Q.
Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_{2} - V_{1} = 20$V. ($ie$, plate 2 is at a higher potential). The plates are separated by $d = 0.1$m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e=1.6$\times 10^{- 19}C,m_{0} = 9.11 \times 10^{- 31}kg)$
Explanation
(a) Since $V_{2} > V_{1},$ so electric field will point from plate 2 to plate 1.
The electron will experience an electric force, opposite to the direction of electric field, and hence move towards the plate 2.
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Use work-energy theorem to find speed of electron when it strikes the plate 2.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{m_{e}v^{2}}{2} - 0 = e(V_{2} - V_{1})$
Where $v$ is the required speed.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{9.11 \times 10^{- 31}}{2}v^{2} = 1.6 \times 10^{- 19} \times 20$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \ \sqrt{\frac{1.6 \times 10^{- 19} \times 40}{9.11 \times 10^{- 31}}} = 2.65 \times 10^{6}{ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.