No.there can be electric field on the Gaussian surface even if the charge enclosed by it is zero.However ,net flux will be zero through the surface.
The question does not specify what z is so this answer could be completely wrong. However, if the question is in the context of the standard Gaussian distribution, the answer is 0.493613 (to 6 dp)
0.5m-1
Perhaps if you read the question properly, you would not have to ask the question!
That's a Gaussian distribution.
Surface area of a sphere: A = 4πr2So, surface area = 4pi52
The question does not specify what z is so this answer could be completely wrong. However, if the question is in the context of the standard Gaussian distribution, the answer is 0.493613 (to 6 dp)
divide 0.108 gm by 1 coulomb of charge because e.c.e equals mass in gm/charge
Surface area equals 864cm2
127.5
0.5m-1
Perhaps if you read the question properly, you would not have to ask the question!
0.6 m-1 is the ratio of surface area to volume for a sphere.
The energy density at the surface of the sphere is given by the formula U = 1/2 * ε0 * E^2, where ε0 is the permittivity of free space and E is the electric field at the surface of the sphere. The electric field at the surface of a conducting sphere is E = σ / ε0, where σ is the surface charge density. Given σ = Q / A, where A is the surface area of the sphere, and Q = 1.1nC, we can calculate the energy density at the surface by substituting these values into the formula for U.
That's a Gaussian distribution.
No.
0.4 m-1 is the ration of surface area 588m2 to volume 1372m3 for a sphere.
Surface area of a sphere: A = 4πr2So, surface area = 4pi52