Sure! Let's go through each part of the problem step-by-step:
### Part (a)
The law in electrostatics that relates the electric flux passing through a surface with the charge enclosed is called Gauss's Law. Gauss's Law states that the electric flux (Φ_E) through any closed surface is equal to the total charge enclosed (Q_enc) by that surface divided by the permittivity of free space (ε_0). Mathematically, it is expressed as:
[tex]\[ \Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0} \][/tex]
where:
- [tex]\(\Phi_E\)[/tex] is the electric flux.
- [tex]\(Q_{\text{enc}}\)[/tex] is the total charge enclosed within the surface.
- [tex]\(\varepsilon_0\)[/tex] is the permittivity of free space.
### Part (b)
Given the charges:
- [tex]\( q_1 = +6 \, \mu C \)[/tex] (microcoulombs)
- [tex]\( q_2 = -4 \, \mu C \)[/tex] (microcoulombs)
To find the ratio of the electric fluxes passing through surfaces S1 and S2, we apply Gauss's Law separately for each surface.
For surface S1:
[tex]\[ \Phi_{E1} = \frac{q_1}{\varepsilon_0} \][/tex]
For surface S2:
[tex]\[ \Phi_{E2} = \frac{q_2}{\varepsilon_0} \][/tex]
Now, we find the ratio of the flux through surfaces S1 and S2:
[tex]\[ \text{Ratio of fluxes} = \frac{\Phi_{E1}}{\Phi_{E2}} = \frac{\frac{q_1}{\varepsilon_0}}{\frac{q_2}{\varepsilon_0}} = \frac{q_1}{q_2} \][/tex]
Substituting the given charges:
[tex]\[ \text{Ratio of fluxes} = \frac{+6 \, \mu C}{-4 \, \mu C} = \frac{6 \times 10^{-6}}{-4 \times 10^{-6}} = -1.5 \][/tex]
So, the ratio of the electric fluxes through surfaces S1 and S2 is [tex]\(-1.5\)[/tex].
### Part (c)
If surface S2 expands to double its area while surface S1 remains the same, we need to consider how Gauss's Law applies to this situation. Gauss's Law states that the electric flux through a closed surface depends only on the enclosed charge and not on the size or shape of the surface.
Since the charge enclosed by surface S2 remains the same (i.e., [tex]\( q_2 \)[/tex] doesn't change) even when its area is doubled, the electric flux passing through S2 will not change.
Therefore, the ratio of the flux passing through surfaces S1 and S2 will remain the same as it was before the area of S2 was doubled.
So, the initial and final ratios of the electric fluxes are the same:
[tex]\[ \text{Ratio of fluxes (final)} = -1.5 \][/tex]
In conclusion:
a) Gauss's Law: [tex]\(\Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0}\)[/tex]
b) Initial ratio of fluxes: [tex]\(-1.5\)[/tex]
c) Final ratio of fluxes (after S2 expands): [tex]\(-1.5\)[/tex]
