1.34 Suppose that the particle in Exercise in 1.33 is an electron projected with velocity = 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.)

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1.34 Suppose that the particle in Exercise in 1.33 is an electron projected with velocity = 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, m_e = 9.1 × 10^–31 kg.)

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alok kumar singh | Contributor-Level 10

5 months ago

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

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

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

Charge of an electron, q = 1.6 $* 10^{- 19}$ C

Mass of an electron, $m_{e}$ = 9.1 $* 10^{- 31}$ kg

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

s = $\frac{q e L^{2}}{2 m {v_{x}}^{2}}$

L = $\sqrt[]{\frac{2 s m {v_{x}}^{2}}{q E}}$ = $\sqrt[]{\frac{2 * 0.005 * 9.1 * 10^{- 31} * ({2.0 * 10^{6})}^{2}}{1.6 * 10^{- 19} * 9.1 * 10^{2}}}$ = $3.64* 10^{- 20} 1.456 * 10^{-16}$

= 0.0158 m = 1.58 cm

Therefore, the electron will strike the upper plate after travelling 1.58 cm.

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