The electric field in a region is given by `vec(E)=(E_(0)x)/l hat(i)`. Find the charge contained inside a cubical volume bounded by the surface `x=0, x=a, y=0, z=0` and `z=a`. `["Take " E_(0)=5xx10^(3)N//C, l=2 cm and a=1 cm]`

Explanation

Given: ` E = (E_0/l)veci , l= 2 cm, a = 1 cm, `
` E_0 = 5 xx (10^3) N/C`
We see that flux passes mainly through
surface area ABDC and EFGH.
As the AEFB and CHGD are parallel to
the flux again in ABCD = 0,
Hence the flux only passes through the
surface area EGFH E=0
flux = `(E_0) a/l xx area `
` 5 xx (10^3) xx a/l xx (a^2) `
` = 5 xx (10^3) xx (a^3/l)`
` = 5 xx (10^5) xx (0.01)^3/ (2 xx (10^-2))`
` = 2.5 xx (10^-1)`
` So, q = epsilon_0 flux `
`= 8.85 xx (10^-12) xx 2.5 xx (10^-1) `
` 22.125 xx (10^-1)`
` = 2.2125 xx (10^-12) C` .