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?

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

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explanation: