First slide

Motion of charged particle in uniform electric field

Question

An electron moving with the speed 5 x $10^{6}$ m/s is shot parallel to the electric field of intensity 1 x $10^{3}$ N/C. Field is responsible for the retardation of motion of electron. Evaluate the distance travelled by the electron before coming to rest for an instant?
(Mass of e = 9 x $10^{- 31}$ kg, charge= 1.6 x $10^{- 19}$ C)

Moderate

A

7 m
B

0.7 mm
C

7 cm
D

0.7 cm

Solution

Electric force, qE = ma ⇒ a = $\frac{QE}{m}$
$\begin{array}{l} ∴ a = \frac{1 .6 \times 10^{- 19} \times 1 \times 10^{3}}{9 \times 10^{- 31}} = \frac{1 .6}{9} \times 10^{15} m / s^{2} \\ u = 5 \times 10^{6} m / s and v = 0 \\ ∵ v^{2} = u^{2} - 2 as \Rightarrow s = \frac{u^{2}}{2 a} \\ ∴ Distance, s = \frac{{(5 \times 10^{6})}^{2} \times 9}{2 \times 1 .6 \times 10^{15}} = 7 cm (approx) \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App