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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.

Q: A Gaussian surface does not enclose a charge Does it mean that E equals 0 on its surface?

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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

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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