The change in surface area of Gaussian surface with radius (r) is 8πr.
<h3>
Electric field from Coulomb's law</h3>
The electric field experienced by a charge is calculated as follows;
where;
- E is the electric field
- Q is the charge
- r is the radius
The electric field reduces by a factor of 
<h3>Surface area of a Gaussian surface;</h3>
The surface area of a sphere is given as;
<h3>Change in area with r</h3>
Thus, the change in surface area of Gaussian surface with radius (r) is 8πr.
Learn more about area of Gaussian surfaces here: brainly.com/question/17060446
Answer:
The number take for the "radius" is normally taken to be the radius from the center to the event horizon, since nothing inside that can be observed, and since it is a point of no return. This radius is directly proportional to the mass of the black hole. From the Wikipedia article on event horizon: "For the mass of the Sun this radius is approximately 3 kilometers and for the Earth it is about 9 millimeters. In practice, however, neither the Earth nor the Sun has the necessary mass and therefore the necessary gravitational force, to overcome electron and neutron degeneracy pressure." The last sentence means that it is unlikely for the Sun to become a black hole, since it has too little mass to become one.
Explanation:
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Answer:
I don't understand your question ❓,the object.....of what experiment
