Charged pendulums

Electricity and Magnetism Level 3

Two pendulums of equal lengths $l$ and different masses $m_1$ and $m_2$ are suspended at the same point. The balls at the end of the pendulum each carry positive electric charges $q_1$ and $q_2,$ respectively, so they repel each other electrostatically. In equilibrium, the second pendulum has about twice the deflection of the first $(\alpha_2 \approx 2 \alpha_1).$

Which statement about the masses or charges of the pendulums applies?

Note: We only consider the case of small deflections $(\alpha \ll \pi)$ so that the height difference of both pendulums can be neglected. Furthermore, $\sin \alpha \approx \tan \alpha \approx \alpha.$ (The angle is given in radians.)

m_1 \approx 2 m_2

m_2 \approx 2 m_1

q_1 \approx 2 q_2

q_2 \approx 2 q_1

None of the above

1 solution

Markus Michelmann
Nov 6, 2017

Both balls experience a Coulomb force with the same absolute value $F_\text{C} = \frac{q_1 q_2}{4 \pi \varepsilon_0 d}$ with the distance $d$ between the balls. The gravitational forces $F_{g1} = m_1 g$ and $F_{g2} = m_1 g$ are different for both pendulum. For the parallelogram of forces we can conclude $\begin{aligned} \frac{F_\text{C}}{m_i g} &\approx \tan \alpha_i \approx \alpha_i\,, \quad i = 1,2 \\ \Rightarrow \quad \frac{m_1}{m_2} &\approx \frac{\alpha_2}{\alpha_1} \approx 2 \end{aligned}$

Charged pendulums

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