Answer:
the speed of the proton is [tex]v = 78.1 km/s[/tex]
Explanation:
Generally Electric potential energy resulting from a proton can be mathematically represented as
[tex]V(z) = \frac{Q(\sqrt{R^2 + z^2} - z )}{2 \pi \epsilon_o R^2}[/tex]
Where
R is the radius of the disc which is given as [tex]R = 2m[/tex]
q is the charge on the proton with a value of [tex]q = 1.602 *10^{-19}C[/tex]
the mass of this proton has a value of [tex]m=1.673*10^{-27}kg[/tex]
z is the distance of the center of the disk from the question
[tex]z_1 =1.0cm = \frac{1}{100} = 0.01m[/tex]
[tex]z_2 = 5cm = \frac{5}{100} = 0.05m[/tex]
[tex]\epsilon_0 = 8.85*10^{-12}F/m[/tex]
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 [tex]\Delta KE = \frac{1}{2} m (v^2 - u^2)[/tex]
Since u = 0 the equation becomes
[tex]\Delta KE = \frac{1}{2} m (v^2)[/tex]
change in electrostatic potential energy is [tex]= q (V(z_1) - V(z_2))[/tex]
Substituting this into the equation we have
[tex]\frac{1}{2} mv^2 +q(V(z_1)- V(z_2)) = 0[/tex]
[tex]\frac{1}{2} mv^2 = -q(V(z_1)- V(z_2))[/tex]
Recalling that [tex]V(z) = \frac{Q(\sqrt{R^2 + z^2} - z )}{2 \pi \epsilon_o R^2}[/tex] we have
[tex]\frac{1}{2} mv^2 = -q(\frac{Q(\sqrt{R^2 +z_2^2}-z_2 )}{2 \pi \epsilon_0 R^2} - \frac{Q(\sqrt{R^2 +z_1^2}-z_1 )}{2 \pi \epsilon_0 R^2} )[/tex]
[tex]\frac{1}{2} mv^2 = q(\frac{Q(\sqrt{R^2 +z_1^2}-z_1 )}{2 \pi \epsilon_0 R^2} - \frac{Q(\sqrt{R^2 +z_2^2}-z_2 )}{2 \pi \epsilon_0 R^2} )[/tex]
Substituting values [tex]\frac{1}{2} mv^2 = 1.602*10^{-19}(\frac{180*10^{-9}(\sqrt{2^2 +0.01^2}-0.011 )}{2 \pi 8.85*10^{-12}* 2^2} - \frac{180*10^{-9}(\sqrt{2^2 +0.05^2}-0.05 )}{2 \pi *8.85*10^{-12} *2^2} )[/tex]
Making v the subject
[tex]v=\sqrt{\frac{1.602*10^{-19}(\frac{180*10^{-9}(\sqrt{2^2 +0.01^2}-0.011 )}{2 \pi 8.85*10^{-12}* 2^2} - \frac{180*10^{-9}(\sqrt{2^2 +0.05^2}-0.05 )}{2 \pi *8.85*10^{-12} *2^2} )}{0.5 *1.673*10^{-27}}}[/tex]
[tex]v = 78.1 km/s[/tex]
