Question

Two charged plates are 2 mm apart. An electron escapes from the negatively charged plate to the positively charged plate in 1.2 * 10^-8 seconds. Find the electric field between the plates.

Answer 1

Two charged plates are $2\mathrm{mm}$ apart. An electron escapes from the negatively charged plate to the positively charged plate in $1.2\cdot 10^{-8}$ seconds. Find the electric field between the plates.

Solution.

Figure 1.

Write Newton's Second Law in projections on the X-axis (see Figure 1):

$$m _ {e} a = E e,$$

where $m_{e} = 9.1 \cdot 10^{-31} \, \mathrm{kg}$ is the mass of the electron, $E$ is the module of the electric field between the plates, $a$ is the module of the acceleration of the electron, $e = 1.6 \cdot 10^{-19} \, \mathrm{C}$ is the elementary charge.

Find $a$ :

$$a = \frac {E e}{m _ {e}}.$$

The initial velocity of the electron equals zero. So using the formula for distance for uniformly accelerated motion we have:

$$d = \frac {a t ^ {2}}{2} = \frac {E e t ^ {2}}{2 m _ {e}},$$

where $d$ is the distance between the plates, $t$ is the time that the electron spends to go from the negative charged plate to the positive charged plate. From the formula above we find $E$ :

$$E = \frac {2 m _ {e} d}{e t ^ {2}} = \frac {2 \cdot 9 . 1 \cdot 1 0 ^ {- 3 1} \cdot 2 \cdot 1 0 ^ {- 3}}{1 . 6 \cdot 1 0 ^ {- 1 9} \cdot 1 . 2 ^ {2} \cdot 1 0 ^ {- 1 6}} = 1 5 8 \mathrm {V / m}.$$

Answer: $158\mathrm{V / m}$