Your Name (Print) Group Members: Date: Group: Energy Conservation: Proton in the

Question

Your Name (Print) Group Members: Date: Group: Energy Conservation: Proton in the Non-uniform Electric Field of an Iron Nucleus Consider a proton (mass 1.67x 1077 kg, art-te) in the potential field of an iron nucleus (mass 9.40 x 10 kg Qron +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. PHYS-112 College Physics II Activities Manual Page 101 of 278

Explanation / Answer

a)We know that

V = kq/r

r = kq/V

q = 26 e ; V = 5 V

r = 9 x 10^9 x 26 x 1.6 x 10^-19/5 = 7.488 x 10^-9 m

Hence, r = 7.488 x 10^-9 m = 7.488 nm

b)r2 = kq/V2

q = 26 e ; V2 = 3 V

r = 9 x 10^9 x 26 x 1.6 x 10^-19/3 = 1.248 x 10^-8 m

Hence, r = 1.248 x 10^-8 m = 12.48 nm

c)r3 = kq/V3

q = 26 e ; V = 1 V

r = 9 x 10^9 x 26 x 1.6 x 10^-19/1 = 3.744 x 10^-8 m

Hence, r = 3.744 x 10^-8 m = 37.44 nm

d)The electrid field lines are always directed away from a positive charge. For this nucles and perpendicular to potential lines, they will be as shown in the attached figure.

e)from conservation of energy

KE = U2 - U1

1/2 m v^2 = q (5 - 1) = 4 q

v = sqrt (8 q/m) = sqrt (8 x 1.6 x 10^-19/(1.67 x 10^-27)) = 2.77 x 10^4 m/s

Hence, v = 2.77 x 10^4 m/s

f)Since the electric field is not constant the acceleration will not constant and we can not unfirmly accelerate.

F = ma and F = qE

a = qE/m

with the variation in E, the acceleration also changes.

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