Question

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

Answer 1

Answer:

2.65 x 10⁶ m/s

Explanation:

E = magnitude of electric field = 20,000 N/C

d = spacing between the plates of the capacitor = 1.0 mm = 0.001 m

ΔV = Potential difference between the plates = ?

Potential difference between the plates is given as

ΔV = E d

ΔV = (20000) (0.001)

ΔV = 20 Volts

q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of the electron as it reach the positive plate

using the conservation of energy

Kinetic energy gained = electric potential energy lost

(0.5) m v² = q ΔV

(0.5) (9.1 x 10⁻³¹) v² = (1.6 x 10⁻¹⁹) (20)

v = 2.65 x 10⁶ m/s

Answer 2

Explanation:

It is given that,

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

Spacing between parallel-plate capacitor, d = 1 mm = 0.001 m

Initial velocity of electron, u = 0

Let v is the electron’s speed when it reaches the positive plate. The force acting on the electron is :

$F=qE$

Also, $ma=qE$

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 20,000}{9.1\times 10^{-31}}$

$a=3.51\times 10^{15}\ m/s^2$

Using third equation of motion as :

$v^2-u^2=2ad$

$v=\sqrt{2ad}$

$v=\sqrt{2\times 3.51\times 10^{15}\times 0.001}$

v = 2649528.2599 m/s

or

$v=2.64\times 10^6\ m/s$

So, the velocity of the electron when it reaches the positive plate is $2.64\times 10^6\ m/s$. Hence, this is the required solution.