Where Is the Electric Field Zero Between Two Charged Particles?

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To find the distance from charge q1 where the electric field is zero between two positive charges q1 and q2 separated by distance s, the correct approach involves setting the magnitudes of the electric fields from both charges equal. The derived equation is x = s / (√(q2/q1) + 1), where x represents the distance from q1. The initial attempt at the solution was incorrect as it solved for s instead of x. The final answer aligns with the calculations provided, confirming its correctness. This method effectively determines the point of zero electric field between the two charges.
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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 Rod and 8008jsmith
yes, i got the same so i guess that its correct.

Thanks for the help
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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