P19B2015 – Plastic Balls & Fur

To determine the final force between the two balls, we must determine the charge of each ball after ball #2 is rubbed with the fur. The situation can be summarized as follows

Ball #1 was not rubbed and therefore its charge is still $+5Q$.

The fur is repelled from ball #1 that has a positive charge. This means that the charge on the fur is positive and that $Q_{fur}=+3Q$.

Using conservation of charge, we then find the final charge $Q_{2f}$ on ball #2.

\begin{aligned}
Q_1+Q_2+Q_{fur}=Q_{1f}+Q_{2f}+Q_{fur\ f}\ \ \ \ \ &\Rightarrow \ \ \ \ \ Q_{2f}=Q_2-Q_{fur\ f} \\
\\
&\Rightarrow \ \ \ \ \ Q_{2f}=Q-3Q=-2Q
\end{aligned}

The two balls will therefore attract each other and the force between them when separated by a distance $2D$ is

\boxed{F_{1\ on\ 2}=F_{2\ on\ 1}=\frac{\left|Q_{1f}Q_{2f}\right|}{4\pi {\varepsilon }_0{\left(2D\right)}^2}=\frac{\left|5Q\cdot \left(-2Q\right)\right|}{4\pi {\varepsilon }_0{\left(2D\right)}^2}=\frac{5Q^2}{8\pi {\varepsilon }_0D^2}}

Answer: a

The cat fur has a positive charge of $+3Q$ after rubbing ball #2. If it is held next to a neutral metal cube, it will induce charge separation: the negative charges will move toward the end closest to the fur, the positive charge will move to the other end. The negative charges being closer to the positively charged fur than the positive charges of the metal, the overall interaction will be slightly attractive.