Question

A uniform electric field of strength “E ” = 22.29 N/C is located in-between two large parallel separated by a distance d = 18.06 cm as per the figure below. A proton (p) and an electron (e) are released from the positive and negative plates (respectively) at the same time. How much time would elapse from the moment of releasing the charges till the moment they pass each other? Express your answer in microseconds. (Neglect effect of charged particles on each other).

Answer 1

Solution :

Given :

E = 22.29 N/C

d = 18.06 cm = 0.1806 m

.

Here, Acceleration of Proton will be :

a_p = F / m = q E / m = {(1.6 x 10^-19)(22.29)} / (1.67 x 10^-27) = 2.14 x 10⁹ m/s²

Similarly, Acceleration of electron will be :

a_e = F / m = q E / m = {(1.6 x 10^-19)(22.29)} / (9.1 x 10^-31) = 3.92 x 10¹² m/s²

.

Now, Using formula : s = u t + (1/2) a t²

Let the distance traveled by proton be x when they passes each other.

Then : x = (1/2) a_p t²

And, For electron : d - x = (1/2) a_e t²

.

Therefore, From above two equations :

∴ d - x = (1/2) a_e t²

∴ d - (1/2) a_p t² = (1/2) a_e t²

∴ d ​​​​​​​= (1/2) a_p t² + (1/2) a_e t² ​​​​​​​

∴ 2 d ​​​​​​​= (a_p + a_e) t² ​​​​​​​

∴ 2(0.1806) ​​​​​​​= { (2.14 x 10⁹ m/s²) + (3.92 x 10¹² m/s²) } t² ​​​​​​​

∴ (0.3612) ​​​​​​​= { 3.92 x 10¹² } t² ​​​​​​​

∴ (9.21 x 10^-14) ​​​​​​​=  t² ​​​​​​​

∴ t = 3.035 x 10^-7 sec = 0.3035 x 10^-6 sec = 0.3 μs