Given a charge distribution that is spherically symmetric and has a charge density defined by ρ=

step 1 So in this question we are given the volume charge density. You can see this is the volume charg

Given a charge distribution that is spherically symmetric...

Key Concepts

Gauss's Law

Gauss’s Law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the total charge enclosed by that surface. In problems with high symmetry, like spherical symmetry, it provides a powerful method to calculate the electric field by reducing the complexity of the integral over the electric field vectors.

Spherical Symmetry

Spherical symmetry indicates that the properties of the charge distribution and the resulting electric field depend only on the radial distance from the center and not on the angular coordinates. This symmetry allows one to use spherical Gaussian surfaces, making the calculations more straightforward since the electric field has the same magnitude at every point on the surface.

Continuous Charge Distribution

A continuous charge distribution describes a charge spread over a volume, area, or line, instead of being concentrated at discrete points. The charge density function specifies how the charge is distributed in space, and integrating this density over the volume of interest yields the total charge enclosed, which is essential for applying Gauss's Law.

Integration in Spherical Coordinates

Integration in spherical coordinates is used to evaluate integrals over volumes where the variables are expressed in terms of radius, polar angle, and azimuthal angle. This method aligns well with spherically symmetric problems, allowing for the effective calculation of quantities like the total charge enclosed by integrating the charge density times the spherical volume element.

Transcript

00:01 So in this question we are given the volume charge density.

00:05 You can see this is the volume charge density when radius goes from zero to capital r.

00:12 And this is the volume charge density when radius goes from capital r to infinity.

00:20 So which implies charge is only confined from zero to capital r.

00:24 So this is a spherical symmetry.

00:26 So how do we generally solve this? we generally find flux then after what we do we find the charge enclosed and after that what we do we simply apply flux is equal to 1 by epsilon times charge enclosed and after that we get the value of electric field so step number one find the flux step number two find the charge enclosed and then step number three apply gas law you will get the electric field so let us begin the discussion.

01:03 So to get the flux, we can assume a gaussian surface, definitely a gaussian surface of radius r.

01:12 There would be flux.

01:15 There would be flux coming out of you.

01:19 And that flux, we know, is equal to e .t.

01:24 D .s integration on the close surface.

01:29 We know the area vector and electric will have perpendicular which implies e will come out of you know integration because e .s.

01:40 Is e .d .s.

01:41 Is equal to eds cause zero which is eds.

01:44 E is constantly will come outside.

01:47 Ds integration over the glue surface will give us the surface area of the square which is a to four by r square.

01:55 So we have done the first step.

01:59 We have calculated the flux which is equal to electric field into 4 pi r square equation number one second step is find the charge enclosed charge enclosed for a spherical symmetry is always equal to row into db integration row is a charge density and what is d b db in case of spherical symmetry it is always equal to 4 pi r square into d r so this is the differential volume basically all right so now we can substitute the value of row here so charge enclosed is equal to we can put the value of row we know the row value of row inside is equal to what is equal to row not times capital r divided by smaller so this value of row is what i'm substituting here so and we can write 4 pi r squared d r all right you can see one r will disappear from you so what are you going to get from here you're going to get ro not 4 pi capital r this value row not 4 pi capital r so what is left here r is left all right so r dr integration so it goes from zero to capital, zero to small r because we are considering the gaussian surface inside the sphere.

03:38 Alright, so the charge enclosed we can get very easily.

03:43 Charge enclosed is equal to row not 4 by r and this would be definitely r square by two because it is r square by two, even goes from zero to r so this is a very simple mac calculation.

04:03 So i have written this charge enclosed.

04:05 So now what we can do is we can apply goslow...

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