Two spherical metal balls of mass $m=10\ g$ each hang from a thread of length $L=0.2\ m$ attached to the ceiling and are separated by a distance $d=0.5\ m$. The balls are initially uncharged and their threads vertical.

1. When both are rubbed with fur, they repel each other. You notice that the angle $\theta $ between each thread and the vertical is equal to $30{}^\circ $. What is the charge of each ball?

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The free-body diagram of each ball is drawn below

Assuming that both balls end up with the same amount of charge $Q$ on them after being rubbed with fur, we conclude that the repulsive force $F$ between them has a magnitude given by Coulomb’s Law that is equal to

F=\frac{kQ^2}{D^2}

where $D$ is the distance between the balls from center to center shown in the figure below.

From geometry, we derive that the distance $D$ is given by

D=d+2L{\mathrm{sin} \left(\theta \right)\ }

Thus, we can apply the equilibrium condition to each one of the balls and derive the charge $Q$ as done below

\begin{aligned} \left\{ \begin{array}{c} F_{net\ x}=0 \\ F_{net\ y}=0 \end{array} \right.\ \ \ \ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} T_1{\mathrm{sin} \left(\theta \right)\ }-F=0 \\ T_1{\mathrm{cos} \left(\theta \right)\ }-mg=0 \end{array} \right. \\ \\ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} T_1{\mathrm{sin} \left(\theta \right)\ }=F \\ T_1{\mathrm{cos} \left(\theta \right)\ }=mg \end{array} \right. \\ \\ &\Rightarrow \ \ \ \ \ {\mathrm{tan} \left(\theta \right)\ }=\frac{F}{mg} \\ \\ &\Rightarrow \ \ \ \ \ mg{\mathrm{tan} \left(\theta \right)\ }=\frac{kQ^2}{{\left(d+2L{\mathrm{sin} \left(\theta \right)\ }\right)}^2} \\ \\ &\Rightarrow \ \ \ \ \ Q=\sqrt{\frac{mg{\mathrm{tan} \left(\theta \right)\ }{\left(d+2L{\mathrm{sin} \left(\theta \right)\ }\right)}^2}{k}} \end{aligned}

Finally, we conclude that the charge $Q$ on each ball is equal to

\boxed{Q=\sqrt{\frac{mg{\mathrm{tan} \left(\theta \right)\ }{\left(d+2L{\mathrm{sin} \left(\theta \right)\ }\right)}^2}{k}}=\sqrt{\frac{10\times {10}^{-3}\cdot 9.8\cdot {\left(0.5+2\cdot 0.2\cdot {\mathrm{sin} \left(30\right)\ }\right)}^2}{9\times {10}^9}}\approx 2.31\ \ \mu C}

2. You repeat the experiment but this time you only rub one ball with saran wrap. The saran wrap creates the same magnitude of charge as the fur did. The balls now attract each other, and their threads make an angle $\mathrm{\Phi }$ with the vertical. Is $\phi $ greater, smaller, or equal to $\theta $? Explain how you know.

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Let’s assume that after being rubbed, the ball ends up with a negative charge $-Q$. Even though the second ball is uncharged, the electric field from the charged ball will induce charge separation in the second metal ball, causing the positive charges to move to its left side and the negative charges to move to its right side.

This induced charge separation causes the two balls to be slightly attracted to each other because the positive charges on the left side of the second ball are slightly closer to the negatively charged ball than the negative charges. Overall, this causes a weak attraction — definitely weaker than if the second ball were also charged — and we conclude therefore that the angle $\phi $ must be less than the angle $\theta $.

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