Question

The electric field strength is 20,000 N/C inside a parallel-plate capacitor with a 1.0 mm spacing. An electron is released from rest at the negative plate. What is the electron’s speed when it reaches the positive plate?

Answer 1

The electron’s speed when it reaches the positive plate is 2.65 X 10⁶m/s

Explanation:

Electric field strength, E = 20,000 N/C

Distance, x = 1  mm = 1 X 10⁻³m

Speed of the electron, s = ?

We know,

Charge of an electron, q = 1.602 X 10⁻¹⁹ C

So the force, F is

20000 N/C x  1.602 X 10⁻¹⁹ C  = 3.2 X 10⁻¹⁵ Newtons

mass of electron is 9.1×10⁻³¹ kg

F = ma

a = 3.2 X 10⁻¹⁵ / 9.1 X 10⁻³¹ = 3.52 X 10¹⁵ m/s²

v = √(2ad) = √(2*3.52 X 10¹⁵ X 0.001) = 2.65 X 10⁶m/s

Therefore, the electron’s speed when it reaches the positive plate is 2.65 X 10⁶m/s