If an electron is accelerated from rest through a potential

If an electron is accelerated from rest through a potential difference of 11 kV, what is its resulting speed? (e = 1.60 × 10-19 C, k = 1/40 = 8.99 × 109 N m2/C2, mel = 9.11 x 10-31 kg). (Give your answer to the nearest km/s).

Solution

We know that the work done by the electric field will be equal to the change in kinetic energy

1/2 mv2 = e*V

Where m is the mass of the electron, V is the final velocity and V is the potential difference

v= sqrt (1.6*10^-19 *11000*2/9.11*10^-31)

v = 62* 10⁶ m/s