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Sagot :
Answer:
(a) The potential near its surface is 45 * 10^6 V.
(b) The distance from which its center is the potential 1.00 MV is 45 m.
(c) Its energy in MeV when the atom is at the distance found in part b is 132 MeV.
Explanation:
Note: This question is not complete. The complete question is therefore provided before answering the question.
A research Van de Graaff generator has a 2.00-m diameter metal sphere with a charge of 5.00 mC on it. (a) What is the potential near its surface?
(b) At what distance from its center is the potential 1.00 MV?
(c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in MeV when the atom is at the distance found in part b?
The explanation of the answer is now provided as follows:
(a) What is the potential near its surface?
Q = Charge on the generator = 5 mC = 5 * 10^(-3)C
r = Sphere radius = 2 / 2 = 1 m
k = Constant of the electric force = 9 * 10^(9) N . m^2 / C^2
Therefore, the electric potential of a point charge can be calculated as follows:
V = kQ / r
V = (9 * 10^9 * 5 * 10^(-3)) / 1 = 45 * 10^6 V
Therefore, the potential near its surface is 45 * 10^6 V.
(b) At what distance from its center is the potential 1.00 MV?
This implies the distance where the potential is 1 MV.
Since the electric potential of a point charge is as follows:
V = kQ / r
Therefore, we can solve for r and estimate it as follows:
R = kQ / V = (9 * 10^9 * 5 * 10^(-3)) / 1 * 10^6 = 45 m
Therefore, the distance from which its center is the potential 1.00 MV is 45 m.
(c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in MeV when the atom is at the distance found in part b?
The link between the potential difference and electrical potential energy can be stated as follows:
ΔV = ΔU / q
Therefore, we have:
ΔU = qΔV = q(Va - Vb) = 3 * (45 – 1) = 132 MeV
Therefore, its energy in MeV when the atom is at the distance found in part b is 132 MeV.
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