Question

An electron has an initial speed of 8.06 x10^6 m/s in a uniform 5.60 x 10^5 N/C strength electic field.The field accelerates the electron in the direction opposite to its initial velocity.

(a) What is the direction of the electric field?
i. opposite
ii. direction to the electron's initial velocity
iii. same direction as the electron's initial velocity
iv. not enough information to decide
(b) How far does the electron travel before coming to rest? m
(c) How long does it take the electron to come to rest? s
(d) What is the electron's speed when it returns to its starting point?

Answer 1

(a) Option (i)

(b) 6.6 x 10^-4 m

(c) 8.2 x 10^-11 s

Step-by-step explanation:

initial velocity, u = 8 .06 x 10^6 m/s

Electric field, E = 5.6 x 10^5 N/C

(a) The direction of field is opposite.

Option (i).

(b) Let the distance is s.

Use third equation of motion

`v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 * (qE)/(m)* s\\\\8.06* 10^6* 8.06* 10^6 = \frac {1.6* 10^(-19)* 5.6* 10^5}{9.1* 10^(-31)} s\\\\s = 6.6* 10^(-4) m`

(c) Let the time is t.

Use first equation of motion.

`v = u + a t \\\\0 = u - * (qE)/(m)* t\\\\8.06* 10^6 = \frac {1.6* 10^(-19)* 5.6* 10^5}{9.1* 10^(-31)} t\\\\t = 8.2* 10^(-11) s`

Answer 2

Step-by-step explanation:

a)

The force on electron acts opposite to the velocity , and direction of force on electron is always opposite to direction of electric field .

Hence direction of electric field must be in the same in which electrons travels.

Hence option iii is correct.

b )

deceleration a = force / mass

= qE / m

= 1.6 x 10⁻¹⁶ x 5.6 x 10⁵ / 9.1 x 10⁻³¹

= .98 x 10²⁰ m /s²

v² = u² - 2 a s

0 = (8.06 x 10⁶ )² - 2 x .98 x 10²⁰ s

s = 64.96 x 10¹² / 1.96 x 10²⁰

= 33.14 x 10⁻⁸ m

c ) time required

= 8.06 x 10⁶ / .98 x 10²⁰

= 8.22 x 10⁻¹² s .

d ) Its speed will be same as that in the beginning ie 8.06 x 10⁶ m/s .