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        What must be the distance between point charge  q1 26.0 mC and point charge  q2 47.0 mC for the electrostatic force between them to have a magnitude of 5.70 N?
2 years ago

							Dear Abdul We can use Coulomb’s law to figure out the distance between the two point charges. Coulomb’s law states:F=k\dfrac{(q_1)(q_2)}{r^2}F=k​r​2​​​​(q​1​​)(q​2​​)​​ (Where FF is the electrostatic force, kk is Coulomb’s constant, q_1q​1​​ is the charge of the first particle, q_2q​2​​ is the charge of the second particle, and rr is the distance between the two charges) Substitute the values we know into our equation.F=k\dfrac{(q_1)(q_2)}{r^2}F=k​r​2​​​​(q​1​​)(q​2​​)​​5.7=(8.99\times 10^9)\dfrac{(26\times 10^{-6})(47\times 10^{-6})}{r^2}5.7=(8.99×10​9​​)​r​2​​​​(26×10​−6​​)(47×10​−6​​)​​ (isolate for r)5.7r^2=(8.99\times 10^9)(26\times 10^{-6})(47\times 10^{-6})5.7r​2​​=(8.99×10​9​​)(26×10​−6​​)(47×10​−6​​)r^2=1.9273298r​2​​=1.9273298r=\sqrt{1.9273298}=1.39r=√​1.9273298​​​=1.39 m

2 years ago
							Use Coulombs law F= k×(q×q)/r^2Substitute the values u will get the answer.Note q and q are different charges they may have same or different magnitude.Regards.
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2 years ago
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