Ncert Exercises & Solutions

1.34 Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx = 2.0 × 106 m s^–1. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate? (|e|=1.6 × 10^–19 C, me = 9.1 × 10^–31 kg.)

Velocity of the particle, $v_{x} = 2.0 \times 10^{6} m / s$

Separation of the two plates, d = 0.5 cm=0.005 m

Electric field between the two plates, E = 9.1 x $10^{2}$ N/C

Charge on an electron, $q = 1 .6 \times 10^{- 19} C$

Mass of an electron, $m_{e} = 9 .1 \times 10^{- 31} kg$

Let the electron strike the upper plate at the end of plate L, when deflection is s. Therefore,

$\begin{array}{l} s = \frac{q E L^{2}}{2 m V_{x}^{2}} \\ \Rightarrow L = \sqrt{\frac{2 s m V_{x}^{2}}{q E}} = \sqrt{\frac{2 \times 0.005 \times 9.1 \times 10^{- 31}}{1.6 \times 10^{- 19} \times 9.1 \times 10^{2}}} \\ = \sqrt{0.00025} = 0.016 m = 1.6 c m \end{array}$

Therefore, the electron will strike the upper plate after traveling at 1.6 cm.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions