Answer:
[tex]3.61581\ Nm^2/C[/tex]
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
V = Volume of cube = [tex]0.04^3[/tex]
[tex]\rho[/tex] = Charge density = [tex]500\ nC/m^3[/tex]
Electric flux is given by
[tex]\phi=\dfrac{Q}{\epsilon_0}\\\Rightarrow \phi=\dfrac{\rho V}{\epsilon_0}\\\Rightarrow \phi=\dfrac{500\times 10^{-9}\times 0.04^3}{8.85\times 10^{-12}}\\\Rightarrow \phi=3.61581\ Nm^2/C[/tex]
The electric flux through the surface of the cube is [tex]3.61581\ Nm^2/C[/tex]
