What is the electric field strength between the disks?
A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?
The electric field between the plates is
E = σ/εo… where σ = area charge distribution
= 10X10^-9/(π*(0.010m)^2) = 3.183×10^-5C/m^2
and εo = 8.854×10^-12C^2/N-m^2
So E = 3.183×10^-5C/m^2/8.854×10^-12C^2/N-m^2 = 3.60×10^6N/C
Using conservation of energy we solve the second part
(K + U) b = (K + U)t…Here Ut – Ub = q*(Vt – Vb)…and Vt – Vb = E*d = 3.60×10^6V/m*1.0×10^-3m = 3595V
Since Kt = 0 we have Kb = q*E*d = 1.60×10^-19C*3595 = 5.75×10^-16J
So 1/2*m*v^2 = 5.75×10^-16J
Therefore v = sqrt(2*5.75×10^-16J/m) = sqrt(2*5.75×10^-16J/1.67×10^-27kg) = 8.30×10^5m/s
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