question_answer

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\[d=0.1\text{ }m\]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? \[(e=1.6\times {{10}^{-19}}C,{{m}_{0}}=9.11\times {{10}^{-31}}kg)\] [AIEEE 2006]

A) \[2.65\times {{10}^{6}}\text{ }m/s\]

B) \[7.02\times {{10}^{12}}\text{ }m/s\]

C) \[1.87\times {{10}^{6}}m/s\]

D) \[32\times {{10}^{-19}}m/s\]

Correct Answer: A

Solution :

[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.

\[\frac{{{m}_{e}}{{v}^{2}}}{2}-0=e({{V}_{2}}-{{V}_{1}})\]             \[(\because \Delta KE=W)\]

where,\[{{m}_{e}}\]and are mass and charge respectively of the electron and v is the required speed.

\[\therefore \] \[\frac{9.11\times {{10}^{-31}}}{2}{{v}^{2}}=1.6\times {{10}^{-19}}\times 20\]

\[\Rightarrow \]   \[v=\sqrt{\frac{1.6\times {{10}^{-19}}\times 40}{9.11\times {{10}^{-31}}}}\]

\[=2.65\times {{10}^{6}}\,m/s\]