing the direction of the field from the positive charge

then find the x and y components of that arrow

(and finally you'll add them to the x and y components of the arrow from the negative charge :wink:)
 
  • #5
so to figure out the x component wouldn't it be Kq/d^2 where k=9X10^9, q is 30X10^9= charge and d would be .05m^2. Or would i have to use cosine and sine to figure out the x and y components and if so could you show me how to set it up.
 
  • #6
no, you must use cos and sin, of the angle the arrow makes
 
  • #7
so would it be Kq/r^2sintheta
and Kq/r^2Costheta. Where theta would be 90 degrees?
 
  • #8
or would theta be 45
 
  • #9
θ is the angle of the line from the charge to the origin
 
  • #10
awesome so 45 degrees.
 
  • #11
So were the equations correct though?
 
  • #12
looks ok :smile:

so what is the total field from both the positive and the negative charge?​
 
  • #13
i got 7.6X10^4 by doing KQ/r^2sin45 and the answer is the same for cos45. BUt i checked the answer and its supposed to be 7.1X10^4. Any suggestions
 
  • #14
i got 7.6X10^4 by doing KQ/r^2sin45 and the answer is the same for cos45. BUt i checked the answer and its supposed to be 7.1X10^4. Any suggestions

your answer looks right to me :confused:

(are you sure it's 5 cm that they're asking about?)
 
  • #15
i think its 5cm because the length of the x and y components is 5cm.
 

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