Flux through a Cube (Eigure 1) A cube has one corner at the origin and the opposite corner at the point (L, L, L). The sides of the cube are parallel to the coordinate planes

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(A) The electric flux through left surface is, ϕ1=E→⋅A→ =((a+bx)i^+cj^)⋅(L2(−i^)) =−L2(a+bx) For left surface x=0. ϕ1=−aL2 Th electric flux through right surface is, ϕ2=E→⋅A→ =((a+bx)i^+cj^)⋅(L2(i^)) =L2(a+bx) For right surface x=L ϕ2=L2(a+bL) =aL2+bL3 The electric flux through top surface is, ϕ3=E→⋅A→ =((a+bx)i^+cj^)⋅(L2(j^)) =cL2 The electric flux through bottom surface is, ϕ4=E→⋅A→ =((a+bx)i^+cj^)⋅(L2(−j^)) =−cL2 Since there is no field along z-direction, the elcetric flux through front and back surface of the the cube must be zero. The total electric flux through surface of the cube is, ϕE=ϕ1+ϕ2+ϕ3+ϕ4 =−aL2+aL2+bL3+cL2−cL2 =bL3 Thus, the electric flux through surface of the cube is bL3 (C) According to Gauss law, ϕE=qϵ0 q=ϵ0ϕE q=ϵ0bl3 Thus, the net charge inside the cube is ϵ0bL3

Flux through a Cube (Eigure 1) A cube has one corner at the origin and the opposite corner at the point (L, L, L). The sides of the cube are parallel to the coordinate planes

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