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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].
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