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A proton moves along the x-axis where some arrangement of charges has produced the potential V(x)=Vosin(2πx/λ),where Vo = 3800 V and λ = 1.0 mm. a) What minimum speed must the proton have at x = 0 to move down the axis without being reflected? b) What is the maximum speed reached by a proton that at x = 0 has the speed you calculated in part A?

User Dharmik
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1 Answer

5 votes

Answer with Explanation:

We are given that


V(x)=Vsin((2\pi x)/(\lambda))


V_0=3800 V,\lambda=1.0 mm

a.At x=0

V(0)=0

Maximum potential,
V_(max)=V_0=3800 V

The proton must have minimum speed when V(x) is maximum

Initial kinetic energy =
q\Delta V


(1)/(2)mv^2=q(V_(max)-V(0))=q(3800-0)=3800q

q=
1.6* 10^(-19) C


m=1.67* 10^(-27) Kg


v^2=(2* 3800 q)/(m)=(3800* 2* 1.6* 10^(-19))/(1.67* 10^(-27))


v=\sqrt{(2* 3800 q)/(m)}=\sqrt{(3800* 2* 1.6* 10^(-19))/(1.67* 10^(-27))}


v=8.5* 10^5 m/s

b.The maximum speed reached by a proton when V(x) is minimum


V_(min)=-3800 V


v=\sqrt{(2q(V_(max)-V_(min))/(m)}


v=\sqrt{(2* 1.6* 10^(-19)(3800-(-3800)))/(1.67* 10^(-27))


v=1.21* 10^6 m/s

User Albin Anke
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