Question

Comment on the corrections of the following two statements based on the given situation.

Consider a line of length ‘l’ and uniform charge density λ. Assume a Gaussian surface S, a cylinder of radius R and height h.

Statement 1: φ=flux through S=λHϵ0

Statement 2: |→E|at all the points on its curved surface=λϵ02πR

True, True

True, false

False, True

False, False

Solution

The correct option is **B**

True, false

Gauss’ law is always valid i.e. the flux of charges through any closed surface is only dependent upon the charge enclosed by the surface, but if there is no symmetry then strength of electric field will be different at different points of the Gaussian surface and thus a common value at any point on the surface is impossible to evaluate.

In this situation, a cylindrical Gaussian surface is chosen. The net charge contained inside the surface=λ∗H , where λ is the linear charge density. Hence, electric flux,φ=λHϵ0 .

Now if the line would have been infinite in length, we could have claimed the same as claimed by statement 2 but the line charge is finite and therefore one should not comprehend the field intensity to be constant over the entire Gaussian surface considered. Therefore statement 2 is not correct.

Suggest corrections

0 Upvotes

View More...

View More...