Where Is the Net Electric Field Zero Between Two Fixed Charges?

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SUMMARY

The net electric field between two fixed charges, -29.7μC and +5.5μC, separated by 4 meters, is zero at a point to the left of the positive charge. The equation used to determine this point is E = k(5.5×10^-6)/d² - k(2.97×10^-7)/(4+d)². After solving for "d," the next step involves calculating the force on a third charge of +48.7μC placed at this location using Coulomb's law. The correct approach requires careful attention to the signs and magnitudes of the charges involved.

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Homework Statement


Two charges, -29.7[tex]\mu[/tex]C and +5.5[tex]\mu[/tex]C, are fixed in place and separated by 4m
(a) At what spot along the line through the charges is the net electric field zero? Locate this spot relative to the position of the positive charge
(b) What would the magnitude of the force on a third charge +48.7[tex]\mu[/tex]C placed at this spot

Homework Equations


E=kq/d[tex]^{2}[/tex]


The Attempt at a Solution



......+5.5microC...-29.7microC
o_________________o______________o_________
|---------d---------|------4m-------|

After drawing out the problem as shown above, I found out that the net electric field would be zero to the left of the positive charge (-x direction). I then set up my equation as follows:

E= k(5.5*10^-6)/d[tex]^{2}[/tex] - k(2.97*10^-7)/(4+d)[tex]^{2}[/tex]

I graphed this formula and came up with -5.2109 or -3.24575. I entered both numbers into the system as positive and negative and came up with the wrong answer every time. Also, once I solve for "d", how do I go about finding the magnitude of force on the third charge?
 
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29.7microC = 2.97*10^-5 C
 

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