Question

Two opposite charged boards (area = 1.0 cm², magnitude of charge = 2.0 uC) are 1.0 cm apart, as shown in the figure. A proton (mass = 6.67x10-27kg, e = 1.6x10-19C) is shot with an initial speed vo from the bottom toward the top vertically. What is the minimum initial speed needed that the proton can reach the top board? Ignore the gravity. [€ = 8.854 x 10-12c2/N] + VO 1.0 cm -0 30 ? ? wi ? A

Answer 1

7 x 10 - 27 xlo kg 6.67 mass of proton, mpa Change of proton, 9-1-6x1096 P -2m Ixlo m. distance between two plates , Y = area of a plate a plate = 1cm² 104 m² 13 A magnitude of charge on plates,q 2 xióc Surface charge density, r = -6 2X10 ele 10 -4 2 x 10 - 2c/m² Electric field between twoplates, E blo FE.V ma = I.V Eo a I. Ve Eo m - 2 2 x 10 az x 1•6x10-19 -13 -27 8.854x10 x 6.GTX10

-21 a = 3.2x 10 59.05 x 10-40 19 x 10 a = 0.0 5419 17 a = 5.419 x 10 2. m/s² Now , v² u²= zas V=0 0² uz 2x5-419 x 10 x 2 IX10 u 2 = 10.838 x 10 105 14 u = 1.0838 x 10 7 u= 1:04 x 107 mis Initial speed of proton, u = 1:04 x107 ms