1. Tardigrade
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  4. Two insulating plates are both uniformly charged in such a way that the potential difference between them is V2 - V1 = 20 V. (i.e., 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=1.6×10-19C, me=9.11×10-31 kg)

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Q. Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2 - V_1 = 20\, V.$ (i.e., 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 ?
$\left(e=1.6\times10^{-19}C, m_{e}=9.11\times10^{-31}\,kg\right)$Physics Question Image

AIEEEAIEEE 2006Electrostatic Potential and Capacitance

Solution:

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.
Use work-energy theorem to find speed of electron when it strikes the plate 2.
$\frac{m_{e}V^{2}}{2}-0=e\left(V_{2}-V_{1}\right)$
where $v$ is the required speed.
$\therefore \frac{9.11\times10^{-31}}{2}v^{2}=1.6\times10^{-19}\times20$
$\Rightarrow v=\sqrt{\frac{1.6\times10^{-19}\times40}{9.11\times10^{-31}}}=2.65\times10^{6}\,m/s$

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