An electron moving with a speed of 5 * 10^8 cm/s is shot parallel to an electric

Question

An electron moving with a speed of 5 * 10^8 cm/s is shot parallel to an electric field of strength 1000 N/C arranged so as to retard its motion. Did I do it correctly?

3. An electron moving with a speed of 5x10 cm/s is shot parallel to an electric field of strength 1000 N/C arranged so as to retard its motion. (5 %) a. How n travel in the field before coming momentarily to rest? b. How much time will elapse before the electron comes momentarily to rest? M 000 1. 44055 E-31 t0006- S0 seconds NE-3

Explanation / Answer

a)

The acceleration is given by

F=qE

ma =qE

a =qE/m =(1.6 x 10-19 Cx 1000N/C)/(9.1 x 10-31kg) =0.1758*1015m/s2=1.758*1014m/s2

The distance travelled is given by

v2-u2=2as

v2 =2as

s =v2/2a

   =(5*106m/s)2/2*1.758*1014m/s2

   =25*1012/3.516*1014

=7.110*10-2m

=0.0711m (Your first part is correct)

b)

The time elapse before the electron come to mometarly rest is given by

v =u+at

v =0

u =5 * 10^8 cm/s =5*106m/s

a =-1.758*1014m/s2 (here we have to consider -a)

t =(v-u)/a =5*106m/s/-1.758*1014m/s2=2.844*10-8s (second part is also correct)

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