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a charge Q is uniformly distributed over a rod of Length 2l. considerncube of edge l with the centre of cube at one end of the rod. the min. possible electric flux through the surface of cube
Given length of rod = edge of cube = ℓ Portion of rod inside the cube = ℓ/2 Total charge = Q. Linear charge density = λ = Q/ℓ of rod. We know: Flux α charge enclosed. Charge enclosed in the rod inside the cube. = ℓ/2ε0 × Q/ℓ = Q/2ε0
Dear student The length of Rod = L Side of cube = LFrom gauss law Phi = q/epselonAs half length is inside cube Charge enclosed by cube = Q/2 phi = Q/2*epselon
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