Question

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 ? $\left(e=1.6\times10^{-19}C, m_{e}=9.11\times10^{-31}\,kg\right)$

• A. $2.65\times10^{6}\,m/s$
• B. $7.02\times10^{12}\,m/s$
• C. $1.87\times10^{6}\,m/s$
• D. $32\times10^{19}\,m/s$

Answer 1

Answer: (A)

$2.65\times10^{6}\,m/s$

Answer 2

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. $\therefore \frac{9.11\times10^{-31}}{2}v^{2}=1.6\times10^{-19}\times20$ $\Rightarrow v=\sqrt{\frac{1.6\times10^{-19}\times40}{9.11\times10^{-31}}}=2.65\times10^{6}\,m/s$