Question

The electric field componets due to a charge inside the cube of side 0.1m are `E_(x) = alpha x,` where `alpha = 500 (N//C) m^(-1)`,
`E_(y) = 0, E_(z) = 0`. Calculate the flux through the cube and the charge inside the cube.

Answer 1

As in clear from Fig. flux is onloy along X-axis.

Through the left face
`phi_(1) = E_(x) . A cos 180^(@)` v
`= 500xx0.1xx10^(-2) (-1) = -0.5`
Through the right face
`phi_(2) = E_(x) . A cos 0^(@)`
`= 500 (0.2) xx10^(-2) xx1 = 1.0`
`:.` Net flux through the cube
`phi = phi_(1) + phi_(2) = -0.5 + 1 = 0.5 Nm^(2) C^(-1)`
Charge inside the cube `= in_(0) phi`
`= 8.85xx10^(-12) xx 0.5 = 4.425xx10^(-12)C`.