An electron with a speed of 797 x 108 cms in the positive di

An electron with a speed of 7.97 x 10^8 cm/s in the positive direction of an x axis enters an electric field of magnitude 1.99 x 10^3 N/C, traveling along a field line m the direction that retards its motion, (a) How far will the electron travel in the held beofre stopping momentanly, and (b) how mu Ch time will have elapsed? (c) If the region containing the electric field is 6.11 mm long (too short ofr the electron to stop within it), what fraction of the electron\'s initial kinetic energy will be last in that region?

Solution

In the electric field the force experienced by the electron

F=eE here e is electronic charge and E is electric field

given E=1.99x10³ N/C

so F= 1.6x10^-19 x1.99x103=3.18x10^-16 N

from F=ma we get acceleration here m is mass of electron = 9.1x10^-31 kg

so a= F/m = 3.18x10^-16 / (9.1x10^-31 ) =0.349x10¹⁵ m/sec²

the initial velocity of the electro=7.97x10⁸ cm/sec=7.97x10⁶ m/sec

final velocity of the electron=0

so from eq of motion v²=u²-2as here v is final velocity,u is initial velocity and s is the distance travelled

0= (7.97x10⁶ )² -2x0.349x10^-15 xs

thus s=(7.97x7.97x10¹²) /(0.699x10¹⁵ )= 63.52x10¹² / (0.699x10¹⁵ )    =90.87x10^-3 m=9.08cm=0.0908m (answer distance)

for time we use v=u-at   here vis 0(final velocity)

thus 0= 7.97x10⁶ -0.349x10¹⁵ xt

time t= (7.97x10⁶) /(0.349x10¹⁵) =22.83x10^-9 sec (answer time)

(c) if region containing the electric field is 6.11mm long

then the speed of electron when it travels a distance 6.11mm=6.11x10^-3 m

from v²=u²-2as

v² =(7.97x10⁶ )² -2x0.349x10¹⁵ x6.11x10^-3 =63.52x10¹² -4.26x10¹² =59.26x10¹² (this is the square of velocity when it travels a distance 6.11mm)

thus at this time the kinetic energy=(1/2) m 59.26x10¹²

initial kinetic energy =(1/2)m 63.52x10¹²

the fraction of kinetic energy lost= (1/2)m(63.52-59.26)x10¹² /(1/2)mx63.52x10¹² = (4.26/63.52) =0.067 = 6.7 percent of kinetic energy is lost