Question

EXERCISE A constant electric field with magnitude 1.50 x 10' N/C is pointing in the positive x-direction. An electron is fired from x = -0.0200 m in the same direction as the electric field. The electron's speed has fallen by half when it reaches x = 0.190 m, a change in potential energy of 5.04 10-1/ J. The electron continues to x = -0.250 m within the constant electric field. If there's a change in potential energy of -1.06 x 10-16 J as it goes from x = 0.190 m to x = -0.250 m, find the electron's speed (in m/s) at x = -0.250 m. Hint Notice the electron turned around at some point. When the electron's speed is half its initial value, you are given the change in electric potential energy. The sum of AU and AK at this point is a conserved quantity, and is equal to zero, so AU + AK will be the same at all points along the electron's trajectory. You are given AU at the point of interest, so write AK in terms of velocities, substitute the velocity at the second point with the initial velocity divided by 2, simplify your equation, and solve for ve Click the hint button again to remove this hint. 1.971e7 Set the kinetic energy (expressed in terms of speed and mass) equal to the work done by the electric field on the particle, and solve for the speed of the particle. m/s Need Help? Read It

A constant electric field with magnitude 1.50 ✕ 10³ N/C is pointing in the positive x-direction. An electron is fired from

x = −0.0200 m

in the same direction as the electric field. The electron's speed has fallen by half when it reaches

x = 0.190 m,

a change in potential energy of 5.04 ✕ 10⁻¹⁷ J. The electron continues to

x = −0.250 m

within the constant electric field. If there's a change in potential energy of −1.06 ✕ 10⁻¹⁶ J as it goes from

x = 0.190 m

to

x = −0.250 m,

find the electron's speed (in m/s) at

x = −0.250 m.

Answer 1

Solution: The work done by the electric field in x direction from x=-0.0200m to x = 0.190 m is, w=-eEAT --(1.6x10-"C)1.50x10 N/C)(0.190m-(-0.0200m)) =-1.6x10-19) 1.50x10 N/C (0210m) =-5.04x10-"J The change in kinetic energy is given by, ake- - *) * 3 mm From work energy theorem, AKE=W AKE+AU = 0 AU =-W _3mv.-5.04x10-J _3mv.-5.04 x10- (5.04x10-17 J) 8 VSN 3m (5.04 x10-458 13(9.1x10-” kg =1.21x10 m/s Now between x=-0.0200 m to x=-0.250m, AU = 5.04 x10- J-1.06x10-48J =-5.56x10-17

AU =-W=-AKE Em (v;- (1.21x10" m/s )° ) = 5.56x10" } v;-(1.21x10?m/s) - 2x5.56x10 m 2x5.56x10"]+(1.21x10 m/s) (9.1x10-*kg) = 1.63x10 m/s The speed of electron is 1.63x10 m/s.