Find the ratio of electric flux passing through two spheres centered at the origin, having radii 4 m and 6 m respectively, with charges \( q_1 = 8 \, \mu C \) and \( q_2 = 2 \, \mu C \) respectively.
Show Hint
In Gauss's law, the electric flux through a closed surface depends only on the charge enclosed inside the surface, not on the size or shape of the surface.
Step 1: Understand the concept of electric flux.
The electric flux \( \Phi_E \) passing through a closed surface is given by Gauss's law:
\[
\Phi_E = \frac{q_{\text{enc}}}{\epsilon_0}
\]
where \( q_{\text{enc}} \) is the charge enclosed within the surface and \( \epsilon_0 \) is the permittivity of free space.
Step 2: Analyze the situation.
For each sphere, the electric flux depends only on the charge enclosed within the sphere. The radius of the sphere does not affect the flux because the electric flux is directly proportional to the enclosed charge, according to Gauss's law.
- For the first sphere (radius \( r_1 = 4 \, \text{m} \)) with charge \( q_1 = 8 \, \mu C \), the electric flux is:
\[
\Phi_{E1} = \frac{q_1}{\epsilon_0}
\]
- For the second sphere (radius \( r_2 = 6 \, \text{m} \)) with charge \( q_2 = 2 \, \mu C \), the electric flux is:
\[
\Phi_{E2} = \frac{q_2}{\epsilon_0}
\]
Step 3: Find the ratio of the fluxes.
The ratio of the electric fluxes is given by:
\[
\frac{\Phi_{E1}}{\Phi_{E2}} = \frac{\frac{q_1}{\epsilon_0}}{\frac{q_2}{\epsilon_0}} = \frac{q_1}{q_2}
\]
Substituting the values of the charges:
\[
\frac{\Phi_{E1}}{\Phi_{E2}} = \frac{8 \, \mu C}{2 \, \mu C} = 4
\]
Thus, the ratio of the electric fluxes is 4.
Final Answer: \( 1.33 \)
