Question

A proton moving at v₀ = 1.70 ✕ 10⁶ m/s enters the region between two parallel plates with charge densities of magnitude σ = 2.60 ✕ 10⁻⁹ C/m² (see the figure below).

A uniform electric field is produced by two charged horizontal plates, each of length d, where the positive plate is above the negative plate. The leftmost side of each plate is at horizontal position x = 0. The upper plate has charge density +σ and the lower plate has charge density −σ. Several parallel vertical arrows representing vector E point from the positive plate down to the negative plate.

A proton is located at x = 0 and vertical position y = 0, to the left of and midway between the plates. Velocity vector v₀ points rightward from the proton into the space between the plates.

(a)

Calculate the magnitude of the electric field (in N/C) between the plates.

293.79 N/C

(b)

Calculate the magnitude of the electric force (in N) acting on the proton.

4.70e-17 N

(c)

Find the y-location of the proton (in m) when it reaches the far edge of the plates, a horizontal distance d = 2.50 ✕ 10⁻² m from where it entered. Assume that the proton does not hit either of the plates. (Assume the +x direction is to the right and the +y direction is up.)

______ m

Answer 1

DY Given 0 = 2.60 xloºc/m2 ♡ = 1.70x106mle (a) Magnitude of electric field is given by 1 El = 5 O Litu le Eo=8.854X10 = 2.60X1o 9 . = 293.652 8.854 x 10 a 293 NIC -12 (b) Magnitude of force - fa q, E = 106 x 10" X293 = 4.688X10' ~ 4.7 x 107N

(c) The deflection of Proton Given by Formula - y = q EL2 a mo ? -2 where q = 1.681019c E= 293Nlc L=d=2. 50X10 M= mass of Preton = 1.67x17??kg Voyez Vo = 1.70x106m/s 4= (1.6x15'9 ) x 293 x 12.50 x102) 2x11.6 7 X10 27) x (1.70x106)2 2 930 X10 -23 - m 9.652 6 x 1015 - 303. 54 xios m = 3.0354 x16m y = - 300354 x1őom The Electric field is in Dowan ward a Direction (- y Direction) Hence The Force is on Proton is in downward direction There fore it will Deflect Downwards ise, -y direction Hence Negative sign is placed.