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Two insulating plates are both  uniformaly 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 )

  • Option 1)

    32\times 10^{-19}m/s

  • Option 2)

    2.65\times 10^{6}m/s

  • Option 3)

    7.02\times 10^{12}m/s

  • Option 4)

    1.87\times 10^{6}m/s

 

Answers (2)

As we learnt in

Parallel Plate Capacitor -

C=\frac{\epsilon _{0}A}{d}

- wherein

Area - A seperation between two plates.

 

 

An electron on plate 1 has electrostatic potentia energy. When it moves, potential energy is converted into kinetic energy

\therefore \: \: \: Kinetic \: energy\: = Electrostatic\: \: potential \: \: energy \:

or\: \: \: \: \: \frac{1}{2}mv ^{2}= e\Delta V

or\:\: \: \: v = \sqrt{\frac{2e\times \Delta V}{m}}= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 20}{9.11\times 10^{-31}}}

or\: \: \: \: \: v = 2.65\times 10^{6}\: \: m/s


Option 1)

32\times 10^{-19}m/s

Incorrect

Option 2)

2.65\times 10^{6}m/s

Correct

Option 3)

7.02\times 10^{12}m/s

Incorrect

Option 4)

1.87\times 10^{6}m/s

Incorrect

Posted by

Sabhrant Ambastha

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