a) To find the net charge on the sphere, we first need to find the total area of the surface. The surface area of a sphere is given by:
A = 4πr^2
where r is the radius of the sphere. In this case, the sphere has a diameter of 1.2 m, so the radius is 0.6 m. Therefore, the surface area of the sphere is:
A = 4π(0.6 m)^2 = 4.52 m^2
The total charge on the surface can be found by multiplying the charge density by the surface area:
Q = σA = (8.1 µC/m^2) (4.52 m^2) = 36.6 µC
Therefore, the net charge on the sphere is 36.6 µC.
b) The electric flux emanating from the surface of the sphere can be found using Gauss's Law, which states that the electric flux through any closed surface is proportional to the charge enclosed within the surface.
Since the sphere has a uniform surface charge density, we can assume that the charge is evenly distributed throughout the entire surface. Therefore, we can imagine a hypothetical sphere of radius r < 0.6 m concentric with the original sphere, and use this to find the electric flux through the surface.
The electric field at a distance r from the center of a uniformly charged sphere is given by:
E = kQr / R^3
where k is Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q is the total charge on the sphere, and R is the radius of the sphere. Using the values Q = 36.6 µC and R = 0.6 m, we can solve for the electric field at a radius r:
E = (8.99 x 10^9 Nm^2/C^2)(36.6 x 10^-6 C) / (0.6 m)^3 * r^2
The electric flux through a spherical surface of radius r is given by:
Φ = E A = E 4π r^2
Substituting the expression for E, we get:
Φ = (8.99 x 10^9 Nm^2/C^2)(36.6 x 10^-6 C) / (0.6 m)^3 * 4π r^2
Simplifying and cancelling terms, we get:
Φ = 4.27 x 10^5 Nm^2/C
Therefore, the total electric flux emanating from the surface of the sphere is 4.27 x 10^5 Nm^2/C.