Source: High school physics (Chinese)
Problem
A positively charged plastic sphere of mass $m = 3 \times 10^{-16}$ kg is stationary between two horizontal metal plates with opposite charges. The distance between the plates is $d = 3.2$ cm. Assume $g = 10$ m/s².
- What is the maximum possible voltage between the two plates?
- What are the three smaller voltage values adjacent to the maximum value?
[Q1] $V_{max} = 600$ V [Q2] $V_2 = 300$ V, $V_3 = 200$ V, $V_4 = 150$ V
For the sphere to be stationary, the upward electric force $F_e$ must balance the downward gravitational force $F_g$.
$$F_e = F_g \implies qE = mg$$The electric field between parallel plates is $E = V/d$, so $q(V/d) = mg$. The voltage is $V = \frac{mgd}{q}$. Charge $q$ is quantized, so $q = ne$, where $n$ is an integer and $e$ is the elementary charge.
$$V = \frac{mgd}{ne}$$For the voltage $V$ to be maximum, the charge $q=ne$ must be at its minimum non-zero value, which corresponds to $n=1$.
$$V_{max} = \frac{mgd}{e}$$The subsequent smaller values correspond to $n=2, 3, 4$.
$$V_n = \frac{V_{max}}{n}$$