Answer: Ranking : a = c > b > d
Explanation:
The amount of electric flux is directly proportional to the amount of charge enclosed.
The greater the charge enclosed, the greater the electric flux through their surface
a)
A cube with sides 10 cm long that contains a + 2.00 micro-coulomb point charge
∅ = Q/∈₀ = (2.00 x 10⁻³) / (8.85 x 10⁻¹²) = 2.26 x 10⁸ Nm²/C
b)
A cube with sides 10 cm long that contains a + 1.00 micro-coulomb point charge
∅ = Q/∈₀ = (1.00 x 10⁻³) / (8.85 x 10⁻¹²) = 1.13 x 10⁸ Nm²/C
c)
A cube with sides 20 cm long that contains a + 2.00 micro-coulomb point charge
∅ = Q/∈₀ = (2.00 x 10⁻³) / (8.85 x 10⁻¹²) = 2.26 x 10⁸ Nm²/C
d) A cube with sides 20 cm long that contains a + 1.00 micro-coulomb point charge
∅ = Q/∈₀ = (1.00 x 10⁻³) / (8.85 x 10⁻¹²) = 1.13 x 10⁸ Nm²/C
Ranking :
a = c > b > d
Answer:
Using Gauss's Law, the total amount of electric flux through a closed surface is proportional to the charge enclosed. (The surface integral of the electric field over the closed surface equals the enclosed charge divided by the constant of proportionality, the permittivity of free space.) So, since surfaces 1 and 3 have the most positive enclosed charge, they have the most positive electric flux (they have the same). Since surfaces 2 and 4 have the least positive enclosed charge, they have the least positive electric flux. and are the same
thus LHS to RHS; a=c > b=d
