(III) Uniform plane of charge. Charge is distributed...

(III) Uniform plane of charge. Charge is distributed uniformly over a large square plane of side $\ell$ , as shown in Fig. $68 .$ The charge per unit area $\left( \mathrm { C } / \mathrm { m } ^ { 2 } \right)$ is $\sigma .$ Determine the electric field at a point $\mathrm { P }$ a distance $z$ above the center of the plane, in the limit $\ell \rightarrow \infty .$ [Hint: Divide the plane into long narrow strips of width $d y ,$ and use the result of Example 11 of "Electric Charge and Electric Field"; then sum the fields due to each strip to get the total field at P.]

Key Concepts

Superposition Principle in Electrostatics

This principle states that the resulting electric field created by multiple charge elements is the vector sum of the individual fields produced by each element. It is a key concept that allows for the integration of infinitesimal contributions to determine the overall field.

Symmetry in Electrostatics

Utilizing symmetry simplifies the evaluation of electric fields. In cases involving uniform charge distributions, symmetry can indicate that certain components of the vector field cancel out, reducing the problem to calculating only the non-canceling components, which often simplifies the integration process.

Uniformly Charged Plane

This concept deals with a two?dimensional sheet where charge is distributed evenly over its surface. In the case of an infinite plane, symmetry dictates that the electric field is uniform and perpendicular to the plane, leading to a well-known result that the field is independent of the distance from the plane.

Electric Field Calculation via Integration

When computing the electric field due to a continuous distribution of charge, the method involves breaking the charged object into small elements. Each element produces a differential field, and the total field at a point is obtained by integrating these contributions over the entire distribution.

Transcript

0:00 Problem number 54.

00:02 You can see the figure on the left.

00:04 We are asked to determine the electric field the z above a large plane of charge, taking advantage of what we proved in example 21 -11 of the textbook.

00:17 So we're advised to divide the plane into long narrow strips.

00:22 Let's say with the y, as you can see here, and length l, and we can write that then dq is going to be equal to the surface charge distribution tons l, dy, being the area, right? so if this dy is really, really narrow, then we can write the charge per unit length as, you know, lambda as dq over l, right? so then if we plug in dq here, sigma l, d -y over l, then lambda is going to be sigma d -y.

01:02 So let's remember from the example 21 -11, we have the electric field, the differential of it as lambda over 2 pi epsilon, y squared plus z squared, adjusting our dimensions for this.

01:21 Problem.

01:22 So what we have is lambda is sigma d y over 2 pi epsilon not y squared plus z squared them, right? so if we look at the figure, take any arbitrary point charge within the strip, you can see that the electric field that it generates is going to have two components, right, the z component and y component.

01:56 But across this, imagine another strip here, purely symmetrical to the other one, right? and it's going to generate a de that goes in the opposite direction.

02:10 So you can see how the horizontals are actually going to cancel out.

02:14 And the only component that we're going to left with is the z component.

02:18 So then it's wise to calculate the z component of the electric field...

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