Your Name (Print): Group Members: Date: - Group: nergy Conservation: Proton in t
ID: 1583067 • Letter: Y
Question
Your Name (Print): Group Members: Date: - Group: nergy Conservation: Proton in the Non-uniform Electric Field of an Iron Nucleus Consider a proton (mass = 1.67x 1027 kg 9m +e) in the potential field of an iron nucleus (mass = 9.40x 1026 kg, qn-+ 26e). Three equipotential surfaces are shown in the diagram as dashed circles; the equipotential lines are NOT equally spaced. Treat the iron nucleus as a point charge. a) Calculate the radial distance from the iron nucleus at which the electric potential is 5.00 Volts. b) Calculate the radial distance from the iron nucleus at which the electric potential is 3.00 Volts c) Calculate the radial distance from the iron nucleus at which the electric potential is 1.00 Volts. Label these three equipotential surfaces in the above diagram.Explanation / Answer
Q = +26*e
= 26*1.6*10^-19
= 4.16*10^-18 C
a) use, V = k*Q/r
r = k*Q/V
= 9*10^9*4.16*10^-18/5
= 7.49*10^-9 m or 7.49 nm
b) use, V = k*Q/r
r = k*Q/V
= 9*10^9*4.16*10^-18/3
= 1.25*10^-8 m or 12.5 nm
c) use, V = k*Q/r
r = k*Q/V
= 9*10^9*4.16*10^-18/1
= 3.74*10^-8 m or 37.4 nm
d) Draw 8 electric field lines radially outward from innere circle to outer circle.
e) use Work-energy theorem,
decrease in potential energy = gain in kinetic energy
q*(Vi - Vf) = (1/2)*m*v^2
==> v = sqrt(2*q*(Vi - Vf)/m)
= sqrt(2*1.6*10^-19*(5 - 1)/(1.67*10^-27) )
= 2.77*10^4 m/s
f) No. Because, Force on proton is not uniform as it moves from inner circle to outer circle.
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