A point charge 5 micro couloumb is at the centre of a cubical gaussian surface 100cm on edge. Calculate the electric flux through the surface.
Huh ??
Gauss!
Does not matter shape or size of surrounding surface. Flux out proportional to charge inside.
The Gauss' Law
flux = integral E dot dA = Q/epsilon
The Gauss law states that electric net flux through any Gauss surface is directly proportional to the net charge enclosed by the surface
where is permittivity of vacuum
I am not going to answer the others because they are really the same as this one.
For example if you have charges 2q and - 2q within a closed surface, the net E flux out is zero.
To calculate the electric flux through the cubical Gaussian surface, we need to use Gauss's law. Gauss's law states that the electric flux through a closed surface is proportional to the electric charge enclosed by the surface.
The formula for the electric flux (Φ) through a closed surface is:
Φ = Σ(E * dA)
Where:
Φ is the electric flux,
E is the electric field,
dA is the differential area, and
Σ represents taking the sum over all small areas on the surface.
Since the point charge is placed at the center of the cubical Gaussian surface, the electric field (E) will have equal magnitudes in all directions and will point radially outward from the charge.
The electric field due to a point charge (q) at a distance (r) is given by:
E = k * (q / r²)
Where:
k is Coulomb's constant (k = 9 × 10^9 Nm²/C²),
q is the charge, and
r is the distance from the point charge to the surface.
In this case, the charge (q) is 5 microcoulombs (5 × 10^-6 C), and the distance (r) from the point charge to any face of the cubic surface is half of the edge length (50 cm or 0.5 m).
Now, let's calculate the electric flux through one face of the cube:
dA = (edge length)²
dA = (0.5 m)² = 0.25 m²
Using the formula for electric field due to a point charge, we can calculate the electric field at the face of the cube:
E = k * (q / r²)
E = (9 × 10^9 Nm²/C²) * (5 × 10^-6 C / (0.5 m)²)
E = (9 × 10^9 Nm²/C²) * (5 × 10^-6 C / 0.25 m²)
E = (9 × 10^9 N × 5 × 10^-6 C) / (0.25 m²)
E = 180 N/C
Now, we can calculate the electric flux through one face (Φ) using Gauss's law:
Φ = E * dA
Φ = (180 N/C) * (0.25 m²)
Φ = 45 Nm²/C
Since the cube has 6 faces, the total electric flux through the surface is:
Total Φ = 6 * Φ
Total Φ = 6 * 45 Nm²/C
Total Φ = 270 Nm²/C
Therefore, the electric flux through the cubical Gaussian surface is 270 Nm²/C.