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Transcript
Motion of charges in uniform fields. [V/m] = [N/C] q1 charged sphere or point charge parallel charged plates E K q1/r2 V Kq1/r ΔV/d F = q2 E ΔV x/d P.E. = q2 V d = do + vot + ½ at2 e = 1.6 x 10-19 C me = 9.11 x 10-31 kg _______________________________________________________ [1] Two plates are separated by 10 cm. and have a potential difference of 200 V between them, positive plate above the negative. An electron enters the field 5 cm above the bottom plate and moving at 10 m/s horizontally (parallel to the plates). Write expressions for the position of the electron in the field. You may neglect gravity here. [2] A 10-6g oil drop with charge -6e is stationary in the electric field between 2 plates separated by 1cm. Determine the potential difference ΔV between the plates needed to keep the drop stationary. [3] An electron is accelerated from rest at the negative plate through the electric field and through a hole in the positive plate. If the plates have a potential difference of 10 Volts, calculate the kinetic energy of the electron as it leaves the hole in the positive plate. Solutions: [1] Force on electron is upward, the electric force F = qE . The electron has horizontal velocity but no horizontal force. This is exactly like our old trajectory problems with an electric field in place of the gravity field. For x(t), we start at xo=0 with vx = 10 m/s and ax = 0 (no force) x(t) = 10 [m/s] t + 0 For y(t), we need to determine the acceleration. Fy = q E upward = e ΔV/d = (1.6E-19 C) (200 V)/(0.1m) = 3.2E-16 VC/m = 3.2E-16 N a = F/m = (3.2E-16 N)/(9.1E-31 kg) = 3.52E14 m/s2 Initial values are yo = 5 cm = .05m and vy0 = o so, y(t) = 0.05m + ½ 3.52E14 t2 [2] Two forces act on the oil drop; electric and gravity. They are equal and opposite and must balance to keep the drop stationary. So, q E = 6e ΔV/d = mg Solving for ΔV gives ΔV = mg d /6e ΔV = (1E-9 kg) (9.8 m/s2)(0.1m)/(6 x 1.6E-19C) = 9.8E-10 J / 1E-18 C = 9.8E8 V [3] This is most easily solved using conservation of energy. The electron begins at rest and is accelerated by the field to give kinetic energy and a lower potential energy. So, PEi = PEf + KE or ΔPE = KE ΔPE = q ΔV = (1.6E-19 C)(10 V) = 1.6E-18 J = KE. Another energy unit is commonly used in electric force problems, the electron volt or eV. In this problem, one electron is accelerated through 10 Volts, giving 10 eV. This means that one eV = 1.6E-19 J.