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.

 

[PJC01]

for your question. The use of infinitesimals in finding the electric field above a square sheet of charge is a common technique in electromagnetism. In order to accurately calculate the electric field at a specific point, we need to divide the sheet into infinitesimally small pieces and integrate their contributions.

In this case, we are using the result for a square loop to find the electric field for a square sheet. This is because the electric field for a square sheet can be seen as the sum of the electric fields created by four equal and opposite square loops. By integrating over these loops, we can find the total electric field at a given point.

As for the change from \lambda to \sigma \frac{da}{2}, this is due to the conversion from a one-dimensional charge density to a two-dimensional surface charge density. In the diagram, we are considering the width of the sheet to be a+da, which is why we use \frac{da}{2} as the width for each loop. If we were to use a+2da as the width, then we would need to use a factor of 2 in the integrand to account for the increased width.

The conversion from \lambda to \sigma \frac{da}{2} is necessary because we are dealing with a two-dimensional surface rather than a one-dimensional line. The surface charge density is defined as the charge per unit area, which is why we need to multiply the one-dimensional charge density by the width of the sheet (da) to get the two-dimensional surface charge density (sigma).

I hope this explanation helps to clarify the reasoning behind the use of infinitesimals and the conversion of charge density in finding the electric field above a square sheet of charge. Let me know if you have any further questions.