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A charge of 5 x109 C is attracted by a charge of –3x107 C with a force of 0.135 N. How far apart are they? (Hint: manipulate the equation to get an expression for d)
 one year ago
 one year ago
I need your help! A charge of 5 x109 C is attracted by a charge of –3x107 C with a force of 0.135 N. How far apart are they? (Hint: manipulate the equation to get an expression for d)
 one year ago
 one year ago

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Iamgmg90Best ResponseYou've already chosen the best response.1
Coulombs law \[F = (Q_1\times Q_2)/(4\pi \times \epsilon_0 \times d^2) \]
 one year ago

dhudz23Best ResponseYou've already chosen the best response.0
Could you help me on how to start it?
 one year ago

Shane_BBest ResponseYou've already chosen the best response.1
You just plug in the values you know into the formula and solve for d
 one year ago

Shane_BBest ResponseYou've already chosen the best response.1
\[\large 0.135N = \frac{5x10^{9}C\times 3x10^{7}C}{4(3.14159) \times 8.85x10^{12} \frac{C^2}{Nm^2}\times d^2}\]As the question states, now you must manipulate this equation to solve for d.
 one year ago

dhudz23Best ResponseYou've already chosen the best response.0
Thank you! I think I got it.
 one year ago
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