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Grade: 12
        
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
 
2 months ago

Answers : (2)

Arun
23832 Points
							
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
2 months ago
Vikas TU
10600 Points
							
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
2 months ago
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