Answer:
The flux through the surface of the cube is [tex]2.314\ Nm^{2}/C[/tex]
Solution:
As per the question:
Edge of the cube, a = 8.0 cm = [tex]8.0\times 10^{- 2}\ m[/tex]
Volume Charge density, [tex]\rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}[/tex]
Now,
To calculate the electric flux:
[tex]\phi = \frac{q}{\epsilon_{o}}[/tex] (1)
where
[tex]\phi[/tex] = electric flux
[tex]\epsilon_{o} = 8.85\times 10^{- 12}\ F/m[/tex] = permittivity of free space
Volume Charge density for the given case is given by the formula:
[tex]\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}[/tex] (2)
Volume of cube, [tex]V = a^{3}[/tex]
Thus
[tex]V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}[/tex]
Thus from eqn (2), the total charge is given by:
[tex]q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}[/tex]
[tex]q = 2.048\times 10^{-11}\ F = 20.48\ pF[/tex]
Now, substitute the value of 'q' in eqn (1):
[tex]\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C[/tex]
