Infinitesimals - finding the E field

[Identity]

The problem is to find the E field a distance $$z$$ above the centre of a square sheet of charge of side length $$a$$ and charge density $$\sigma$$.

The solutions use the result for a square loop and integrate to get the result for the square sheet, however it's the change $$\lambda \to \sigma \frac{da}{2}$$

In the diagram on the right, since we have $$a+da$$, widths of $$\frac{da}{2}$$ make sense. But what if we instead chose to do $$a+2da$$, so that we would have widths of $$da$$? That would increase the integrand by a factor of 2.

Also, I still can't get my head around how the 1-dimensional charge density $$\lambda$$ could be converted into a 2-dimensional surface charge density $$\sigma$$. Could someone please explain the reasoning behind it?


Thanks

 

[Doc Al]

Also, I still can't get my head around how the 1-dimensional charge density $$\lambda$$ could be converted into a 2-dimensional surface charge density $$\sigma$$. Could someone please explain the reasoning behind it?

You can view the charge of a strip in terms of a line charge (λ*length) or a surface charge (σ*area). Either way, the charge is the same:

λa = σa da/2 → λ = σ da/2

(ignoring higher powers of infinitesimals)

 

[Identity]

Oh thanks, so multiplying "σa" by the infinitesimal "da/2" makes it look like approximately like a line distribution, i get that

But the da/2 still seems too arbitrary... operations on infinitesimals is tricky

What if we started out by assigning the widths "da" to the left and right sides. Then the bottom would simply be "a+2da", wouldn't it?
I mean, it changes the whole answer by a factor of 2, but it still seems pretty reasonable to me.

 

[Doc Al]

What if we started out by assigning the widths "da" to the left and right sides. Then the bottom would simply be "a+2da", wouldn't it?
I mean, it changes the whole answer by a factor of 2, but it still seems pretty reasonable to me.

But that would mean that your variable of integration is no longer the length of the side, but twice the length of the side. You'd have to divide by 2, which brings you back to where you started.