Electric field on each midpoint of square of side L

[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 :)