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

2.1406 ×
`10^6` m/sec

Step-by-step explanation:

we know that energy is always conserved

so from the law of energy conservation

`qV=(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=((1)/(2)* 1.67* 10^(-27)50000^2)/(1.6* 10^(-19))=13.045`

now mass of electron =9.11×
`10^(-31)`

so for electron

`(1)/(2)* 9.11* 10^(-31)v^2=1.6* 10^(-19)* 13.045=2.1406* 10^6 m/sec`

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