Charge Is Quantized — Question 6

Question 6

Discrete charge states constrain a levitated droplet mass

Use $e=1.602176634\times10^{-19}$ C, exact. A droplet rests under gravity and a calibrated upward electric force $F_e=\kappa Q$, with $\kappa=1.00\times10^4$ N/C and $g=9.80$ m/s$^2$. Its positive charge is $ne$ with integer $n$. Independent mass bounds are $3.20\times10^{-15}\le m\le3.34\times10^{-15}$ kg.

1. Use force balance and the mass interval to infer the only possible integer $n$, then the equilibrium mass.

2. The droplet captures one electron while $\kappa$ remains fixed. Neglect the electron mass. Find its immediate vertical acceleration, taking upward as positive.

Question 6 - Solution

Discrete charge states constrain a levitated droplet mass

(a) Method and calculation

$$ne\kappa=mg\Rightarrow \frac{m_{\min}g}{e\kappa}\le n\le\frac{m_{\max}g}{e\kappa}.$$$$19.5734\lesssim n\lesssim20.4297\Rightarrow n=20,\quad m=20e\kappa/g\approx3.26975\times10^{-15}\,\mathrm{kg}.$$

(b) Interpretation and further calculation

$$Q\prime=19e,\quad F_{\mathrm{net}}=19e\kappa-mg=-e\kappa.$$$$a_y=-e\kappa/m=-g/20=-.490\,\mathrm{m/s^2}.$$The electric support drops by one twentieth while weight remains essentially unchanged.

Physical check

The measured equilibrium selects a charge state only because the allowed mass interval is narrower than the spacing of adjacent supported masses.