Question

A sphere of radius a is centered at the origin. If pv = {5r¹/², 0 < r < a

{ 0, otherwise

Determine E everywhere.

Answer 1

To find the electric field at a point outside the sphere, we can use Gauss's Law. The electric field at a point outside the sphere is given by E = (a^3p)/(3ε_0r^2). To find the electric field at a point inside the sphere, we can again use Gauss's Law. The electric field at a point inside the sphere is given by E = (r^2p)/(3ε_0).

Step-by-step explanation:

To find the electric field at a point outside the sphere, we can use Gauss's Law. Since the sphere is symmetric, we can consider a Gaussian surface in the shape of a sphere with radius r, centered at the origin. The total charge enclosed by this surface is q = (4/3)πa^3p, where p is the charge density of the sphere. By applying Gauss's Law, we have E(4πr^2) = q/ε_0, where E is the electric field and ε_0 is the permittivity of free space.

Simplifying the equation, we have E = (qa)/(4πε_0r^2) = (4/3)πa^3p/(4πε_0r^2) = (a^3p)/(3ε_0r^2). Therefore, the electric field at a point outside the sphere is given by E = (a^3p)/(3ε_0r^2).

To find the electric field at a point inside the sphere, we can again use Gauss's Law. Since the sphere is symmetric, we can consider a Gaussian surface in the shape of a sphere with radius r, centered at the origin. The total charge enclosed by this surface is q = (4/3)πr^3p, where p is the charge density of the sphere. By applying Gauss's Law, we have E(4πr^2) = q/ε_0, where E is the electric field and ε_0 is the permittivity of free space.

Simplifying the equation, we have E = (qr)/(4πε_0r^2) = (4/3)πr^3p/(4πε_0r^2) = (r^2p)/(3ε_0). Therefore, the electric field at a point inside the sphere is given by E = (r^2p)/(3ε_0).