Exercise EE1-01.04 – The floating table tennis ball

Prof. Dr. Thorsten Jungmann (last updated 2025-09-12)

Relates to EE1-01 Introduction

Task

A table tennis ball (mass $m=2.70\mathrm{g}$) is charged to $Q_1=-15\mathrm{nC}$ by friction.

a) Calculate the weight-force of the ball.
b) At what distance $r$ above the ball must a second electric charge $Q_2=+20\mathrm{nC}$ be located so that the attractive Coulomb force exactly compensates the weight-force?
c) Is this arrangement stable? Give reasons for your answer.

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Solution

a) Weight-force

with the gravitational acceleration $g=9.81m/{\mathrm{s}}^2$

$$F_g=m\cdot g=2.70\cdot 10^{-3}\mathrm{kg}\cdot 9.81\frac{\mathrm{m}}{{\mathrm{s}}^2}=\underline{26.5\mathrm{mN}}$$

Note on precision

The precision with which the results are given depends on the precision of the source data, here: three significant digits for $m$ and $g$. The calculations were carried out with the calculator values, but the results were always given with no more than three significant digits.

b) Equilibrium condition

$$F_{\mathrm{Coulomb}}=F_g$$ $$\frac{1}{4\pi {\epsilon }_0}\cdot \frac{|Q_1\cdot Q_2|}{r^2}=m\cdot g$$

Solving for $r$:

$$r=\sqrt{\frac{1}{4\pi {\epsilon }_0}\cdot \frac{|Q_1\cdot Q_2|}{m\cdot g}}=\sqrt{8.99\cdot 10^9\cdot \frac{15\cdot 10^{-9}\cdot 20\cdot 10^{-9}}{26.5\cdot 10^{-3}}}$$ $$\underline{r\approx 3.3\mathrm{mm}}$$

where:

• $m$ the mass in kilograms ($\mathrm{kg}$)
• $g$ the acceleration of free fall with $g=9.81m/{\mathrm{s}}^2$
• $F$ the force in newtons ($\mathrm{N}$)
• $Q_1,Q_2$ the electric charges in coulombs ($\mathrm{C}$)
• $r$ the distance in metres ($\mathrm{m}$)

Note on precision

The precision with which the results are given depends on the precision of the source data, here: two significant digits for $Q_1$ and $Q_2$. The calculations were carried out with the calculator values, but the results were always given with no more than two significant digits.

c) Considering stability

The arrangement is unstable. If the ball is displaced slightly upwards, the Coulomb force increases ($r^2$ is in the denominator), while the weight-force remains constant. The ball would continue to accelerate upwards.