Question

A non-uniformly charged sphere of radius R has a total charge Q. The electric field inside this charge distribution is described by E=Emax(r4 /R4 ), where Emax is a known constant. Using the differential form of Gauss’s law, find volume charge density as a function of r. Express your result in terms of r, R and Emax.

Answer 1

Answer 2

To find the volume charge density (ρ) as a function of r for the non-uniformly charged sphere, we'll start with Gauss's law in differential form, which states:

∇ · E = ρ / ε₀

where ∇ · E is the divergence of the electric field E, ρ is the volume charge density, and ε₀ is the permittivity of free space.

Given the electric field inside the charged sphere as:

E = Emax * (r^4 / R^4)

Now, we need to find the divergence of the electric field (∇ · E). Since the electric field is spherically symmetric, we can write it in terms of the radial coordinate (r) only:

E = E(r) * r̂

where r̂ is the radial unit vector.

The divergence in spherical coordinates is given by:

∇ · E = (1/r^2) * (∂ / ∂r) (r^2 * E(r))

Substituting the expression for E(r) into the divergence formula:

∇ · E = (1/r^2) * (∂ / ∂r) (r^2 * Emax * (r^4 / R^4))

∇ · E = (1/r^2) * Emax * (∂ / ∂r) (r^6 / R^4)

∇ · E = (1/r^2) * Emax * (6r^5 / R^4)

Now, equating this to ρ / ε₀, we get:

(1/r^2) * Emax * (6r^5 / R^4) = ρ / ε₀

Solving for ρ:

ρ = (6ε₀ / Emax) * (r^5 / R^4)

Expressing the result in terms of r, R, and Emax, the volume charge density (ρ) inside the non-uniformly charged sphere is:

ρ(r) = (6ε₀ / Emax) * (r^5 / R^4)