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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

Arun
25763 Points
one year ago
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
Vikas TU
14149 Points
one year ago
Dear student
The length of Rod = L
Side of cube = L
From gauss law
Phi = q/epselon
As half length is inside cube
Charge enclosed by cube = Q/2
phi = Q/2*epselon