Question

NOTE: FOR AN INSULATING SHEET THE FIELD GOES IN BOTH DIRECTIONS. E-a/(20) 1. At time t = 0 a proton is a distance of 0.360 nm from a very large insulating sheet of charge and is moving parallel to the sheet with a speed 9.70 x 10° m/s. The sheet has a uniform charge density of 2.34 x 10° C/m2. What is the speed of the proton att 5.001093 $

Answer 1

Electric field due to insulating sheet(E) = $\sigma$/(2*epsilon)

given, $\sigma$ = 2.23*10^-9 C/m^2

epsilon = 8.85*10^-12

So, E = (2.23*10^-9)/(2*8.85*10^-12)

E = 126 N/C

Since electrostatic force(Fe) = q*E

charge on proton = q= 1.60*10^-19 C

Fe = (1.60*10^-19)*(126)

Fe = 2.02*10^-17 N

acceleration of proton(a) = Fe/(mass of proton)

a = (2.02*10^-17)/(1.67*10^-27)

a = 1.21*10^10 m/sec^2

By kinematics law,

Vfy = Viy + a*t

here, Vfy = final speed of proton perpendicular direction of plate = ??

Viy = initial speed of proton perpendicular direction of plate = 0

t = 5*10^-8 sec.

So, Vfy = 0 + (1.21*10^10)*(5.00*10^-8)

Vfy = 605 m/sec.

final speed of proton in x direction remains same ,

So, Vfx = 9.70*10^2 m/sec. = 970 m/sec.

therefore net velocity(Vf) = sqrt(Vfy^2 + Vfx^2)

Vf = sqrt(605^2 + 970^2)

Vf = 1143.21 m/sec.

Vf = 1.14*10^3 m/sec.

Please upvote.