Question

A proton moving at speed 100 km/s flies into a homogeneous electric field of width 1 m and field intensity 100 V/m at an angle 60° to the electric field lines. In what time will the proton leave the area of homogeneous electric field? Neglect the effects of gravity. Take the ratio of proton charge to its mass equal to 9.6 x (10)^7 C/kg. Give your answer in microseconds and round it to the whole.

Answer 1

The force that the proton will experience:

$$F=Eq\text{ cos}60°.$$

The deceleration the proton will experience:

$$a=\frac{F}{m}=\frac{Eq\text{ cos}60°}{m},\\\space\\ a=-100\cdot(9.6\cdot10^7)\text{ cos}60°=4.8\cdot10^9\text{ m/s}^2.$$

So, we have initial velocity of 100 km/s, deceleration, and final velocity, which is

$$v_f=\sqrt{v_i^2+2ad}.$$

The time it will take:

$$t=\frac{v_f-v_i}{a}=\frac{\sqrt{v_i^2+2ad}-v_i}{a},\\\space\\ t=\frac{\sqrt{(10^5)^2+2(-4.8\cdot10^9)1}-10^5}{-4.8\cdot10^9}=17\cdot10^{-6}\text{ s}.$$

The answer is 17 μs.