First slide

Electric field

Question

A charged particle of radius $5 \times 10^{- 7} m$ is located in a horizontal electric field of intensity $6.28 \times 10^{5} V / m$ . The surrounding medium has the coefficient of viscosity $η = 1.6 \times 10^{5} N - s / m^{2}$ . The particle start moving under the effect of electric field and finally attains a uniform horizontal speed of $0.02 m / s$ . Find the $_______\times 10^{8}$ number of electrons on it. Assume gravity free space

Moderate

Solution

$W h e n e l e c t r i c f o r c e a n d d r a g f o r c e b a l a n c e s e a c h o t h e r, p a r t i c l e m o v e s w i t h u n i f o r m s p e e d . ∴ q E = 6 π η r v \Rightarrow (N e) E = 6 π η r v \Rightarrow N = \frac{6 π η r v}{e E} \Rightarrow N = \frac{6 π \times 1.6 \times 10^{5} \times 5 \times 10^{- 7} \times 0.02}{1.6 \times 10^{- 19} \times 6.28 \times 10^{5}} \Rightarrow N = 3000 \times 10^{8}$

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