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Question

Consider a cube of side a=0.1 m placed such that its six faces are given by equation x=0,x=+a,y=0,y=+a,z=0 and z=+a. If its placed in a electric field given by E=x2^i+y^j N/C, the electric flux crossing out of the cube in the unit of 104Nm2/C is

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Solution


Flux through any surface us given by ϕ=E.dA

Flux through face ABCD(x=0,y=0 to a)E=y^jdA=dA(^i)dϕ=E.dA=0 (^j.^i=0)ϕABCD=0Flux through face EFGH(x=a,y=0 to a)E=a2^i+y^jdh=dA^idϕ=E.dA=a2dAϕEFGH=a4=104 Nm2C1Flux through face CDGH(y=0,x=0 to )E=x2^idA=dA(^j)dϕ=0(^i.^j=0)ϕCDGH=0Flux through face ABFE(x=0 to a,y=a)E=x2^i+a^jdA=dA^jdϕ=E.dA=adAϕABFE=adA=a3=103 Nm2C1Flux through face AEGC and BFHD (x=0 to a,y=0 to a)E=x2^i+y^jdA=dA(±^k)dϕ=E.dA=0 (^i.^k=^j.^k=0)ϕAEGC=ϕBFHD=0Total flux through cube ϕnet=ϕEFGH+ϕABFE=104+103=11×104 Nm2C1

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