Where Is the Electric Field Zero Between Two Charged Particles?

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SUMMARY

The electric field between two charged particles, q1 and q2, is zero at a distance x from charge q1, determined by the equation x = s / (√(q2/q1) + 1). This solution arises from equating the magnitudes of the electric fields produced by both charges, leading to a relationship that incorporates the distance s and the charges q1 and q2. The derived formula allows for calculating the specific point where the electric fields cancel each other out, confirming the correctness of the approach through algebraic manipulation.

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Homework Statement



Two particles with positive charges q1 and q2 are separated by a distance s.

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero?


Homework Equations


Express your answer in terms of some or all of the variables s , q1 , q2 and k=1/(4pi epsilon)


The Attempt at a Solution



I got s=sqrt(q1(d-s)^2 / q2), but i think its wrong.

Anyone who knows?
 
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You should probably wait for someone more knowledgeable on this subject to post an answer, but here is what I came up with:

It looks like you solved for s instead of the distance from q1 where the field is 0.

If your problem looks like this:

Code:
            |----------s----------|
-----------(q1)-----0------------(q2)---------
            |---x---|-----s-x-----|

I figured that the magnitude of E from each (+) charge would be equal so I made the equation:

<br /> {k}_{e}\frac{{q}_{1}}{x^2} = {k}_{e}\frac{{q}_{2}}{(s-x)^2}<br />

Divide both sides by ke:

<br /> \frac{{q}_{1}}{x^2} = \frac{{q}_{2}}{(s-x)^2}<br />

Cross multiply, then divide by x^2:

<br /> \frac{(s-x)^2}{x^2} = \frac{{q}_{2}}{q}_{1}<br />

Square root both sides, then distribute the x denominator:

<br /> \frac{s}{x} - \frac{x}{x} = \sqrt\frac{{q}_{2}}{q}_{1}<br />

Simplify x/x to 1 and add to other side then multiply by x:

<br /> s = (\sqrt{\frac{{q}_{2}}{q}_{1}} + 1)x<br />

Finally divide to get x alone:

<br /> x = \frac{s}{\sqrt{\frac{{q}_{2}}{q}_{1}} + 1}<br />

So that's my final answer, which I'm pretty sure is correct but then I usually make some careless error. (Sorry if I showed too many baby steps with the algebra there.)

Anyways, hope that helps.
 
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Likes   Reactions: Rod and 8008jsmith
yes, i got the same so i guess that its correct.

Thanks for the help
 

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