A cubical volume is bounded by the surfaces x = 0 , x = a , y = 0 , y = a , z = 0 , z = a . The electric field in the region is given by E → = E 0 x i ^ . Where E 0 = 4 × 10 4 NC - 1 m - 1 . If a = 2 cm , the charge contained in the cubical volume is Q × 10 – 14 C . The value of Q is ______. (Take ϵ 0 = 9 × 10 - 12 C 2 N - 1 m - 2 )

Step-by-step solution

Given here, electric field, E → = E 0 x i ^ Here, the flux passes mainly through surface areas, A B C D   and   E F G H . As the surfaces AEFB   and   CGHD are parallel to the electric field, so flux for these surfaces are zero. Again in E F G H , a = 0 , thus, electric field is zero. Hence, the flux only passes through the surface are A B C D . The net flux is ϕ net = ϕ ABCD = E 0 a · a 2 Using Gauss's law, q en ϵ 0 = E 0 a 3 ⇒ q en = E 0 ε 0 a 3 = 4 &#