Net Electric Field for 4 Equidistant Charges

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SUMMARY

The net electric field at the center of a square formed by four charges (q1 = 6.35 nC, q2 = -17.9 nC, q3 = 17.9 nC, q4 = -6.35 nC) with an edge length of 6.10 cm is calculated to be 7.29E4 N/C. The x-components of the electric fields cancel each other out, leaving only the y-components to contribute to the net field. The correct formula for the net electric field is ΣEp = k/r^2 * (-q1 + q2 + q3 - q4), where k is the electrostatic constant and r is the distance from the charges to the center of the square.

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  • Understanding of Coulomb's Law and electric fields
  • Familiarity with vector components and their summation
  • Knowledge of the electrostatic constant (k)
  • Basic geometry of squares and distances in two dimensions
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CARNiVORE
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Homework Statement


In the figure the four particles form a square of edge length a = 6.10 cm and have charges q1 = 6.35 nC, q2 = -17.9 nC, q3 = 17.9 nC, and q4 = -6.35 nC. What is the magnitude of the net electric field produced by the particles at the square's center?

96b64812a6cb9587e046628e4fcc2f73.png


Homework Equations


E=kq/r^2
The distance from each point to the center of the square = a/√2
The x-components cancel each other out. Only the E-fields' y-components will end up affecting the net electric field.

The Attempt at a Solution


Since the x-components cancel each other out, I believe that ΣE = -E14 + E23 + E13y + E24y. Let me justify this:

E14 is negative because it is the only field vector among these four that has downward influence. All of the others have upward influence. In addition to this, the influence both E14 and E23 points in only one direction (downward and upward, respectively), so they do not have to be converted to components.

On the other hand, E13y and E24y must be converted to their y-component to solve this problem, because their influences are pointed upward and to the left and upward to the right, respectively. The angle in question is 45 degrees, and this reduces the two fields by a factor of √2. However, I'm not completely sure how to express E13y and E24y in terms of charge.

For example, is this correct?

E13y = k(q1 - q3)/r13^2

If not, I believe I may be approaching this problem incorrectly. Thanks so much for your time.
 
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Not sure I understand what your E14 etc. represent. Is that the field in the centre due to q1 in the direction q1 to q4?
Why not just sum the y components of the four fields in the obvious way?
 
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Hey, you're right about summing the y-components - I completed the problem correctly using the obvious manner that you were talking about:

ΣEp = E1 + E2 + E3 + E4
ΣEp = k/r^2 * (-q1 + q2 + q3 - q4)

Where q1 and q4 are negative because they cause downward influence in the y-axis, whereas q2 and q3 provide the opposite.

ΣEp = 7.29E4 N/C

I guess I got confused by the initial complexity I felt when I saw this problem. This problem becomes a breeze when you find that the x-components cancel each other out, but I didn't see that. Thanks for your help, haruspex.
 

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