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advanced higher 2002 Q6c


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#1 sachy

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Posted 15 May 2006 - 08:29 PM

ive been doing question 6c in the 2002 paper and im constantly getting the wrong answer. does anyone know how to do the working?

#2 Pringles

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Posted 15 May 2006 - 09:41 PM

To get the formula for distance of closest approach you equate kinetic energy with electrostatic energy.

 \frac{1}{2}m{v^2} = \frac{Q_1*Q_2}{4\pi\epsilon_o*r}

manipulating the formula to make  r the subject of it.

This becomes  r = \frac{Q_1*Q_2}{2\pi\epsilon_o*mv}

From this you simply put the numbers into the equation.

For the the charge on Gold = 79 x 1.6 x 10 -19





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