Question

A proton moving horizontally at a speed of 2.88x10^5 m/s enters a region of space between two

square metal plates of side length 8.77 mm. The proton's path is deflected an angle of 12º upward. Protons have a mass of 1.67x10^-27 kg.

a) Determine the electric field between the plates.

b)Determine the amount and sign of electric charge on each plate.

Answer 1

horizontal speed v= 2.88x10 ⁵ m/s

Side length L =8.77 mm

                    = 8.77 x10 ^-3 m

deflected angle =12º

Protons mass m= 1.67x10 ^-27 kg.

Proton charge q = 1.6 x10 ^-19 C

a) The electric field between the plates E = ?

We know tan = qEL /(mv ²)

                     E = mv ² tan / qL

                        = [(1.67x10 ^-27 )( 2.88x10 ⁵ ) ² tan 12] / [1.6 x10 ^-19 x 8.77 x10 ^-3 ]

                        = (2.944 x10 ^-17 ) / [1.4 x10 ^-21]

                        = 20980.6 N/C

(b). Charge on each plate Q = CV      Where C = Capacitance = A/d and V = potential difference =Ed

                                         = (A/d) (Ed)

                                         = AE

Where A = area of crosssection = L ²

   = 7.69 x10 ^-5 m ²

         = permitivity of free space = 8.85 x10 ^-12 C ² / Nm ²

Substitute values you get , Q = 1.428 x10 ^-11 coulomb

upper plate is negative and bottom plate is positive.