Question

An electron initially has a speed 16 km/s along the x-direction and enters an electric field of strength 27 mV/m that points in the y-direction and has a width of 22.5 cm. (a) How long does it take the electron to pass through the field? μs (b) How far is the deflection?

Answer 1

Answer:

(a). The time is 14.0 μs.

(b). The deflection is 0.47 m.

Explanation:

Given that,

Speed = 16 km/s

Electric field strength = 27 mV/m

Width = 22.5 cm

(a). We need to calculate the time

Using formula of velocity

$v=\dfrac{d}{t}$

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{22.5\times10^{-2}}{16\times10^{3}}$

$t=0.0000140625\ sec$

$t=14.0\times10^{-6}\ sec$

$t=14.0\ \mu\ s$

(b). We need to calculate the deflection

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times\dfrac{qE}{m}\times t^2$

Here, s = deflection

q = charge of electron

m = mass of electron

Put the value in the equation

$s=\dfrac{1}{2}\times\dfrac{1.6\times10^{-19}\times27\times10^{-3}}{9.1\times10^{-31}}\times(14.0\times10^{-6})^2$

$s=0.47\ m$

Hence, (a). The time is 14.0 μs.

(b). The deflection is 0.47 m.

Answer 2

Answer:

a)$t=1.4\times 10^{-5}\ s$

b)S= 46.4 cm

Explanation:

Given that

Velocity = 16 Km/s

V= 16,000 m/s

E= 27 mV/m

E=0.027 V/m

d= 22.5 cm

d= 0.225 m

a)

lets time taken by electron is t

d = V x t

0.225 = 16,000 t

$t=1.4\times 10^{-5}\ s$

b)

We know that

F = m a = E q                    ------------1

Mass of electron ,m

$m=9.1\times 10^{-31}\ kg$

Charge on electron

$q=1.6\times 10^{-19}\ C$

So now by putting the values in equation 1

$a=\dfrac{E q}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 0.027}{9.1\times 10^{-31}}\ m/s^2$

$a=4.74\times 10^{9}\ m/s^2$

$S= ut+\dfrac{1}{2}at^2$

Here initial velocity u= 0 m/s

$S= \dfrac{1}{2}\times 4.74\times 10^{9}\times (1.4\times 10^{-5})^2\ m$

S=0.464 m

S= 46.4 cm

S is the deflection of electron.