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!
Surface area of a sphere: A = 4πr2So, surface area = 4pi52
That's a Gaussian distribution.
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
the point would be the kolwea-4 where the positive charge is q3. xoxoxo
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 a charged conductor is the surface charge density squared , divided by 2 x the permittivity of free space. The surface charge density is the charge divided by the area it sits on. So if, e = permittivity = 8.85 x 10^-12 CC/Nmm and D = surface charge density, and U = energy density and R = radius of sphere and q = charge on sphere, then; U = (1/2e) x D^2 where D = q/4piR^2 = 1.1 x 10^-9/(4 x 3.14 x 1) = 8.76 x 10^-11 , where 4piR^2 is the surface area of a sphere. So; D^2 = 76.7 x 10^-22 then ; U = (76.7 x 10^-22)/(17.7 x 10^-12) = 4.33 x 10^-10 Joules/mmm
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