Optimal Placement for Net Zero Electrical Force in a System of Charged Spheres

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SUMMARY

The optimal placement for a third sphere with a charge of 3.0 x 10^-6 C, in a system of two charged spheres (1.6 x 10^-5 C and 6.4 x 10^-5 C) separated by 2.0 m, is approximately 0.67 m from the first sphere. To achieve a net electrical force of zero, the fields generated by the two spheres must cancel each other out. The charge of the third sphere is irrelevant in determining its position, as the focus is on the balance of electric fields from the first two spheres.

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Quantum Fizzics
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Homework Statement


Two small spheres with charges 1.6*10^-5 C & 6.4*10^-5 C, are 2.0m apart. The charges have the same sign. Where, relative to these two spheres, should a third sphere, of opposite charge 3.0*10^-6 C, be placed if the third sphere is to experience no net electrical force? Do we really need to know the charge or sign of the third object?
Given: q1 =1.6*10^-5 C
q2 = 6.4*10^-5 C
q3 = 3.0*10^-6 C

Homework Equations


F= kq1q2/d^2

The Attempt at a Solution


F31 = kq3q1/d^2
= 9x10^9*3.0x10^-6*1.6x10^-5/d^2

9x10^9*3.0x10^-6*1.6x10^-5 = d^2
0.657267069m= d

What I want to know is how will I know? If this will be the right distance where the 3rd sphere won't experience no net electrical force? That is the answer well ( 0.67m ) & yeah, because honestly we never learned this stuff in terms of coulombs law but I know that its the same as Newtons Law also I just got to substitute some variables & such. An explanation would be great, thanks ( I got a quiz tomorrow so)
 
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So you want the net force to equal zero. The net force = ?

By the way, it looks like you assumed F31 = 1 when you were solving for d.
 
elegysix said:
So you want the net force to equal zero. The net force = ?

By the way, it looks like you assumed F31 = 1 when you were solving for d.
yeah, damn. Well I looked through my notes & I did it all wrong, I have to I guess make it into a complex trinomial & complete the square?(This is in terms Newtons law, so I don't know how to transition that into coulombs law
 
Quantum Fizzics said:
F31 = kq3q1/d^2
= 9x10^9*3.0x10^-6*1.6x10^-5/d^2

9x10^9*3.0x10^-6*1.6x10^-5 = d^2
As elegysix points out, to get that you seem to have assumed F31=1. You might have realized this made no sense if you had paid attention to the units or their dimensions. In the second equation above, the right hand side has dimension length squared, while the LHS has dimension time-squared * length / mass. It might also have occurred to you that you have not used any information regarding q2.

You are right that the charge of q3 is irrelevant. All you are looking for is where the fields due to q1 and q2 cancel. Write an equation for that.
 
haruspex said:
As elegysix points out, to get that you seem to have assumed F31=1. You might have realized this made no sense if you had paid attention to the units or their dimensions. In the second equation above, the right hand side has dimension length squared, while the LHS has dimension time-squared * length / mass. It might also have occurred to you that you have not used any information regarding q2.

You are right that the charge of q3 is irrelevant. All you are looking for is where the fields due to q1 and q2 cancel. Write an equation for that.

alright I got it now, is it okay to leave it as a decimal instead of keeping it into an exponent? or will that mess up everything
 
Quantum Fizzics said:
is it okay to leave it as a decimal instead of keeping it into an exponent? or will that mess up everything
Since the answer looks likely to of the order of a metre, I would give the answer as a decimal.
 

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