In Fig, electric field is dirceted 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 dimenstions. 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

Here, `A = 10 NC^(-1) m^(-1), B = NC^(-1)`
`E_(x) = 5 Ax + 2B`
As electric field is directed along `=q` direction along `+x` direction, therefore electric flux is asscociated with faces P and Q only.
Area of each face, `ds = 10xx10 cm^(2) = 10^(-2) m^(2)`.
For face `P : x = 0`
From (i) `E_(x) = 2B = 2xx5 = 10 NC^(-1)`,
Electric flux through face `P , phi_(1) = E_(x) ds cos 180^(@) = 10xx10^(-2) xx (-1) = 0.1 NC^(1) m^(2)`
For face Q : `x = 10 cm = 10^(-1) m`
From (i) `E_(x) = 5xx10 (10^(-1)) + 2xx5 = 15 NC^(-1)`
`:.` Electric flux through face `Q = phi_(2) = E_(x) ds cos theta^(@) = 15xx10^(-2) + 15xx10^(-2) = 5xx10^(-2) = 5xx10^(-2) NC^(-1) m^(2)`
From Gauss's therorem, `phi = (q)/(in_(0))`
`q = in_(0) phi = 8.85xx10^(-12)xx5xx10^(-2) = 0.4425xx10^(-12)C`