An electron is moving east in a uniform electric field of 1.45 N / C directed to

Question

An electron is moving east in a uniform electric field of 1.45 N/C directed to the west. At point A, the velocity of the electron is 4.50×105 m/s pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.390 m east of point A? ANSWER: 6.34*10^5 PLEAS ANSWER PART B A proton is moving in the uniform electric field of part A. At point A, the velocity of the proton is 1.93×104 m/s again pointed towards the east. What is the speed of the proton at point B?

again pointed towards the east. What is the speed of the proton at point B?8

Explanation / Answer

Force acting on the electron is F = q*E = 1.6*10^-19*1.45 = 2.32*10^-19 N along east

but F = m*a = 2.32*10^-19

accelaration is a = (2.32*10^-19)/(9.11*10^-31) = 2.54*10^11 m/sec^2 along east

displacement is S = 0.39 m

initial speed is Vo = 4.5*10^5 m/sec

Using kinematic equations

V^2-Vo^2 = 2*a*S

V^2- ( 4.5*10^5)^2 = 2*2.54*10^11*0.39

V = 6.34*10^5 m/sec

B) for proton

a = F/m = (2.32*10^-19)/(1.67*10^-27) = 1.38*10^8 m/sec^2 along west

Using kinematic equations

V^2 -Vo^2 = 2*a*S

V^2 - (1.93*10^4)^2 = -2*1.38*10^8*0.39

V = 1.63*10^4 m/sec is the speed of the proton at B

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