Question

An electron is released from rest in uniform electric field of 106 NC-1. Computer its acceleration. Also find the time taken by the electron in attaining a speed of 0.1c, where c = 3 x 108 ms-1.

Answer 1

qE=ma
a=qE/m
a=18.637*10^12
v=u+at
u=0
v=at
t=a/v
t=18.637*10^12/3*10^7
= 6.21*10^5 m/s



Hope it will help you

Answer 2

The acceleration of electron, a = 1.76 × 10^17 m/s^2

The time taken by the electron, t = 1.7 × 10^(-10) s

Given,

uniform electric field, E = 10^6 N/C

speed, v = 0.1 c

{c = 3 × 10^8 m/s}

To Find,

Acceleration, a

Time, t

Solution,

We know that Force = Mass × Acceleration

$\implies Eq = ma$

{where q is the charge of an electron, q = 1.6 × 10^(-19) C

m is the mass of an electron, m = 9.1 × 10(-31)}

$\implies a = \frac{Eq}{m}\\\\\implies a = \frac{10^6 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}\\\\\implies a = 1.76 \times 10^{17} \hspace{0.1 cm} m/s^2$

Therefore, the acceleration of the electron is 1.76 × 10^17 m/s^2

Now, to find the time taken by the electron, we use the equations of motion

$\implies v = u + at$

Here, the initial velocity, u = 0, since the body is released from rest

$\implies v = at\\\\\implies t = \frac{v}{a}\\\\\implies t = \frac{0.1 \times 3 \times 10^8}{1.76 \times 10^{17}}\\\\\implies t = 1.7 \times 10^{-10} s$

Therefore, the time taken by the electron is 1.7 × 10^(-10) s

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