A uniform electric field E = 3 i + 6 j + k passes through a closed cuboidal surface. One face of the cuboid has an area 4 m^2 and an outward unit normal given by 2 i +2 j +3 k 17 . If the electric flux through the remaining 5 faces is zero, the charge enclosed by the cuboid is:

Step-by-step solution

The electric flux through the given face of the cuboid is calculated using the dot product of the electric field vector and the area vector. The area vector is given by A = A n , where A = 4 and n = 2 i + 2 j + 3 k 17 . A = 4 ( 2 i + 2 j + 3 k 17 ) = 8 i + 8 j + 12 k 17 The flux through this face is: ₁ = E A = (3 i + 6 j + k ) ( 8 i + 8 j + 12 k 17 ) ₁ = (3 8) + (6 8) + (1 12) 17 = 24 + 48 + 12 17 = 84 17 It is given that the electric flux through the remaining 5 faces is zero. Therefore, the total electric flux th