An electric field of strength E = 2100 N/C is directed along the +x-axis as show

Question

An electric field of strength E = 2100 N/C is directed along the +x-axis as shown. A proton is initially at rest at point A. What is the magnitude of the force on the proton due to the electric field? What is the magnitude of the acceleration of the proton? What is the work done by the electric field on the proton after it has moved a distance of 1 cm from point A to point B? What is the kinetic energy of the proton at this instant? What is the speed (magnitude of the velocity) of the proton at this instant?

Explanation / Answer

a)

F=qE=1.6*10^-19*2100

F=7.62*10^-23 N

b)

a=F/m =7.62*10^-23/1.67*10^-27

a=45623.04 m/s^2

c)

W=Fd =7.62*10^-23*0.01

W=7.62*10^-25 J

d)

kE=W=7.62*10^-25 J

c)

KE=(1/2)mV^2

7.62*10^-25 =(1/2)*1.67*10^-27*V^2

V=30.21 m/s

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