Question

An electron moving in a electric potential field $V_1$ enters a higher electric potential field $V_2$ then the change in kinetic energy of the electron is proportional to

Answer 1

The higher potential is given as $V_2$

The lower potential is given as $V_1$

Hence, the potential difference is calculated as $(V_2-V_1)$

The kinetic energy gained by the electron due to the potential difference is the product of the charge of electron with the potential difference.

Mathematically kinetic energy K.E $=e(V_2-V_1)$ [1]

We are asked to calculate the velocity of electron.

Let the velocity of the electron is $v$ and mass is $m$

Hence, the kinetic energy of the electron is calculated as -

$K.E=\frac{1}{2}m{v}^2$ [2]

From 1 and 2 we get,

$⟹\frac{1}{2}m{v}^2=e(V_2-V_1)$

$⟹v^2=\frac{2e}{m}{V}_2-V_1)$

$⟹v=\sqrt{\frac{2e}{m}{V}_2-V_1)}$