The electric field in a region is given by `vecE=E_0x/iota veci`. Find the charge contained inside a cubical volume bounded by the curfaced `x=0, x=alpha, y=0, y=alpha, z=0 and z= alpha`. Take `E_0= 5 xx 10^3 N C^(-1)`, `iota= 2cm and alpha=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` .