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The electrical force of attraction between two point charges is F. The charge on one of the objects is quadrupled and the charge on the other object is doubled. The new force between objects is

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Jacob193idn

Answer:

[tex]8\, F[/tex].

Explanation:

By Coulomb's Law, the magnitude of electrostatic force between two point charges is proportional to the product of the two charges:

[tex]\displaystyle F = \frac{k\, q_{1}\, q_{2}}{r^{2}}[/tex],

Where:

  • [tex]k[/tex] is Coulomb's Constant,
  • [tex]q_{1}[/tex] and [tex]q_{2}[/tex] are the magnitude of the electrostatic charge on each point charge, and
  • [tex]r[/tex] is the distance between the two point charges.

In this question, it is given that the magnitude of one charge is quadrupled while the other is doubled. The magnitude of the electrostatic force between the two point charges would become:

[tex]\begin{aligned}\frac{k\, (4\, q_{1})\, (2\, q_{2})}{r^{2}} = 8\, \left(\frac{k\, q_{1}\, q_{2}}{r^{2}}\right) = 8\, F\end{aligned}[/tex].

In other words, the magnitude of the electrostatic force between the two point charges would become eight times the initial amount.

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