In Fig, electric field is directed along `+X` direction and is given by `E_(x) = 5 Ax + 2 B`, where E is in `NC^(-1)` and x is in meter, A and B are constants having dimensions. Taking `A = 10 NC^(-1) m^(-1) and B = 5 NC^(-1)`, calculate (i) the electric flux through the cube and (ii) net charge enclosed within the cube.

Explanation

(i)The electric field is along X-axis, thus electric flux through all the surfaces except M and N will be zero
Electric field through surface M
`E_M=5Ax+2B`
`rArr E_M=5xx10xx0+10=10 N C^(-1)`
Electric flux through surface M
`phi_M=E_MA cos 180^@ =10xx(0.1)^2xx(-1)`
`=-0.1 N m^2 C^(-1)`
Electric field through surface N
`E_N=5Ax+2B`
`rArr E_N = 5xx10xx0.1+10=15 N C^(-1)`
Electric flux through surface N
`phi_N=E_N A cos 0^@ =15xx(0.1)^2xx(1)`
`=0.15 N m^2 C^(-1)`
Net flux through the cube is 0.15 -0.1
`=0.05 N m^2 C^(-1)`
(ii) Net charge enclosed within the cube is
`q=phiepsilon_0`
`rArr q=0.05xx8.85xx10^(-12) C =0.4425xx10^(-12)` C