Question

Two 2.6 cm -diameter disks face each other, 3.0 mm apart. They are charged to ±20nC. 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? Express your answer to two significant figures and include the appropriate units.

Answer 1

  By C = ɛo x A/d
=>& C = q/V
=>V = q/C
=>V = qd/ɛoA
=>By E = V/d
=>E = q/ɛoA
=>E = (20 x 10^-9)/(8.85 x 10^-12 x πr^2)
=>E = (20 x 10^-9)/[8.85 x 10^-12 x 3.14 x (0.013)^2]
=>E = 4.25 x 10^6 N/C
The force on proton (F) = eE
=>The work done (W) = F x s = 1.6 x 10^-19 x 4.25 x 10^6 x 3 x 10^-3 = 2.04 x 10^-15
Thus By work energy relation:-
=>W = ∆KE
=>2.04 x 10^-15 = 1/2 x 1.67 x 10^-27 x v^2
=>v = √[2.443 x 10^12]
=>v = 15.63 x 10^5 m/s