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A charge of 5 x10-9 C is attracted by a charge of –3x10-7 C with a force of 0.135 N. How far apart are they? (Hint: manipulate the equation to get an expression for d)

Question

I need your help!

A charge of 5 x10-9 C is attracted by a charge of –3x10-7 C with a force of 0.135 N. How far apart are they? (Hint: manipulate the equation to get an expression for d)

anonymous · Answer

Coulombs law $$F = (Q_1\times Q_2)/(4\pi \times \epsilon_0 \times d^2) $$

anonymous · Answer

Could you help me on how to start it?

shane_b · Answer

You just plug in the values you know into the formula and solve for d

shane_b · Answer

$$\large 0.135N = \frac{5x10^{-9}C\times -3x10^{-7}C}{4(3.14159) \times 8.85x10^{-12} \frac{C^2}{Nm^2}\times d^2}$$As the question states, now you must manipulate this equation to solve for d.

anonymous · Answer

Thank you! I think I got it.

shane_b · Answer

np :)