A spherical charge distribution given byρ = ρ0 (1 −r2a2) if r ≤ aρ = 0 if r > aIf a is radius of sphere. Find the electric field at a distance r from the center when r>a and r<a. Alsoshow that electric field will be maximum when r =√5 a3

The problem involves a spherical charge distribution and we are asked to find the electric field at a distance r from the center for both r>a and r<a. We are also asked to show that the electric field will be maximum when r = √5 a/3.

Step 1: Gauss's Law We will use Gauss's law to solve this problem. Gauss's law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity of free space.

Step 2: Electric Field for r<a For r<a, we consider a Gaussian surface of radius r inside the sphere. The charge enclosed by this surface is obtained by integrating the charge density over the volume of this smaller sphere:

Q = ∫ρ dV = ∫ρ0 (1 - r^2/a^2) dV = 4πρ0 ∫r^2 (1 - r^2/a^2) dr from 0 to r = 4πρ0 [r^3/3 - r^5/5a^2] from 0 to r = 4πρ0 [r^3/3 - r^5/5a^2]

The electric field E inside the sphere is then given by Gauss's law:

E = Q / 4πε0r^2 = ρ0 [r/3ε0 - r^3/5ε0a^2]

Step 3: Electric Field for r>a For r>a, the Gaussian surface includes the entire sphere, so the total charge Q is just the total charge of the sphere:

Q = 4πρ0 ∫r^2 (1 - r^2/a^2) dr from 0 to a = 4πρ0 [r^3/3 - r^5/5a^2] from 0 to a = 4πρ0 [a^3/3 - a^5/5a^2] = 4πρ0 [2a^3/15]

The electric field E outside the sphere is then given by Gauss's law:

E = Q / 4πε0r^2 = ρ0 [2a^3/15ε0r^2]

Step 4: Maximum Electric Field To find the maximum electric field, we take the derivative of E with respect to r and set it equal to zero. This gives us the value of r for which the electric field is maximum. For r<a, the derivative is complicated, but for r>a, we find:

dE/dr = -2ρ0 [2a^3/15ε0r^3] = 0

Solving for r gives r = √5 a/3, as required.