Answer:
The Electric flux will be
Explanation:
Given
Strength of the Electric Field at a distance of 0.158 m from the point charge is
We know that the flux of the Electric Field can be calculated by using Gauss Law which is given by
Let consider a sphere of radius 0.158 m as Gaussian Surface at a distance of 0.158 m from the point charge and Let be the flux of the Electric Field coming out\passing through it which is given by
It can be observed that same amount of flux which is passing through the Gaussian sphere of radius 0.158 is also passing through the Gaussian sphere of radius 0.142 m at a distance of 0.142 m from its centre.
Also it can be observed that the charge inside the two Gaussian Sphere mentioned have same value so the Flux of electric field through them will also be same.
So the electric flux through the surface of sphere that has given charge at its centre and that has radius 0.142 m is
Acoustics is the common name ....
To solve this problem we will apply the concept related to the electric field defined from the laws of Coulomb. For this purpose we will remember that the electric field is equivalent to the product of the Coulomb constant due to the change of the charge over the squared distance, mathematically this is
Here,
k = Coulomb's constant
r = Distance from center of terminal to point where electric field is to found
q = Excess charge placed on the center of terminal of Van de Graff's generator
Replacing we have that,
Therefore the electric field is
Mass and distance
If mass is doubled, the force of gravity between the objects is doubled