<p>I got some weird answer to a hw question. Can someone check if this is right?</p>
<li>What is the repulsive electrical force between two protons in a nucleus that are 5.0 x 10^-15 m apart from each other?</li>
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<p>F = kQ1Q2/r^2
= 9.0x10^9Nm^2((1.60x10^-19C)(1.60x10^-19C)/(5.0x10^-15cm))
= 4.6x10^-32N</p>
<p>Did you remember to square the radius?</p>
<p>Both Q's = 1.602 </p>
<p>Yeah I forgot to sqaure it. I have another problem.</p>
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<li>A +5.7 uC and a -3.5 uC charges are placed 25 cm apart. Where can a third charge be placed so that it experiences no net force?</li>
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<p>Simple. You need to place the third charge in between them where the repulsive force between the new one and the +5.7 uC charge is equivalent to the repulsive force between the new one and the -3.5 uC charge. Therefore set the two equations equal to eachother... Since they both contain this new charge, it can be disregarded in the equation. Therefore Kq1/^2 = Kq2/(d-x)^2 where K = K, q1 = 5.7 and q2 = -3.5. Solve for x. Too bad I am too lazy to do the math. But thats how I would solve it (and to think, I took standard physics last year and forensic science this year...). I need to take the physics SAT II.</p>
<p>Yes, Jimmy is right on with the approach.</p>