Question

An isolated free electron is traveling through an electric field from some initial point where it's coulombic potential energy per unit charge (voltage) is 17KJ/C and velocity = 93Mm/s to some final point where it's coulombic potential energy per unit charge is 6KJ/C. Determine the charge in velocity of the electron . Neglect gravitational forces. What is the solution please.

Answer 1

We have

$$\phi_1=17000 (V); \phi_2 =6000 (V); v_1=93\cdot10^6 (m/s)$$

And we need: light velocity; electron rest mass; electron charge

$$c=3\cdot 0^8 (m/s); m_0=9.1\cdot 10^{-31} (kg); q=-1.6\cdot 10^{-19} (C)$$

Comparing the speed of the electron with the speed of light

$$\frac{93\cdot 10^6 }{3\cdot 10^8}\cdot 100 = 30$$

We see 30 percents of the light velocity, Then this is a relativistic electron.

Kinetic energy in relativistic theory

$$W=m_0c^2(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)$$

For initial condition kinetic energy of electron

$$W_1=9.1\cdot 10^{-31}\cdot (3\cdot 10^8)^2\cdot(\frac{1}{\sqrt{1-\frac{(93\cdot 10^6)^2}{(3\cdot 10^8)^2}}}-1)=4.244*10^{-15} (J)$$

When electron travelling trough electric field, the energy change for

$$\Delta W=q(\phi_1-\phi_2)$$ $$\Delta W=-1.6\cdot 10^{-19} \cdot (17000-6000)=-1.76\cdot 10^{-15} (J)$$

The kinetic energy decreases

$$W_2=W_1-\Delta W$$ $$W_2=4.244\cdot 10^{-15} - 1.76 \cdot 10^{-15} = 2.484\cdot 10^{-15} (J)$$

And electron velocity from kinetic energy equal

$$v_2=c\sqrt{1-\frac{1}{(\frac{W_2}{m_0\cdot c^2}+1)^2}}$$

$$v_2=72.25\cdot 10^6 (m/s)$$

Change velocity

$$\Delta v=v_2-v_1$$ $$\Delta v= (72.25-93)\cdot 10^6=-20.75 (Mm/s)$$

Answer: velocity decreases on 20.75 Mm/s