Electric field of a sheet of charge

[bigevil]

Homework Statement

(This is a truncated question.)

The electric field of a circular sheet of charge of radius a and surface charge density sigma and distance x away from the centre of the sheet is

$$E = \frac{\sigma}{2 \epsilon_0} [1 - \frac{x}{\sqrt{x^2 + a^2}}]$$

Prove that for x > 0

$$E = \frac{\sigma}{2\epsilon_0}$$ when x << a
$$E = \frac{Q}{4\pi \epsilon_0 x^2}$$ when x >> a

The sheet resembles an infinite sheet and a point charge in each case and I'm required to prove this mathematically.

The Attempt at a Solution

For the first case, I note that for x << a, x/a approaches 0. I factor out x from the square roots to get the answer required.

However, for the second case, I try the same thing, x >> a, now a/x approaches 0, but in this case the expression of E becomes E = 0. I've tried several methods and obtained the same thing. Someone help...

 

[rl.bhat]

You can write E = sigma/2epsilon(not)[sqrt(x^2+a^2) - x]/sqrt(x^2+a^2)]
Multiply and dived [sqrt(x^2+a^2) + x] and simplify. Neglect the term a/x.

 

[nlightner]

I would approach this problem by first understanding the physical concept behind the electric field of a sheet of charge. A sheet of charge can be thought of as a collection of point charges, with each point charge contributing to the overall electric field at a given point.

To prove the given expressions for the electric field in the two cases, we can use the concept of superposition, which states that the total electric field at a point is the vector sum of the individual electric fields due to each point charge.

For the first case, x << a, we can consider a small element of charge on the sheet, with charge dq. The electric field at a point P, located at a distance x from the centre of the sheet, due to this element of charge can be written as dE = \frac{dq}{4\pi\epsilon_0 (x^2+a^2)^{3/2}}.

Since x << a, we can approximate the term (x^2+a^2)^{3/2} as a^3, and thus dE = \frac{dq}{4\pi\epsilon_0 a^3}.

Integrating this expression over the entire sheet, we get the total electric field as E = \frac{\sigma}{2\epsilon_0} (since \sigma = \frac{Q}{\pi a^2}), which is the required result.

For the second case, x >> a, we can consider a small element of charge on the sheet, with charge dq. The electric field at a point P, located at a distance x from the centre of the sheet, due to this element of charge can be written as dE = \frac{dq}{4\pi\epsilon_0 x^2}.

Integrating this expression over the entire sheet, we get the total electric field as E = \frac{Q}{4\pi\epsilon_0 x^2}, which is the required result.

In summary, by using the concept of superposition and approximations for the distance terms, we can mathematically prove the expressions for the electric field in the two cases. This also highlights the importance of understanding the physical concept behind the equations, rather than just manipulating mathematical expressions.