Electric field on each midpoint of square of side L

[ap123]

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

 

[rude man]

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 ...

 

[ap123]

? 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

 

[rude man]

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.

 

[ap123]

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

 

[rude man]

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.

 

[ap123]

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

I hope the OP has absorbed this long thread.

 

[rude man]

How do you get that?
I'm just taking the x-component of the vector, eg
E3x = E3 * cosθ

Edit :

OK

I hope the OP has absorbed this long thread.

So at least you & I agree, right?

 

[ap123]

So at least you & I agree, right?

Yes :)