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# capa3 - Phys 1120 CAPA#3 solutions 1 This problem is...

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Phys 1120, CAPA #3 solutions. 1) This problem is basically solved for you as an Example of the text. I HIGHLY recommend that you work it out for yourself - just taking an example from the book doesn't teach you much. But the solution is nicely written up, so I won't repeat it here. So, E(z) = k z Q / (z^2 + R^2)^(3/2). They gave you ALL these quantities, just plug in. Remember to convert all distances to meters, and charges to Coulombs, before plugging in!! 2) We want to maximize E(z) , that means set dE/dz = 0. Taking the derivative of that function is a little nasty (Remember, k, Q, and R are constants. I use the "(bottom times deriv of top - top times deriv of bottom) / bottom^2" rule, maybe you have your own! You need to use the chain rule, the derivative of (z^2 + R^2)^(3/2) all by itself is (3/2) (z^2+R^2)^(1/2)*(2z). This is about the worst derivative I think we'll have to take all semester!? I won't give you ones quite this nasty on exams (!) but I *do* expect you to be able to use the chain rule for simple cases, so this is good practice. I get k Q [ (z^2 + R^2)^(3/2) - z*(3/2)*(z^2+R^2)^(1/2)*(2z) ] / (z^2+R^2)^3] = 0 Looks like a mess. But it's not so bad, you just have to believe you can clear the dust! There really is some pleasure to be found in "simplifying" nasty expressions like this, it gives you some faith in the orderliness and simplicity of the universe :-) Multiply both sides by the denominator (z^2_R^2)^3, it just goes away!!! Divide by kQ, gone.! Lastly: divide through by the common (z^2+R^2)^(1/2). Here's what's left: (z^2+R^2)^(1) - z*(3/2)*(2z) = 0, which means z^2 + R^2 - 3 z^2 =0, or R^2 = 2 z^2. z = R/Sqrt[2]. That's it. (Don't you love it when nasty algebra simplifies away to almost nothing?) If you have trouble with this sort of exercise, my best suggestions are to take a deep breath, use a nice big clean piece of paper, write neatly, keep yourself organized... It's a great skill to develop. 3) Recall, from #6, E(z) = k z Q / (z^2 + R^2)^(3/2). Near the origin, where z very small compared to R, R^2 + z^2 becomes R^2, so E becomes E = k z Q / (R^2)^(3/2). You can neglect z in the denominator because when you ADD something big to something small, the small thing is neglected. You don't neglect it in the numerator, because it's MULTIPLYING things, it isn't small *compared* to anything... (Think about this! It's an idea that crops up everywhere in science and engineering, and is worth wrapping your head around) So now we have E(z) = k z Q / R^3. Then, the force on an electron is F = q*E = -eE F = -( e *k Q / R^3) * z Just as advertised, that's of the form -c*z, where c = e*k*Q/R^3. Plug in your numbers. (e = 1.6E-19 C, k = 8.99E9 N m^2/C^2, R is given. You have to work out the units, they're N/m, as they should be. Try it to see!)

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