Calculating E Field Magnitude: Contributions from + and - Charges 5cm Above Axis

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The discussion focuses on calculating the electric field (E field) magnitude at the origin due to both positive and negative charges positioned 5 cm above the horizontal axis. Participants emphasize the importance of determining the contributions from each charge and breaking them down into x and y components using trigonometric functions. The positive charge's contribution is calculated as 5.4 x 10^4, while the user attempts to compute the total E field, initially arriving at 7.6 x 10^4 but expecting a correct answer of 7.1 x 10^4. The conversation highlights the need for clarity on the angle used in calculations and the verification of the distance involved. Overall, the thread illustrates the collaborative effort to resolve the complexities of electric field calculations.
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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
 

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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:
 
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.
 
sasuke07 said:
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:)
 
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.
 
no, you must use cos and sin, of the angle the arrow makes
 
so would it be Kq/r^2sintheta
and Kq/r^2Costheta. Where theta would be 90 degrees?
 
or would theta be 45
 
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
sasuke07 said:
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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