Respuesta :
Explanation:
(a) It is known that relation between charge and volume is as follows.
[tex]q_{enclosed} = \rho V_{cube}[/tex]
= [tex](509 \times 10^{-9} C/m^{3}) \times (0.04 m)^{3}[/tex]
= [tex]509 \times 10^{-9} \times 6.4 \times 10^{-5}[/tex]
= [tex]3.26 \times 10^{-11} C[/tex]
Now, according to Gauss's law
[tex]\phi = \frac{q_{enclosed}}{\epsilon_{o}}[/tex]
= [tex]\frac{3.26 \times 10^{-11} C}{8.85 \times 10^{-12}C^{2}N^{-1}m^{-2}}[/tex]
= 3.68 [tex]N m^{2}/C[/tex]
Hence, the electric flux through this cubical surface if its edge length is 4.00 cm is 3.68 [tex]N m^{2}/C[/tex].
(b) Similarly, we will calculate the electric flux when edge length is 16.8 cm as follows.
[tex]q_{enclosed} = \rho V_{cube}[/tex]
= [tex](509 \times 10^{-9} C/m^{3}) \times (0.168 m)^{3}[/tex]
= [tex]509 \times 10^{-9} \times 4.74 \times 10^{-3}[/tex]
= [tex]2.41 \times 10^{-11} C[/tex]
Now, according to Gauss's law
[tex]\phi = \frac{q_{enclosed}}{\epsilon_{o}}[/tex]
= [tex]\frac{2.41 \times 10^{-11} C}{8.85 \times 10^{-12}C^{2}N^{-1}m^{-2}}[/tex]
= 2.72 [tex]N m^{2}/C[/tex]
Therefore, the electric flux through this cubical surface if its edge length is 4.00 cm is 2.72 [tex]N m^{2}/C[/tex].
