Two parallel horizontal plates are spaced 0.60 cm apart in air.You introduce an oil droplet of mass 7.4 × 10^-¹⁷ kg between the plates.If the droplet carries five electronic charges and if there were no air buoyancy,you could hold the droplet motionless between the plates if you kept the potential difference between them at A)5.4 V B)27 V C)3.0 V D)0.54 V E)0.27 kV

Question

Quizplus · Accepted Answer

The electric force on the droplet is equal to the gravitational force on the droplet. The electric force is given by F = qE, where q is the charge on the droplet and E is the electric field between the plates. The gravitational force is given by F = mg, where m is the mass of the droplet and g is the acceleration due to gravity. Setting these two forces equal to each other, we get:qE = mgSolving for E, we get:E = mg/qSubstituting the given values, we get:E = (7.4 × 10^-17 kg)(9.8 m/s^2)/(5e) = 2.94 × 10^4 V/mThe potential difference between the plates is given by V = Ed, where d is the distance between the plates. Substituting the given values, we get:V = (2.94 × 10^4 V/m)(0.60 cm) = 5.4 VTherefore, the correct answer is A) 5.4 V.

Two Parallel Horizontal Plates Are Spaced 0