A 13 pC droplet is in an upward electric field of 12,000 N/C under a thunderstorm. Compare the electric force Fe to the gravitational weight Fg; Fe is directed upward. Which statement is true?

• A. Electric force is about 2.2×10^-3 times the weight
• B. Electric force is about equal to the weight
• C. Electric force is larger by a factor of 10
• D. Electric force is zero

Answer: (A)

The main idea is to compare the electric force on the droplet to its weight. The electric force is Fe = qE, while the weight is Fg = mg. We can estimate how big Fe is relative to Fg by computing Fe and Fg for a typical raindrop.

Fe = (13 × 10^-12 C)(12,000 N/C) ≈ 1.56 × 10^-7 N upward.

For a water droplet, use density ρ ≈ 1000 kg/m^3. If the droplet is about 1 mm in radius, its volume V ≈ (4/3)π(1×10^-3 m)^3 ≈ 4.19×10^-9 m^3, so mass m ≈ ρV ≈ 4.19×10^-6 kg and Fg ≈ mg ≈ (4.19×10^-6 kg)(9.8 m/s^2) ≈ 4.1×10^-5 N.

The ratio Fe/Fg ≈ (1.56×10^-7 N) / (4.1×10^-5 N) ≈ 3.8×10^-3, i.e., Fe is about a few ×10^-3 of the weight. This matches the given statement that Fe is about 2.2×10^-3 times the weight. Direction being upward fits, but gravity dominates so the droplet still falls overall.

So the electric force is not zero, not equal to the weight, and not larger by a factor of 10; it is much smaller than the weight by roughly a few thousandths.