Question

A parallel-plate capacitor is constructed of two disks spaced 2.00 mm apart. It is charged to a potential difference of 500. V. A proton is shot through a small hole in the negative plate with a speed of 2.0 × 105 m/s. What is the farthest distance from the negative plate that the proton reaches?

Answer 1

Answer:

$0.000835\ \text{m}$

Explanation:

d = Distance between plates = 2 mm

V = Potential difference = 500 V

v = Velocity of proton = $2\times 10^5\ \text{m/s}$

a = Acceleration

m = Mass of proton = $1.67\times 10^{-27}\ \text{kg}$

Electric field is given by

$E=\dfrac{V}{d}\\\Rightarrow E=\dfrac{500}{2\times 10^{-3}}\\\Rightarrow E=250000\ \text{V/m}$

Force balance is given by

$ma=qE\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times 250000}{1.67\times 10^{-27}}\\\Rightarrow a=2.395\times 10^{13}\ \text{m/s}^2$

We have the relation

$v^2=u^2+2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(2\times 10^5)^2-0}{2\times 2.395\times 10^{13}}\\\Rightarrow s=0.000835\ \text{m}$

The farthest distance from the negative plate that the proton reaches is $0.000835\ \text{m}$.