Electric field derivation question

[afromanbob]

Hey, if someone can show me the solution to this I'd be very thankful. Here's the question:

An infinitely long sheet of charge has width L and surface charge density n. The sheet lies in the xy-plane between x=-L/2 and x=+L/2

a) Derive an expression for the electric field E along the x-axis for a point outside the sheet a distance d away from the edge of the sheet (the distance from the origin to the point is x=d+L/2)
Hint: How does n relate to the linear charge density, lambda, of a narrow strip of the sheet?

b) Derive an expression for the electric field E at a point on the z-axis a distance z=a above the center of the plane.Thanks.

By the way, I basically have no starting work of mine to show, I'm really quite stumped by this problem. It was actually a question on our test that almost everyone in the class missed, so the professor is allowing us to turn in a solution for a few more points added to our test score. I don't even really understand if the shape described is basically a line... or what...

 

[Nate810]

a) The linear charge density, λ, of a narrow strip of the sheet is related to the surface charge density, n, by λ = n*L. The electric field at the point outside the sheet, a distance d away from the edge of the sheet would then be given by E = λ/(2πε0*(d+L/2))b) The electric field at a point on the z-axis a distance z=a above the center of the plane is given by E = λ/(4πε0*(z+L/2))

 

[m2003]

Sure, I'd be happy to help with this problem. First, let's start by visualizing the problem. We have an infinitely long sheet of charge with width L and surface charge density n. This means that the charge is spread out evenly across the sheet, with a linear charge density of n. The sheet is located in the xy-plane, with the edges at x=-L/2 and x=+L/2. We are asked to find the electric field at two different points: a point on the x-axis a distance d away from the edge of the sheet, and a point on the z-axis a distance z=a above the center of the sheet.

a) To find the electric field at a point on the x-axis, we can use the principle of superposition. This means that we can break the sheet into small strips of charge and calculate the electric field at the point due to each strip, and then add them all together to get the total electric field.

Let's consider a small strip of width dx at a distance x from the origin. The linear charge density of this strip would be n*dx, since n is the surface charge density and dx is the width of the strip. The electric field at our point (x=d+L/2) due to this strip can be calculated using Coulomb's Law:

dE = k*(n*dx)*(d+L/2-x)/((d+L/2-x)^2 + a^2)

Here, k is the Coulomb's constant. We can now integrate this expression from x=-L/2 to x=+L/2 to get the total electric field at our point:

E = ∫dE = ∫ k*(n*dx)*(d+L/2-x)/((d+L/2-x)^2 + a^2)

= k*n*∫(d+L/2-x)/((d+L/2-x)^2 + a^2) dx

= k*n*ln((d+L/2-x)^2 + a^2)|-L/2 to +L/2

= k*n*ln((d+L/2-x)^2 + a^2)|-L/2 to +L/2

= k*n*(ln((d+L/2+L/2)^2 + a^2) - ln((d+L/2-L/2)^2 + a^