Question

The electric field strength is 4.60×104 V/m inside a parallel-plate capacitor with a 2.1 mm spacing. A proton is released from rest at the positive plate. What is the proton's speed when it reaches the negative plate? Please give answer in m/s and show all work!

Answer 1

Answer:

1.36 x 10^5 m/s

Explanation:

E = 4.6 x 10^4 V/m

d = 2.1 mm = 2.1 x 10^-3 m

The kinetic energy of proton is due to the potential difference between the plates. Let v be the speed.

1/2 mv^2 = q V

v^2 = 2 e V / m

V = E d = 4.6 x 10^4 x 2.1 x 10^-3 = 96.6 V

v^2 = (2 x 1.6 x 10^-19 x 96.6) / (1.67 x 10^-27)

v = 1.36 x 10^5 m/s

Answer 2

Answer:

$v=1.36\times 10^5\ m/s$

Explanation:

It is given that,

Electric field strength inside a parallel plate capacitor, $E=4.6\times 10^4\ V/m$

Spacing between plates, d = 2.1 mm = 0.0021 m

A proton is released from rest at the positive plate, u = 0

We need to find the speed of proton when it reaches the negative plate of the capacitor. Let it is given by v.

$qE=ma$

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

Using third equation of motion as :

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

$v^2=2ad$

$v=\sqrt{2ad}$

$v=\sqrt{2\dfrac{qE}{m}{d}$

Where,

q is the charge on proton

m is the mass of the proton

$v=\sqrt{2\times \dfrac{1.6\times 10^{-19}\times 4.6\times 10^4}{1.67\times 10^{-27}}\times {0.0021}}$

v = 136052.12 m/s

or

$v=1.36\times 10^5\ m/s$

So, the speed of proton when it reaches the negative plate is $1.36\times 10^5\ m/s$. Hence, this is the required solution.