Question

A proton initially at rest is accelerated by a unifor electric field. The proton moves 6.40 cm in 1.13 10-6 s. Find the voltage drop through whiclh the proton moves.

Answer 1

So voltage drop will be 16.7125 volt

Step-by-step explanation:

We have distance traveled by proton = 6.40 cm = 0.064 m

Time =
`T=1.13* 10^(-6)sec`

So velocity
`v=(distance)/(time )=(0.064)/(1.13* 10^(-6))=0.0566* 10^6m/sec`

Kinetic energy of proton is given by
`E=(1)/(2)mv^2=(1)/(2)* 1.67* 10^(-27)* (0.0566* 10^6)^2=2.674* 10^(-18)j`

And potential energy of proton is given by
`E=eV=1.6* 10^(-19)* V`

Both potential energy and kinetic energy will be same

So
`1.6* 10^(-19)* V=2.674* 10^(-18)`

`V=16.7125Volt`