Question

6. A proton is released from rest 1.0 cm off a center of a uniformly charged disk of a 2.0 m radius carrying a +180 nC charge. What is the speed of the proton when it is 5 cm from the disk?

Answer 1

the speed of the proton is
`v = 78.1 km/s`

Step-by-step explanation:

Generally Electric potential energy resulting from a proton can be mathematically represented as

`V(z) = (Q(√(R^2 + z^2) - z ))/(2 \pi \epsilon_o R^2)`

Where

R is the radius of the disc which is given as
`R = 2m`

q is the charge on the proton with a value of
`q = 1.602 *10^(-19)C`

the mass of this proton has a value of
`m=1.673*10^(-27)kg`

z is the distance of the center of the disk from the question

`z_1 =1.0cm = (1)/(100) = 0.01m`

`z_2 = 5cm = (5)/(100) = 0.05m`

`\epsilon_0 = 8.85*10^(-12)F/m`

According the law of conservation of energy

The change in kinetic energy of the proton + change in electrostatic = 0

potential energy

The change in kinetic energy is
`\Delta KE = (1)/(2) m (v^2 - u^2)`

Since u = 0 the equation becomes

`\Delta KE = (1)/(2) m (v^2)`

change in electrostatic potential energy is
`= q (V(z_1) - V(z_2))`

Substituting this into the equation we have

`(1)/(2) mv^2 +q(V(z_1)- V(z_2)) = 0`

`(1)/(2) mv^2 = -q(V(z_1)- V(z_2))`

Recalling that
`V(z) = (Q(√(R^2 + z^2) - z ))/(2 \pi \epsilon_o R^2)` we have

`(1)/(2) mv^2 = -q((Q(√(R^2 +z_2^2)-z_2 ))/(2 \pi \epsilon_0 R^2) - (Q(√(R^2 +z_1^2)-z_1 ))/(2 \pi \epsilon_0 R^2) )`

`(1)/(2) mv^2 = q((Q(√(R^2 +z_1^2)-z_1 ))/(2 \pi \epsilon_0 R^2) - (Q(√(R^2 +z_2^2)-z_2 ))/(2 \pi \epsilon_0 R^2) )`

Substituting values
`(1)/(2) mv^2 = 1.602*10^(-19)((180*10^(-9)(√(2^2 +0.01^2)-0.011 ))/(2 \pi 8.85*10^(-12)* 2^2) - (180*10^(-9)(√(2^2 +0.05^2)-0.05 ))/(2 \pi *8.85*10^(-12) *2^2) )`

Making v the subject

`v=\sqrt{(1.602*10^(-19)((180*10^(-9)(√(2^2 +0.01^2)-0.011 ))/(2 \pi 8.85*10^(-12)* 2^2) - (180*10^(-9)(√(2^2 +0.05^2)-0.05 ))/(2 \pi *8.85*10^(-12) *2^2) ))/(0.5 *1.673*10^(-27))}`

`v = 78.1 km/s`