At a particular instant an electron is traveling with speed 9000 m/s. What is th

Question

At a particular instant an electron is traveling with speed 9000 m/s. What is the kinetic energy of the electron? If a proton were traveling at the same speed (9000 m/s), what would be the kinetic energy of the proton? You move from location i at m to location f at m. All along this path there is a nearly uniform electric field i = N/C. Calculate Delta V = V_f - V_if including sign and units. Locations A, B, and C are in a region of uniform electric field, as shown in the diagram above. Location A is at m. Location B is at m. In the region the electric field 1 = N/C. For a path starting at A and ending at B, calculate: (a) The displacement vector Delta r (b) the change in electric potential: (c) the potential energy change for the system when a proton moves from A to B: (d) the potential energy change for the system when an electron moves from A to B:

Explanation / Answer

v = 9000 m/s, mass of electron m =9.1*10^-31 kg

Kelectron = 0.5*mv^2 = 0.5*9.1*10^-31*9000^2

Kelectron = 3.7*10^-23 J

mass of proton m = 1.67*10^-27 kg

Kproton = 0.5*1.67*10^-27*9000^2

Kproton = 6.76*10^-20 J

i = <4,3,3 > , j = <8,9 ,12>

r = (8i +9j+12k) - (4i+3j+3k) = 4i +6j +9k

E = <1000, 220, -520 >

DV = Vf - Vi

DV = E.r = (1000 i+220 j -520 k).(4i +6j+9k)

DV = (4000 +1320 - 4680) = 640 V

A = < -0.5, 0, 0 > , B = < 0.5, 0 ,0 >

E = <-650, 450 ,0 >

(a) Dr = - 0.5i - 0.5i = - 1 i

(b) DV = (-650i +450 j) .(- 1i)

DV = 650 V

(c) DU = q*DV = 1.6*10^-19(650)

DU= 1.04*10^-16 J

(d) DU = - 1.6*10^-19(650)

DU= - 1.04*10^-16 J

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