A spherical surface completely surrounds a collection of charges. Find the electric flux (with its sign) through the surface if the collection consists of (a) a single +1.80 × 10-6 C charge, (b) a single -8.20 × 10-6 C charge, and (c) both of the charges in (a) and (b
flux= EA= QA/permitivity free space*area
the area cancel out we left with flux=charge/permitivity
so try to solve it by replacing the charges wit Q given
permitivity= 8.85 *10^-12
To find the electric flux through a spherical surface due to a collection of charges, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space (ε₀).
The formula to calculate electric flux (Φ) is:
Φ = Qenc / ε₀
where Qenc is the total charge enclosed by the surface.
Let's calculate the electric flux for each case:
(a) Single +1.80 × 10^-6 C charge:
Since there is only one charge, the total charge enclosed is equal to the charge itself (Qenc = +1.80 × 10^-6 C). The sign of the charge determines the direction of the electric field lines, which determines the sign of the flux. In this case, the charge is positive, so the flux will be positive.
Φ(a) = Qenc(a) / ε₀
Φ(a) = +1.80 × 10^-6 C / ε₀
(b) Single -8.20 × 10^-6 C charge:
Similar to case (a), the total charge enclosed is equal to the charge itself (Qenc = -8.20 × 10^-6 C), and the sign of the charge determines the direction of the electric field lines and, consequently, the sign of the flux. In this case, the charge is negative, so the flux will be negative.
Φ(b) = Qenc(b) / ε₀
Φ(b) = -8.20 × 10^-6 C / ε₀
(c) Both charges combined:
To find the total charge enclosed by the surface when both charges are present, we need to sum the individual charges.
Qenc(c) = Q(a) + Q(b)
Qenc(c) = +1.80 × 10^-6 C - 8.20 × 10^-6 C
Φ(c) = Qenc(c) / ε₀
Φ(c) = (-6.40 × 10^-6 C) / ε₀
In all cases, we will have to multiply the flux by a factor of 4π to account for the spherical symmetry of the surface.
So, the electric flux for each case is:
(a) Φ(a) = (4π * 1.80 × 10^-6 C) / ε₀
(b) Φ(b) = (4π * -8.20 × 10^-6 C) / ε₀
(c) Φ(c) = (4π * -6.40 × 10^-6 C) / ε₀
Note: To obtain a numerical value, you need to know the value of the permittivity of free space (ε₀), which is approximately 8.854 × 10^-12 C²/N·m².
To find the electric flux, we need to use Gauss's law and calculate the electric field passing through the surface. The electric flux through a closed surface is given by:
Φ = ∮ E ⋅ dA
where Φ is the electric flux, E is the electric field, and dA is a small vector representing the area element normal to the surface.
For a closed surface, the electric flux is given by:
Φ = Q / ε₀
where Q is the total charge enclosed by the surface and ε₀ is the permittivity of free space (ε₀ ≈ 8.85 × 10^-12 C^2/(N*m^2)).
Let's calculate the electric flux for each case:
(a) Single +1.80 × 10^-6 C charge:
In this case, we have a single positive charge enclosed by the surface. Therefore, the electric flux through the surface is equal to the charge divided by the permittivity of free space:
Φ(a) = Q / ε₀ = (1.80 × 10^-6 C) / (8.85 × 10^-12 C^2/(N*m^2))
(b) Single -8.20 × 10^-6 C charge:
In this case, we have a single negative charge enclosed by the surface. The electric flux produced by the negative charge passing through the surface is negative. So, the flux is given by:
Φ(b) = -Q / ε₀ = (-8.20 × 10^-6 C) / (8.85 × 10^-12 C^2/(N*m^2))
(c) Both charges (a) and (b):
In this case, we can simply add the individual fluxes to obtain the total flux:
Φ(c) = Φ(a) + Φ(b)
Now you can substitute the values of the charges and permittivity of free space into the above equations to calculate the electric flux for each case.