The Levitating Droplet · AI Generator #GEN-5EFD9C06

Physics · electromagnetism

Problem

A small water droplet with mass m=2.0×10−12  kgm = 2.0 \times 10^{-12}\;\text{kg} carries an electric charge q=+8.0×10−16  Cq = +8.0 \times 10^{-16}\;\text{C}. The droplet is suspended motionless in air between two horizontal parallel plates separated by a distance d=5.0  mmd = 5.0\;\text{mm}. The upper plate is positively charged and the lower plate is negatively charged, creating a uniform electric field pointing downward. Find the magnitude of the electric field required to keep the droplet in equilibrium against gravity. Take g=9.8  m/s2g = 9.8\;\text{m/s}^2.
Intuitive explanation

For a charged object to float in an electric field, the electric force must exactly balance gravity—this is an equilibrium problem where two forces point in opposite directions. Since the droplet is positively charged and the field points downward, the electric force pushes it upward; you need to set this equal to the weight and solve for the field strength EE. The key is recognizing that at equilibrium, Felectric=FgravityF_{\text{electric}} = F_{\text{gravity}}, so qE=mgqE = mg.

Intuitive explanation

For a charged object to float in an electric field, the electric force must exactly balance gravity—this is an equilibrium problem where two forces point in opposite directions. Since the droplet is positively charged and the field points downward, the electric force pushes it upward; you need to set this equal to the weight and solve for the field strength EE. The key is recognizing that at equilibrium, Felectric=FgravityF_{\text{electric}} = F_{\text{gravity}}, so qE=mgqE = mg.

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