An infinite slab of charge of thickness 2𝒵₀ lies in the xy-plane...

Question

An infinite slab of charge of thickness 2𝒵₀ lies in the xy-plane between 𝒵 = -𝒵₀ and 𝒵 = +𝒵₀ . The volume charge density p (C/m³) is a constant. (b) Find an expression for the electric field strength above the slab (𝒵 ≥ 𝒵₀).

Accepted Answer

Hey, everyone. Let's go through this practice problem. A uniformly charged plate having infinite dimensions in the YZ plane and a thickness of four D from X equals negative two D to X equals positive two D exhibits a constant volume charge density of row with units of coons per cubic meter. Using Gauss's law, derive an expression for the electric field strength at a position X greater than or equal to two D from the plate. We have four multiple choice options. A row D divided by four multiplied by epsilon knot with the I hat unit vector. Option B with the same I hat unit vector. But now it's two row D divided by epsilon knott, option D four row D divided by epsilon knot and option D row D divided by two epsilon knot. All of them have the same I hat unit vector. So pointing along the X direction, first off note that the problem specifically asks us to use Gauss's Law and recall that Gauss's law states that the surface integral of the electric field. So E dot D A is equal to the charge enclosed within some Gaussian surface divided by epsilon. Nas in order to use Gaussian's law, we first have to set up the problem and understand what's going on here and choose something to be our Gaussian surface. So let's take a look at what's happening here. We have a uniformly charged plate and we know it has a thickness of four D. So I'm going to draw kind of a side view of the plates. So this is kind of what a side view would look like. Uh that it stretches out into infinity and has a thickness of four D since we're looking at it at a side of you, and the plate is entirely in the YZ plane. That means that the X axis is going to be what's perpendicular to it. The diagram I've drawn here, the problem asks us to use Gauss's law to estimate the field strength at a distance outside of the plate. So we can choose our Gaussian surface to be something that kind of encloses around the walls of the plate. So something like this and note that because the plate is charged, there are electric field lines coming out of the surface of the plate. So they're gonna be some pointing down like this and they're gonna be some pointing up like this because this is a conductor, there are going to be no electric field lines within the plate. So there aren't gonna be any field lines passing through these side walls here. So notice that with the way we've set up our Gaussian surface. The electric field lines point straight through perpendicular to the walls of the Gaussian surface, which means that the left hand side of our Gauss's law formula. This complex dot product surface integral can be more simply written as a sum of the products of the electric field strength and the surface area of the, of the the of the walls of the Gaussian surface. So E A top, for example, plus E A bottom and that's what's equal to this and closed, divided by epsilon. Not term, I'm gonna write this further underneath the diagram for room. And since the surface areas of the two sides of the Gaussian surface are going to be the same area. That means that A is that the A S are the same, which means that we can write the left hand side of the Gauss's law formula as simply two multiplied by E A and that's equal to the enclosed charge divided by epsilon knot. We don't know the magnitude of the enclosed charge, especially since our Gaussian surface was drawn more or less arbitrarily. But we can relate the magnitude of that enclosed charge to the surface dens to the volume charge density of the plate, which we were given as a in the problem as being represented by row. So that means that the enclosed charge is going to be equal to the charge density multiplied by the volume of the conductor enclosed by the Gaussian surface. So row multiplied by whatever that volume is divided by epsilon knot. Now, the volume isn't given to us either. But we do know that the volume of this, of this uh rectangular cubic area here is going to be equal to the thickness of that area four D multiplied by the cross sectional surface area of it which we've discussed as being a. But what you'll notice now is that we have an a on both sides of the equation which will cancel out. So two E equals row four D divided by epsilon knot. And we can also simplify this two and this four as we can divide them both by two. And so we find E as being equal two, just two row D divided by epsilon knots. And so that is the magnitude of the, of the E field that is the magnitude of the strength of the electric field outside of the plates. Of course, all the multiple choice options have this written as a vector. But since we discussed, since we showed in the diagram that these field lines are all pointing along the x axis, that means it makes sense for us to write out this expression as having the I hat unit vector because it is along the X axis. And so that is our answer to this problem. And if we look at our multiple choice options, we can see that this expression we have found agrees with option B is the problem. So option B is the correct answer to this problem and that is it for this problem. I hope this video helped you out and please consider checking out some of our other tutoring videos which will give you more experience with these types of problems, but that's all for now and I hope you all have a lovely day. Bye bye.

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An infinite slab of charge of thickness 2𝒵₀ lies in the xy-plane between 𝒵 = -𝒵₀ and 𝒵 = +𝒵₀ . The volume charge density p (C/m³) is a constant. (b) Find an expression for the electric field strength above the slab (𝒵 ≥ 𝒵₀).

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