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Net Electric Field Equals Zero

  • Thread starter AFRaven
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
277
1
29.7microC = 2.97*10^-5 C
 

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