Question

An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accelerated by the presence of the electric field. 50% Part (a) What is the change in the electron’s kinetic energy, in kiloelectron volts, if it travels over a distance of 0.25 m in this field? ΔK = - 4.63 * 105|

Answer 1

`W = 462.5 keV`

Step-by-step explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

`F = qE`

`F = (1.6 * 10^(-19))(1.85 * 10^6)`

`F = 2.96 * 10^(-13) N`

now the distance moved by the electron is given as

`d = 0.25 m`

so we have

`W = F.d`

`W = (1.6 * 10^(-19))(1.85 * 10^6)(0.25)`

`W = 7.4 * 10^(-14) J`

now we have to convert it into keV units

so we have

`1 keV = 1.6 * 10^(-16) J`

`W = 462.5 keV`