A very large thin plane has uniform surface charge density σ. Touching it on the right (Fig. 45)

A very large thin plane has uniform surface charge density...

A very large thin plane has uniform surface charge density $\sigma .$ Touching it on the right (Fig. 45$)$ is a long wide charge density $\rho_{\mathrm{E}} .$ Determine the electric field (a) to the left of the plane, (b) to the right of the slab, and $(c)$ everywhere inside the slab.slab of thickness $d$ with uniform volume charge density $\rho_{\mathrm{E}} .$ Determine the electric field $(a)$ to the left of the plane, $(b)$ to the right of the slab, and $(c)$ everywhere inside the slab.

Key Concepts

Gauss's Law

Gauss's Law is an essential principle in electromagnetism stating that the total electric flux through a closed surface is proportional to the enclosed charge. It is particularly useful in problems with high symmetry, allowing for the calculation of electric fields using an appropriately chosen Gaussian surface that simplifies the integration of the electric field over the surface.

Superposition Principle

The superposition principle in electrostatics states that the net electric field created by multiple charge distributions is the vector sum of the fields due to each individual charge distribution. This concept is crucial when calculating the overall field resulting from both a surface charge and a volumetric charge, as each contributes independently to the total field.

Electric Field Due to an Infinite Plane of Charge

An infinite plane with uniform surface charge density produces an electric field that is constant in magnitude and direction at all points in space (ignoring edge effects), with the field magnitude given by ?/(2??). This result is derived using symmetry and Gauss's Law, and it serves as a fundamental model in electrostatics.

Electric Field Inside and Outside a Uniformly Charged Slab

A uniformly charged slab generates different electric field profiles inside and outside its volume. Inside the slab, the field varies linearly with distance from the center due to the gradual increase in enclosed charge, while outside, the field behaves as if all the charge were concentrated at the center, often leading to discontinuities at the boundaries. Calculating the field in these regimes is a classic application of Gauss's Law and symmetry arguments.

Transcript

00:01 So we have a non -conducting material which has charge density row and we have a conducting material with surface charge density sigma.

00:16 Now we have as a gaussian surface, let's consider a cylinder that looks like this.

00:28 Before we go ahead, remember that as an infinite plane or charge has field that is perpendicular to it's perpendicular to it because of symmetry.

00:40 Similarly, in this case, we expect the charge to be horizontal.

00:46 Also, notice that because of this charge sigma, the field on both directions will be on both sides of it will be the same.

00:55 And similarly, the field due to this charge distribution row will also be the same on both of its side, which means the charge, the electric field will be symmetric across this charge distribution.

01:11 Now let's find the electric field on the left side.

01:21 So let's say the electric field here is e.

01:23 We expect the electric field to be e over here too.

01:26 Now this is our origin zero.

01:32 Now the width of this non -conducting charge distribution is d.

01:39 So, and we know that the electric field is perpendicular to this curved surface.

01:45 So the flux there is zero.

01:46 Hence the flux will only will only be.

01:47 Be through the two circular surfaces that is two five circular and this will be equal to charge enclosed divided by epsilon not now here the charge enclosed has two contributions one from the surface charge density and one from the volume charge density the surface charge density will give you a charge of sigma a where a is the area of this cylindrical surface of the circular surface plus we have charge distribution charge from the volume distribution that will correspond to row times volume of this cylinder which will be d into a divided by epsilon not now this is equal to e into the flux which is now we know that this is equal to two into five through the circular surface and in order that the flux through circular surface will simply be e times the area so that is equal to sigma a plus row d a by epsilon not, which means e will be equal to sigma plus row d by 2x0.

02:59 Now for part b, we want to find the field on the other side and using the symmetry condition, as we've already argued, the electric field will be the same except for the direction.

03:13 In the first case, it will be leftwards whereas in the second case, it will be rightwards...

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