Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Answer:
Approximately [tex]1.69 \times 10^{-23}\; {\rm N}[/tex].
Explanation:
Look up the value of Coulomb's Constant: [tex]k \approx 8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}}[/tex].
Consider point charges of magnitude [tex]q_{1}[/tex] and [tex]q_{2}[/tex]. If the distance between these charges is [tex]r[/tex], the magnitude of the electrostatic force between them would be [tex](k\, q_{1}\, q_{2}) / (r^{2})[/tex].
In this question, the two [tex](+q)[/tex] charges are [tex]5\; {\rm cm}[/tex] and [tex]3\; {\rm cm}[/tex] away from the center [tex](-2\, q)[/tex] charge, respectively. Convert units to standard unit of distance (meters, [tex]{\rm m}[/tex]) and charge (coulombs, [tex]{\rm C}[/tex]):
[tex]q = 1.15 \; {\rm nC} = 1.15 \times 10^{-9}\; {\rm C}[/tex].
[tex]\begin{aligned} 5\; {\rm cm} = 5\; {\rm cm} \times \frac{1\; {\rm m}}{100\; {\rm cm}} = 0.05\; {\rm m} \end{aligned}[/tex].
[tex]\begin{aligned} 3\; {\rm cm} = 3\; {\rm cm} \times \frac{1\; {\rm m}}{100\; {\rm cm}} = 0.03\; {\rm m} \end{aligned}[/tex].
The magnitude of the electrostatic forces on the [tex](-2\, q)[/tex] charge would be:
[tex]\begin{aligned}\frac{k\, q_{1}\, q_{2}}{r^{2}} &\approx \frac{1}{(0.05\; {\rm m})^{2}} \\ &\quad \times (8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}})\\ &\quad \times ((-2) \, (1.15\times 10^{-9}\; {\rm C}))\, (1.15\times 10^{-9}\; {\rm C})) \\ &\approx 9.509\times 10^{-24}\; {\rm N}\end{aligned}[/tex].
[tex]\begin{aligned}\frac{k\, q_{1}\, q_{2}}{r^{2}} &\approx \frac{1}{(0.03\; {\rm m})^{2}} \\ &\quad \times (8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}})\\ &\quad \times ((-2) \, (1.15\times 10^{-9}\; {\rm C}))\, (1.15\times 10^{-9}\; {\rm C})) \\ &\approx 2.641\times 10^{-23}\; {\rm N}\end{aligned}[/tex].
Since the charges are of opposite sign, the [tex](-2\, q)[/tex] charge would attract both of the [tex](+q)[/tex] charges. In particular, the (approximately) [tex]9.509\times 10^{-24}\; {\rm N}[/tex] force would point to the left. The (approximately) [tex]2.641 \times 10^{-23}\; {\rm N}[/tex] force would point to the right.
As a result, the net force on the [tex](-2\, q)[/tex] charge would point to the right. The magnitude of the net force on this charge would be approximately [tex]2.641 \times 10^{-23}\; {\rm N} - 9.509\times 10^{-24}\; {\rm N} \approx 1.69 \times 10^{-23}\; {\rm N}[/tex].
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
