Answer to Question #88045 in Electric Circuits for Ahmed abdo

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