A spherically symmetric charge distribution has a charge...

A spherically symmetric charge distribution has a charge density given by $\rho=a / r,$ where $a$ is constant. Find the electric field within the charge distribution as a function of $r .$ Note: The volume element $d V$ for a spherical shell of radius $r$ and thickness $d r$ is equal to $4 \pi r^{2} d r .$

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. This law is particularly effective in calculating electric fields for charge distributions with high symmetry, such as spherical symmetry, by reducing complex integrals to simpler expressions.

Spherical Symmetry

Spherical symmetry implies that the physical properties of a system, such as the charge density and electric field, depend solely on the radial distance from the center, not on the angular coordinates. This symmetry allows for significant simplifications in the mathematical treatment of the problem, including the use of spherical Gaussian surfaces.

Volume Element in Spherical Coordinates

The volume element in spherical coordinates, expressed as 4?r²dr for a thin spherical shell, is crucial for integrating quantities like charge density over a spherical volume. This element accounts for the increasing surface area of spherical shells as the radius increases and is essential in computing the total enclosed charge.

Integration of Charge Density

Integration of the charge density over a specified volume is necessary to determine the total charge enclosed within a given radius. When the charge density is provided as a function of the radial coordinate, integrating it with the appropriate volume element allows one to apply Gauss's Law and calculate the resultant electric field.

Transcript

00:01 Okay, so we have a spherically symmetric charge distribution and we have to calculate the electric flux in this charge distribution.

00:11 We know that using the electric flux we can say that the close integral of the field, the electric field da is going to be equal the charge inside the gaussian surface divided by absolute so on zero.

00:33 Okay, so since the electric field is constant, we can say that this is going to be e that multiplies the integral of the area, which is the surface area of this sphere.

00:51 So it's 4 pi r square.

00:55 This is the gaussian surface.

00:56 Remember, this is going to be equal the charge inside the gaussian surface and the charge inside the gaussian surface can be described as the density dv actually the integral of the density dv divided by epsilon zero so what is this integral this is the integral of the volume therefore we can say this is going to be one moment let's put in here we have e that multiplies four r square that is equal 1 divided by epsilon 0 the close the integral not close integral of the density and the problem says that the density is just a divided by r and the volume the volume is just the area let's put in here the volume dv is just the area multiply by the r therefore the surface area this is going to be just 4 pi r square d r and the radius goes from 0 to r therefore we just need to calculate this so let's continue here we have that the electric field is 4 pi r square is going to be equal 1 divided by epsilon 0 a goes out of the integral so this is just a divided by epsilon 0 we have 4 pi in here integral of r d r from 0 to r therefore let me see this is going to to be let's continue here it is going to be e 4 pi r square it's going to be equal 4 pi a epsilon 0 r squared divided by 2 from r to 0 so this is just 4 pi a divided by 2, absalom 0, r...

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