lectric field: Force, Potential and Kinetic Energy electric field between two pa

Question

lectric field: Force, Potential and Kinetic Energy 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. 2.0 nC located half-way a) ute the force (magnitude and direction) acting on charge Q between the two plates. What is the potential energy of the charge If we plate? b) c) rel release the charge Qi from rest, what is its kinetic energy when it reaches the lower Compute the force (magnitude and direction) acting on chargeOr -4.0 nC located half- way between the two plates. e) What is the potential energy of the charge 0?

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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