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Q.

In a uniform electric field, a cube of side 1 cm is placed. The total energy stored in the cube is  8.85μJ. The electric field is parallel to four of the faces of the cube. The electric flux through any one of the remaining two faces is k2  SI units. The value of k is

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Detailed Solution

Energy = 12ε0E2×volume
8.85×10−6=12×8.85×10−12E2(10−6)E=2×106 V/mflux  (ϕ)=EA=2×10+6×10−4=1002(V−m)

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