A hemispherical surface of radius r is located in a uniform electric f

1.	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» 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 !

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»

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 !