Question

a nonuniform, but spherically symmetric, distribution of charge has a charge density rho(r) given as follows:rho(r)=rho0(1−4r/3r)for r≤rrho(r)=0for r≥rwhere r and rho0 are positive constants.

Answer 1

The question requires using Gaussian surfaces and integrating the given nonuniform charge density to find the electric field inside and outside a spherically symmetric charge distribution.

Step-by-step explanation:

The problem presents a scenario involving a nonuniform charge density within a spherically symmetric object. To calculate the electric field at any point in relation to this object, one needs to employ the concept of a Gaussian surface. This mathematical construct helps in applying Gauss's Law, which relates the electric field to the charge enclosed within the Gaussian surface. For a spherical charge distribution, a spherical Gaussian surface is ideal. Charge density here is given by ρ(r) = ρ0(1 − 4r/3R) for r ≤ R and ρ(r) = 0 for r > R, which suggests that charge density decreases linearly with distance from the center up to a distance R.

For regions inside this distribution (r ≤ R), the integration of the charge density over the volume enclosed by the Gaussian surface must be performed. Outside the charge distribution (r > R), the electric field is the same as if all the charge Q were concentrated at the center of the sphere. The total charge Q can be found by integrating the charge density ρ(r) over the volume of the sphere with radius R.

In summary, this question exemplifies physical principles of electrostatics and how they apply to nonuniform, but spherically symmetric, charge distributions.