# The electric filed in a region is given by Find the charge contained inside a cubical volume bounded by the surfaces x = 0, x = a, y = 0, y = a, z = 0 and z = a. Take E0 = 5 × 103 NC–1, ℓ = 2 cm and α = 1 cm.

Given:

E = E0=5× 103N/C

l=2cm=2× 10-2m

a=1cm=1× 10-2m We know that,

Flux of electric field through a surface of area ΔS is given by dot product of electric field E with surface area ΔS From the fig. we can see that electric field lines are parallel to the normal of the surfaces ABCD and EFGH therefore flux is nonzero through these surfaces, whereas for other faces of the cube electric field is perpendicular to normal of faces therefore flux through these surfaces is zero

For face ABCD:

Area vector ΔS =a2î

Electric field E = flux from (i) is given by  For face EFGH

Area vector ΔS = -a2î

Electric field E = flux from (i) is given by Total flux =  Nm2/C Nm2/C

We know that,

By Gauss’s law, flux of net electric field through a closed surface equals the net charge enclosed by the surface divided by ϵ0 qin=ϕ × ϵ0 C C

Therefore the net charge contained inside the cubical volume is given by 2.2125× 10-12 C

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