Delta function integral and deriving Coulomb's law from Gauss's law

[jlmac2001]

1. What is (integral over all space) delta^3(r-r0)r^2d^3r where d^3r=dxdydz=dV and r0=(0,1,2)?

answer: I'm not quite sure what the intergral over all space means.How can r0 be integrated if it's just given as points?

2. Obtain Coulomb's law for the electric field of a point charge from Gauss's law, by considering a spherical surface S of radius r with a point charge +q at its center.

answer: I don't know how to do this. Can someone help? Please!

 

[HallsofIvy]

Originally posted by jlmac2001
1. What is (integral over all space) delta^3(r-r0)r^2d^3r where d^3r=dxdydz=dV and r0=(0,1,2)?

answer: I'm not quite sure what the intergral over all space means.How can r0 be integrated if it's just given as points?

2. Obtain Coulomb's law for the electric field of a point charge from Gauss's law, by considering a spherical surface S of radius r with a point charge +q at its center.

answer: I don't know how to do this. Can someone help? Please!

1. r0 can't be integrated- its a constant! The integration is over space. delta(r-r0) is DEFINED as having integral, over all space, equal to 1. r0= (0, 1, 2) but that's really not relevant. The integral is still 1. More important would be integral delta^3(r- r0)f(x,y,z) dV= f(0,1,2).

 

[M98Ranger]

For question 1, the integral over all space means integrating over all three dimensions (x, y, and z) to cover the entire space. In this case, it is asking for the integral of the function delta^3(r-r0)r^2 over all space, where r0 is the point (0,1,2). This integral can be solved by first finding the limits of integration, which would be from -infinity to +infinity for all three dimensions. Then, solving the integral would involve plugging in the limits and solving for the value.

For question 2, Coulomb's law states that the electric field of a point charge is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge. Gauss's law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of the medium. By considering a spherical surface S with a point charge +q at its center, we can use Gauss's law to derive Coulomb's law. The electric flux through the surface S is equal to the charge enclosed by the surface, which in this case is +q. Therefore, using Gauss's law, we can write:

Electric flux = Enclosed charge / permittivity of the medium

E * 4πr^2 = q / ε

Solving for the electric field E, we get:

E = q / (4πεr^2)

This is the same as Coulomb's law, where q is the charge, ε is the permittivity of the medium, and r is the distance from the charge.