Question

One-fourth of a sphere of radius R is removed
as shown in fig. An electric field E exists
parallel to x-y plane. Find the flux through the
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Answer 1

your complete question is ----> One fourth of a sphere of radius R is removed as shown in figure . an electric current E exists parallel to the xy plane . Find the flux through the remaining curved part...

A very important concept : Electric flux through closed surface is zero.

i.e., electric flux through curved surface + electric flux through plane surface = 0

see diagram, there are two plane surface and each plane is half of circle of radius R.

so, area of each plane surface is πR²/2

now, electric flux through plane surface parallel to x-axis, $\phi_1$ = (πR²/2)(-j).Ecos45°(j) = -πR²E/2√2

similarly, electric flux through the plane surface parallel to y-axis , $\phi_2$ = (πR²/2)(-i).Esin45° i = -πR²E/2√2

then, $\phi_1+\phi_2$ + electric flux through remaining curved part = 0

or, -πR²E/2√2 - πR²E/2√2 + electric flux through remaining curved part = 0

so, electric flux through remaining curved part = πR²E/√2

Answer 2

Answer:

Electric flux through closed surface is zero.

electric flux through curved surface + electric flux through plane surface =0

There are two plane surface and each plane is half of circle of radius R.

so, area of each plane surface is πR²/2

now, flux through plane surface parallel to x-axis,  = (πR²/2)(j).Ecos45°(j) =                        -πR²E/2√2

flux through the plane surface parallel to y-axis ,  = (πR²/2)(-i).Esin45° i = -πR²E/

+ electric flux through remaining curved part = 0

or, -πR²E/2√2 - πR²E/2√2 + electric flux through remaining curved part = 0

Hence electric flux through remaining curved part = πR²E/√2