Question

What potential difference is needed to accelerate an electron from rest to a speed of 1.6×106 ?

Answer 1

What potential difference is needed to accelerate an electron from rest to a speed of $1.6 \cdot 10^{6} \frac{m}{s}$?

**Solution.**

$$|q| = 1.6 \cdot 10^{-19} C, \quad m = 9.1 \cdot 10^{-31} kg, \quad v_1 = 0, \quad v_2 = 1.6 \cdot 10^{6} \frac{m}{s};$$ $$U - ?$$

Change the kinetic energy of the electron is equals the work of the electric field on the electron transport:

$$\Delta W = A.$$

Change the kinetic energy of the electron:

$$\Delta W = W_2 - W_1;$$ $$W_1 = \frac{m v_1^2}{2};$$

$v_1 = 0$, then $W_1 = 0$ - an electron at rest.

$$W_2 = \frac{m v_2^2}{2};$$ $$\Delta W = \frac{m v_2^2}{2} - 0 = \frac{m v_2^2}{2}.$$

The work of the electric field on the electron transport:

$$A = |q| U.$$ $$\Delta W = A.$$ $$\frac{m v_2^2}{2} = |q| U;$$

The potential difference:

$$U = \frac{m v_2^2}{2 |q|}.$$ $$U = \frac{9.1 \cdot 10^{-31} \cdot (1.6 \cdot 10^{6})^2}{2 \cdot 1.6 \cdot 10^{-19}} = 7.28 (V).$$

Answer: $U = 7.28 V$.