Answered

Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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

Sagot :

jacob193

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.

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.