How Do Electric Fields Affect Particle Movement?

[Tariq]

Homework Statement

1. What is the magnitude and direction of the electric field at point Z in the diagram?

q = - 2.0 x 10-5 C q = + 8.0 x 10-6 C
60 cm 30 cm
X Y Z

2. An electron is released from rest adjacent to the negative plate in a parallel plate apparatus. A potential difference of 500 V is maintained between the plates, and they are in a vacuum. With what speed does the electron collide with the positive plate?

Homework Equations

Electric Field type questions

The Attempt at a Solution

I attempted them but I don't remember how to do them, it was a long time. I need some assistance. Your help will be greatly appreciated. Thank you.

 

[Tariq]

nvm I got the answer for the first one though.

 

[Moetasim]

1. The magnitude of the electric field at point Z can be calculated using the formula E = kq/r^2, where k is the Coulomb's constant, q is the charge and r is the distance between the point and the charge. In this case, the distance between point Z and the negative charge is 60 cm, and the distance between point Z and the positive charge is 30 cm. Plugging in the values, we get:

E = (9 x 10^9 Nm^2/C^2) x (-2.0 x 10^-5 C) / (0.6 m)^2 + (9 x 10^9 Nm^2/C^2) x (8.0 x 10^-6 C) / (0.3 m)^2

= -1.5 x 10^4 N/C + 8 x 10^4 N/C

= 6.5 x 10^4 N/C

The direction of the electric field at point Z will be towards the positive charge, as it has a higher magnitude compared to the negative charge.

2. The speed of the electron can be calculated using the formula v = √(2qV/m), where q is the charge of the electron, V is the potential difference and m is the mass of the electron. Plugging in the values, we get:

v = √(2 x (-1.6 x 10^-19 C) x 500 V / (9.11 x 10^-31 kg)

= 4.2 x 10^7 m/s

Therefore, the electron will collide with the positive plate at a speed of 4.2 x 10^7 m/s.