Question

Two 3.0-cm-diameter disks are parallel to each other, 1.0 mm apart. They are charged to ± 10 nC .

A. What is the electric field strength between the disks?

B. A proton is placed somewhere between the two disks. What is the magntiude of the electric force acting on that proton due to the electric field between the discs?

C. 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? Assume the weight of the proton is negligible.

Answer 1

a] Electric field E = sigma/e0

= [q/area]/e0

= [10e-9/(pi*0.015^2)]/8.85e-12

= 1598543 N/C

b] Force = qE = 1.6e-19*1598543 = 2.558*10^-13 N

c] work done by force = change in KE

-F*d = - 0.5mv^2

- 2.558*10^-13*1e-3 = -0.5*1.673e-27*v^2

v = 553000 m/s answer