Asked by Megan
                Two spheres with charges of 4.00nC and 3.00nC are placed 0.500m apart.  At what point between the two charges must the third charge of 2.00cN be placed so that the net electrostatic force acting on this charge is zero?
            
            
        Answers
                    Answered by
            Elena
            
    k •q1 •q3/x^2 = k •q2 •q3/(d-x)^2  
q1 •q3 (d-x)^2 = k •q2 •q3• x^2
Solve for x
    
q1 •q3 (d-x)^2 = k •q2 •q3• x^2
Solve for x
                    Answered by
            Elena
            
    The second line
q1 •q3 (d-x)^2 = q2 •q3• x^2
    
q1 •q3 (d-x)^2 = q2 •q3• x^2
                    Answered by
            Megan
            
    Elena, for some reason I'm getting an answer of -3
(8.00 x 10^-12) x (d-x)^2 = (6.00 x 10^-12) x (x^2)
(d-x)^2/x^2 = 7.5 x 10^-1
(d-x)(d-x)/x^2 = 7.50 x 10^-1
x^2-x+0.25/x^2= 7.50 x 10^-1
then I get x = -3
Can you help! I'm not sure where i am going wrong.
    
(8.00 x 10^-12) x (d-x)^2 = (6.00 x 10^-12) x (x^2)
(d-x)^2/x^2 = 7.5 x 10^-1
(d-x)(d-x)/x^2 = 7.50 x 10^-1
x^2-x+0.25/x^2= 7.50 x 10^-1
then I get x = -3
Can you help! I'm not sure where i am going wrong.
                    Answered by
            Elena
            
    q1 • (d-x)^2 = q2 •  x^2
4•10^-9•(d-x)^2 = 3•10^-9•x^2
4(d^2 - 2•d•x + x^2) = 3•x^2,
x^2 - 4x + 1 = 0
x =2 ± sqrt(4-1) =2 ± 1.73.
x1 = 0.27 m, x2 =3.73m (falls out)
    
4•10^-9•(d-x)^2 = 3•10^-9•x^2
4(d^2 - 2•d•x + x^2) = 3•x^2,
x^2 - 4x + 1 = 0
x =2 ± sqrt(4-1) =2 ± 1.73.
x1 = 0.27 m, x2 =3.73m (falls out)
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.