Solve this:
2. Electric field at point P is given by  E = rE 0 . The total flux through  the  given cylinder of radius R and height h is

a .   E 0 π R 2 h b .   2 E 0 π R 2 h c .   3 E 0 π R 2 h d .   4 E 0 π R 2 h

Dear student
Let us consider a small area dA=2πRdzcurved surface area in vector form is dA=2πRdz(ur^) where ur^ is a unit radial vectorElectric field in vector form isE=E0(zk^+Ru^r)electric flux linked isdϕ=E0(zk^+Ru^r).(2πRdz(ur^))=E0×2πR2dzintegrating both sidesdϕ=0HE0×2πR2dzϕ=E0×2πR2HRegards

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