Answer:
P=3.42×10^-6 J/s
Explanation:
From the kinematics of motion with constant acceleration we know that :
vf^2=vi^2+2*a(xf-xi)
Where :
• vf , vi, are the the final and the initial velocity of the electron
• a is the acceleration of the electron
• xf , xi are the final and the initial position of the electron .
Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.
Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m
vf^2 =vi^2+2*a(xf-xi)
vf^2-vi^2=2*a(xf-xi)
2*a(xf-xi)= vf^2-vi^2
a = (vf^2-vi^2)/2(xf-xi)
Pluging known information to get :
a = (vf^2-vi^2)/2(xf-xi)
= 1.411 × 10^17
From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m
so,
vf^2 =vi^2+2*a(xf-xi)
vf^2 =5.312× 10^7
From the following Eq. we can calculate the time elapsed in this motion .
xf =xi+vi*t+1/2*a*t
xf =xi+vi*t+1/2*a*t
t=√2(xf-xi)/a
t=3.765×10^-10 s
now we can use the power P Eq.
P=W/Δt => ΔK/Δt
Where: the work done W change the kinetic energy K of the electron ,
ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2
P=1/2*m*vf^2-1/2*m*vi^2/Δt
P=3.42×10^-6 J/s
. Using the value of q and E given by the problem, we find