Electric field in the space is given as →E=2x^i+y^j, then charge enclosed inside cube of side length 2m as shown in figure will be
A
12ϵ0
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B
24ϵ0
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C
−6ϵ0
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D
8ϵ0
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Solution
The correct option is B24ϵ0
Net flux associatated with cube is ϕnet=ϕABCD+ϕEFGH+ ϕADEH+ϕBCFG+ϕABGH+ϕCDEF
where, ϕBCFG=E.A=2xA=2(2)A=4A ϕEFGH=E.A=yA=2A
The flux through all other surfaces is zero. ϕnet=6A
So, ϕnet=6A or ϕnet=qenϵ0 qen=6Aϵ0=24ϵ0
as, A=2×2=4