PART A) A pair of oppositely charged parallel plates are separated by 5.54 mm. A potential difference of 550 V exists between the plates. What is the…

PART A) A pair of oppositely charged parallel plates are separated by 5.54 mm. A potential difference of 550 V exists between the plates. What is the strength of the electric field between the plates? Answer in units of V/m.PART B) What is the magnitude of the force on an electron between the plates? Answer in units of N.PART C) How much work must be done on the electron to move it to the negative plate if it is initially positioned 3.08 mm from the positive plate? Answer in units of J.

Electric field is given byE = V/dE = 550 / (5.54 x 10-3)E = 9.927 x 104 V/m Force, F = q * EF = 1.6 x 10-19 x 9.927 x 104F = 1.588 x 10-14 N Work done, W = F . xW = 1.588 x 10-14 x 2.46 x…

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