Electric field on each midpoint of square of side L

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SUMMARY

The electric field at the midpoint of each side of a square with point charges of +Q at diagonal corners and -Q at the opposite corners is calculated using Coulomb's Law. The resulting electric field is not zero, as initially assumed, but rather 1.15 x 10^10 Nm^2/C^2 in the x-direction for two midpoints and -1.15 x 10^10 Nm^2/C^2 for the other two. The calculations involve determining the x and y components of the electric fields produced by each charge and recognizing that the fields do not cancel out due to their directions. Proper vector analysis is crucial for accurate results.

PREREQUISITES
  • Coulomb's Law for electric fields (E = kQ/r^2)
  • Vector decomposition of forces (Ex = Ecosθ, Ey = Esinθ)
  • Understanding of electric field directionality (away from positive charges, towards negative charges)
  • Basic geometry of squares and distances between points
NEXT STEPS
  • Review vector addition in electric fields to understand component cancellation.
  • Practice calculating electric fields for different charge configurations using Coulomb's Law.
  • Learn about superposition principle in electric fields for multiple charges.
  • Explore graphical methods for visualizing electric field vectors and their interactions.
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Students studying electrostatics, physics educators, and anyone interested in understanding electric fields in multi-charge systems.

  • #31
OK - this is the same as I've calculated then.

For E3x + E4x, I'm getting (8√5)/25 for the coefficient.
This multiplied by k gives me 6.43e9 - quite a lot different from your value.

Edit : the difference between the 2 values is √5
 
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  • #32
ap123 said:
OK - this is the same as I've calculated then.
? You said you agreed with to the magnitudes of post 17 ...
For E3x + E4x, I'm getting (8√5)/25 for the coefficient.
This multiplied by k gives me 6.43e9 - quite a lot different from your value.

E3x = E4x = +(8/5)kQ/L^2
The entire Ex field at the bottom mid-point is -6.4kQ/L^2 = -5.76e10Q/L^2.
Edit : the difference between the 2 values is √5
Don't understand that ...
 
  • #33
? You said you agreed with to the magnitudes of post 17 ...
I do.
You wrote that E1x = E2x = -8kQ/L^2 rather than E1x + E2x = -8kQ/L^2 which is what I was unsure about.
This is also the magnitude that is in post #17

E3x = E4x = +(8/5)kQ/L^2
You mean
E3x + E4x = +(8/5)kQ/L^2
I have an additional √5 in the denominator

Don't understand that ...
See above
 
  • #34
OK. Right on both counts. I meant "+" when I said "=".

Where did you get the extra sqrt(5)? The distance squared from the bottom midpoint to either upper corner is r^2 = L^2 + (L/2)^2 = 5L^2/4 by Pythagoras.
 
  • #35
Where did you get the extra sqrt(5)? The distance squared from the bottom midpoint to either upper corner is r^2 = L^2 + (L/2)^2 = 5L^2/4 by Pythagoras.

From the cosθ when I took the x-component of the field.
cosθ = 1 / √5
 
  • #36
ap123 said:
From the cosθ when I took the x-component of the field.
cosθ = 1 / √5

If you're going to do it that way, isn't it cos2θ?

Never mind, I just woke up. Yes, the cosθ needs to be incorporated in my Ex3 and Ex4, making it Ex3 + Ex4 = 0.447(8/5)Q/L^2. Pardon my obtuseness.
 
Last edited:
  • #37
If you're going to do it that way, isn't it cos2θ?
How do you get that?
I'm just taking the x-component of the vector, eg
E3x = E3 * cosθ

Edit :
Pardon my obtuseness
OK :smile:

I hope the OP has absorbed this long thread.
 
  • #38
ap123 said:
How do you get that?
I'm just taking the x-component of the vector, eg
E3x = E3 * cosθ

Edit :

OK :smile:

I hope the OP has absorbed this long thread.

So at least you & I agree, right?
 
  • #39
So at least you & I agree, right?
Yes :)
 

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