Electric field in a region is given by E→=−4xi^+6yj^N/C. If the charge enclosed in the cube of side 1 m as shown in the diagram is found to be n∈0C, find the value of n.

From the given function of electric field strength, we can see that at planex:0 and the plane y = 0, E = 0 thus no flux will enter in the cube through these two planes but flux will come out of the cube from the planes x = 1 and y = 1 which is given asϕout =−4(1)(1)2y=1+6(1)⋅(1)2x=1⇒ϕout =2V−mUsing Gauss's law we can find the net enclosed charge in the cube. If q is the charge then we use for the cube surfaceϕout =qϵ0⇒q=2ϵ0