A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radius r with r < R is used to calculate the magnitude of the electric field E at a distance r from the
center of the sphere. Which of the following equations results
from a correct application of Gauss’s law for this situation? And Why?
1. E (4 p r^2) = (Q r^3)/( epsilon sub0 * R3)
2. E(4pr^2)= (Q 3r^3)/ (epsilon sub0 * 4 p R)
3. E (4 p r^2) = (Q) / (epsilon sub0)
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ =
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
We know that Gauss's law states that the Flux enclosed by a Gaussian surface is given by
Here , E is electric field and S is surface are and q is charge enclosed by the surface and e is electrical permeability of the medium.
Here the Gaussian is of radius r<R so area of surface is
Also, charge enclosed by the surface is
Here Q is total charge,
Insert values in Gauss's law
on further solving
This is the required form.
when earth is in rotational motion around the sun,it displaces from its place.
when a ball is placed on an inclined surface it does both the rotational and circular motion.
scientific laws are the result of testing theories and proving them to be facts.