Problem 3- Uniform electric field: Force, Potential and Kinetic Energy The elect

Question

Problem 3- Uniform electric field: Force, Potential and Kinetic Energy The electric field between two parallel plates that are separated by a distance d 5.0 mm is directed vertically downward and has a magnitude E-300 V/m. a) Compute the force (magnitude and direction) acting on charge Q1 -2.0 nC located half-way between the two plates. b) What is the potential energy of the charge Q1? c) If we release the charge Qh from rest, what is its kinetic energy when it reaches the lower plate? d) Compute the force (magnitude and direction) acting on charge Q -4.0 nC located half- way between the two plates. What is the potential energy of the charge 02? e)

Explanation / Answer

d = 5*10^-3 m

E = 300 v/m

a) F = Q1*E = 2*10^-9*300 = 6*10^-7 N

direction is vertically downwards

b) work done by the electric field is stored as potential energy

W = F*S = 6*10^-7*(5*10^-3)/2) = 1.5*10^-9 J is the potential energy of charge

C) KE = W = F*S = 6*10^-7*5*10^-3 = 3*10^-9 J

D) Force is F = Q2*E = 4*10^-9*300 = 1.2*10^-6 N

Direction is vertically upwards

E) PE = F*S = 1.2*10^-6*(2.5*10^-3) = 3*10^-9 J

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