Net Electric Field for 4 Equidistant Charges

In summary, the conversation discusses the calculation of the net electric field produced by four charged particles arranged in a square. The x-components of the electric fields cancel each other out, leaving only the y-components to be considered. The solution is found by summing the y-components of the four fields in the appropriate direction, resulting in a net electric field of 7.29E4 N/C.
  • #1
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?

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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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  • #2
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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  • #3
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.
 

FAQ: Net Electric Field for 4 Equidistant Charges

1. What is the net electric field for 4 equidistant charges?

The net electric field for 4 equidistant charges is the vector sum of all the individual electric fields produced by each charge at a given point in space. It is a measure of the overall strength and direction of the electric field at that point.

2. How is the net electric field calculated for 4 equidistant charges?

To calculate the net electric field for 4 equidistant charges, you must first determine the magnitude and direction of the electric fields produced by each charge at the given point. Then, you can use vector addition to find the total electric field by adding the individual electric fields together.

3. Can the net electric field for 4 equidistant charges be zero?

Yes, it is possible for the net electric field to be zero for 4 equidistant charges. This can occur if the charges are arranged in a specific way such that their individual electric fields cancel each other out at the given point.

4. How does the distance between the charges affect the net electric field?

The distance between the charges has a significant impact on the net electric field. As the distance increases, the strength of the electric fields produced by each charge decreases, resulting in a decrease in the net electric field. Similarly, as the distance decreases, the net electric field increases.

5. What is the significance of having 4 equidistant charges in this scenario?

The significance of having 4 equidistant charges is that it allows for a symmetrical arrangement, making it easier to calculate the net electric field. Additionally, having equidistant charges ensures that the individual electric fields produced by each charge are equal, simplifying the vector addition process.

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