Final answer:
In Physics, if the second ball is moved away without grounding and the rod is removed from the first ball afterwards, both balls will have induced charges due to electrostatic induction. The first ball will have a net charge opposite to the removed rod, and the second ball's charge will depend on its interaction with the first ball.
Step-by-step explanation:
The scenario described involves a concept in Physics known as electrostatic induction. When one conducts the 'Prepare for Movement', and if the second ball is moved away (without grounding) and the rod is subsequently removed from the first ball, the two balls will indeed have induced charges.
Initially, when a charged rod is brought near the first (initially uncharged) ball, it induces a separation of charges within the ball due to electrostatic induction, attracting opposite charges to the near side and repelling like charges to the far side.
If you then move the second ball away while the rod is still inducing charges on the first ball, the second ball will have a net charge based on the charges it has gained or lost during its proximity to the first charged ball. Removing the rod afterward will leave the first ball with an induced charge opposite to that of the rod.
This is because when the rod is removed, there's no external electric field causing a separation of charges, and the imbalance of charges within the first ball remains. As a result, the first ball will retain a net charge opposite to the rod's charge, and the second ball will have a net charge depending on its interaction with the first ball.