Answer:
q=0.06 C
Explanation:
Given that
Net flux [tex]\phi = 6.78\times 10^{9}\ Nm^2/C[/tex]
Lets take charge inside Gaussian surface = q C
We know that
[tex]\phi=\dfrac{q}{\varepsilon _o}[/tex]
[tex]\varepsilon _o=8.854\times 10^{-12} farads\ per\ meter[/tex]
Now by putting the values in the above equation we get
[tex]6.78\times 10^{9}=\dfrac{q}{8.854\times 10^{-12}}[/tex]
[tex]q=6.78\times 10^{9}\times 8.854\times 10^{-12}\ C[/tex]
q=0.06 C
Therefore the net charge inside the surface will be 0.06 C.
q=0.06 C
