Question

A hemispherical surface of radius r is located in a uniform electric field that is parallel to the axis of the hemisphere what is the magnitude of the electric flux through the hemisphere surface

Answer 1

1. Question⇒   A hemispherical surface of radius r is located in a uniform electric field that is parallel to the axis of the hemisphere what is the magnitude of the electric flux through the hemisphere surface?

Explanation:

• Since number of field lines entering from circular side is equal to number of field lines leaving the hemispherical surface, net flux is 0.
• therefore (Flux)h + (flux)c = 0 ——→ 1[* (flux)h means flux from hemispherical surface and (flux)c means flux from circular surface]
• Therefore, (flux)h = - (flux)c ———→ 2
• Flux = E.A [ Where E and A are vectors ]
• Flux = E A cos θ
• (Flux)c = EA cos π [ Because field and surface area vector are anti parallel]
• (flux)c = E πR^2 (-1) ———→ 3 [ Because area of circular surface is πR^2 ]
• From equation 2 and 3
• (flux)h = - (- EπR^2)
• => (Flux)h = EπR^2
• Hope this helped !

Answer 2

Answer:

πR2E

Explanation:

As the field intensity E is parallel to the axis of the circular plane so only this circular surface will contribute the flux through the hemisphere.

By Gauss's law, flux , ϕ=∫E.ds=E.πR2