Question

Two small spheres have mass $m_1$ and $m_2$ and hanging from massless insulating threads of lengths $\ell_1$ and $\ell_2$. Two sphere carry charges $q_1$ and $q_2$ respectively. The spheres hang such that they are on the same horizontal level and the threads are inclined to the vertical at angle $\theta_1$ and $\theta_2$ respectively. Which of the following condition is true if $m_1 = m_2$.

• A. $\theta_1 = \theta_2$
• B. $q_1 = q_2$
• C. $\frac{\ell_1}{tan\theta_1} = \frac{\ell_2}{tan\theta_2}$
• D. $\frac{q_1}{tan\theta_1} = \frac{q_2}{tan\theta_2}$

Answer 1

Answer: (A)

$\theta_1 = \theta_2$

Answer 2

The equilibrium of each sphere requires the horizontal component of tension to balance the electrostatic force and the vertical component of tension to balance the gravitational force. This leads to $\tan \theta_i = \frac{F_e}{m_i g}$ for sphere $i$. Since the electrostatic force $F_e$ between the two spheres is equal in magnitude for both, we have $m_1 g \tan \theta_1 = m_2 g \tan \theta_2$. Given $m_1 = m_2$, this equation simplifies to $m_1 g \tan \theta_1 = m_1 g \tan \theta_2$, which implies $\tan \theta_1 = \tan \theta_2$. For angles between 0 and $\pi/2$, this means $\theta_1 = \theta_2$.