Question

an electron aquires 3.45 e-16j of kinetic energy when it is accelerated by an electric field from plate a to b in a computer monitor what is the potential difference between the plates and which plate has the higher potential?

Answer 1

Potential difference will be equal to 2156.25 volt

Step-by-step explanation:

We have given kinetic energy of electron
`KE=3.45* 10^(-16)J`

Charge on electron
`e=1.6* 10^(-19)C`

Let the potential difference is V

So energy electron will be equal to
`E=qV`, here q is charge on electron and V is potential difference

This energy will be equal to kinetic energy of the electron

So
`1.6* 10^(-19)* V=3.45* 10^(-16)`

V = 2156.25 volt

So potential difference will be equal to 2156.25 volt