Question

A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negativeplate with a speed of 50,000 m/s. What will be the final speed ofan electron released from rest at the negative plate

Answer 1

Answer:

2.1406 ×$10^6$ m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

$qV=\frac{1}{2}mv^2$

here V is the potential difference  

we know that mass of proton = 1.67×$10^{-27}$ kg

we have given speed =50000m/sec

so potential difference $V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045$

now mass of electron =9.11×$10^{-31}$

so for electron

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec$

so the velocity of electron will be 2.1406×$10^6$ m/sec