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Electric field in a region is given by `vec(E)=-4xhat(i)+6yhat(j)`. The charge enclosed in the cube of side `1 m` oriented as shown in the diagram is given by `alpha in_(0)`. Find the value of `alpha`.

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The correct Answer is:
2

`phi=(6y) "Area"-(4x) "Area" =6xx1xx(1)^(2)-4xx1xx(1)^(2)=2` therefore `q/epsilon_(0)=2 rArr q=2epsilon_(0)`
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