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#1 Cooldude

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Posted 26 October 2007 - 07:53 PM

Right i'm completely stuck on this question sad.gif...

Calculate the point between the charges Q1 and Q2 where the electric field strength 'E' is zero. Q1 is 0.35C and Q2 is 0.7C. They're 250m apart.



I might be being stupid but i'm very confused anyway!!

#2 Cooldude

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Posted 28 October 2007 - 04:13 PM

anyone? i'm still stumped...

#3 ad absurdum

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Posted 28 October 2007 - 06:37 PM

The first step is to write the distance at which the electric field strength is zero from Q_1 to be r, then the distance from Q_2 is 250-r. And we're trying to find the point at which the resultant electric field strengths is zero, so this is going to be the point at which the magnitude of the electric field strength due to Q_1 is the same as the magnitude of the electric field strength due to Q_2. So, mathematically,

 \frac{Q_1}{4\pi\epsilon r^2} =  \frac{Q_2}{4\pi\epsilon (250-r)^2}.

 \frac{Q_1}{r^2} =  \frac{Q_2}{(250-r)^2}.

Then solve this for r. You should get two solutions, one of which satisfies 0<r<250, and the other one should be ignored (I think this solution for r will be negative, because since Q_2 > Q_1, there will be a point on the opposite side of Q_1 from Q_2 at which the magnitude of the electric field strengths are equal. Not that the resultant electric field strength is not zero here, as both the electric fields are acting in the same direction.)

Hope that helps.
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#4 Cooldude

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Posted 29 October 2007 - 08:25 PM

ohhh ok.... already handed my homework in but i shall remember that.... thanks biggrin.gifbiggrin.gif





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