Answer:
a)The sign of energy is negative, b) we can approach the two charges, c) d ⇒ 0
Explanation:
a) The elective potential energy is
U = k q₁q₂ / r
Applied to this case.
U = -k q 9 / d
U = -9 k q / d
The sign of energy is negative, which corresponds to a stable system
b) c) for the energy to be more negative we can approach the two charges
c) To increase the energetic of the system, that is to say less negative, we can separate the discharges whereby U approaches zero or we can decrease the value of the charge q.
d ⇒ 0
