Question

An electron moving parallel to the x axis has an ini- AMT tial speed of 3.70 3 106 m/s at the origin. Its speed is M reduced to 1.40 3 105 m/s at the point x 5 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. (b) Which point is at the

higher potential?

Answer 1

Answer 2

a) conservation of energy dKE = -dPE

0.5*9.11E-31*(1.403E5^2 - 3.703E6^2) = -1*-1.6E-19*dV

dV = -38.98 V

b) the initial point is at the higher potential since dV<0

Answer 3

Answer 4

mass of electron = m= 9.1*10^-31 kg

u=3.703*10^6 m/s

v=1.403*10^5 m/s

change in kinetic energy = 0.5*m*(v^2-u^2)= -6.23*10^-18 J

this energy is stored as the potential energy.

E=q*(U1-U2)

=>-6.23*10^-18 = (-1.6*10^-19)*(U1-U2)

=>U1-U2= 38.9375 V = ans

as U1-U2>0 , U1 is at higher potential . hence the potential at origin is more.

Answer 5

Answer 6

Not entirely sure what your values mean, but the change in potential is the same as the change in kinetic energy for this question...

=difference in potential energy = qV where q is charge on electron(which is the same as the change in kinetic energy) =    \((m/2).(v_2^2-v_1^2)\)    Therefore potential difference is:

   $+\(V=(m/2).(v_2^2-v_1^2)/q\)+$   

Kinetic energy has been lost, so potential energy has been gained.

However since the charfe on the electron is negative it must have gone from high potential to low potential.

So the initial point is the point of higher potential.

- Sam