Question

Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2-V_1=20V.$ (i.e., plate 2 is at a higher potential). The plates are separated by $d=0.1m$ 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 ?
$\left(e=1.6\times 10^{-19}C,m_e=9.11\times 10^{-31}kg\right)$

• A. $2.65\times 10^{6}m/s$
• B. $7.02\times 10^{12}m/s$
• C. $1.87\times 10^{6}m/s$
• D. $32\times 10^{19}m/s$

Answer 1

Answer: (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.
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.
$\therefore \frac{9.11\times 10^{-31}}{2}{v}^2=1.6\times 10^{-19}\times 20$
$\Rightarrow v=\sqrt{\frac{1.6\times 10^{-19}\times 40}{9.11\times 10^{-31}}}=2.65\times 10^{6}m/s$