what is the electric flux through a sphere that has a radius of 2.2m and carries charge of 5.3 micro couloumb at its centre?
solution
this is discussed here:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html
To find the electric flux through a sphere, 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 that surface divided by the permittivity of free space (ε₀).
The formula for electric flux (Φ) through a closed surface is given by:
Φ = Q / ε₀
where:
Φ is the electric flux,
Q is the total charge enclosed by the closed surface, and
ε₀ is the permittivity of free space (ε₀ ≈ 8.854 x 10^-12 C²/N·m²).
In this case, we have a sphere with a radius of 2.2m and a charge of 5.3 microcoulomb (5.3 x 10^-6 C) at its center.
Step 1: Calculate the total charge enclosed by the sphere.
Since the sphere has a charge of 5.3 microcoulomb at its center, the total charge enclosed by the sphere is also 5.3 microcoulomb.
Step 2: Substitute the values into the formula.
Φ = Q / ε₀
= (5.3 x 10^-6 C) / (8.854 x 10^-12 C²/N·m²)
Step 3: Calculate the electric flux.
Φ ≈ 5.99 x 10^5 N·m²/C
Therefore, the electric flux through the sphere is approximately 5.99 x 10^5 N·m²/C.
Note: Make sure to keep track of the units throughout the calculation to ensure accurate results.