Finding Zero Electric Field Points on the x-axis Due to Two Point Charges

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SUMMARY

The discussion centers on determining the regions along the x-axis where the electric field is zero due to two point charges, +q at x=0 and +2q at x=3d. The analysis reveals that the zero electric field point exists between the two charges, specifically closer to +q. The quadratic equation derived from the electric field equations, E = k(q/r), leads to the solution for the exact location of the zero electric field point in terms of d.

PREREQUISITES
  • Understanding of electric fields and point charges
  • Familiarity with the equation E = k(q/r)
  • Basic algebra skills for solving quadratic equations
  • Knowledge of the concept of charge repulsion
NEXT STEPS
  • Study the principles of electric fields and forces between point charges
  • Learn how to solve quadratic equations in physics contexts
  • Explore the concept of superposition in electric fields
  • Investigate the effects of varying charge magnitudes on electric field distribution
USEFUL FOR

Students studying electromagnetism, physics enthusiasts, and anyone seeking to understand electric field interactions between point charges.

Davetty
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Homework Statement



Two point charges +q and +2q are held along the x-axis, +q at x=0, +2q at x=3d.
i) which region(s) on the x-axis where the electric field due to the two charges is zero?
ii) Find the exact location in terms of d

+q_____P_______+2q
x=0 x=d x=3d

Homework Equations



E = k(q/r) Hint: solve the quadratic equation: K(q/x2)=K(2q/(3d-x)2

The Attempt at a Solution


For part i, the choices were: a) to the right of +2. b) between +2q and point P. c) between +q and point P. d) to the left of +q. e) both a and c. f) both b and d.

Since both charges are positive they would be repulsing each other so I believe that there would be a charge both to the left of +q and to the right of +2q so this cancels out a and d and consequently e and f. I chose c because I figured that +2q is a stronger charge so zero would be closer to +q.

for part ii, I attempted to solve the quadratic equation cancelling out K on both sides leaving: q/x2= 2q/(3d-x)2. I took the square of both side of the equation and ended up with: square root q/x=square root 2q/3d-x and now I'm stuck.

I'm really bad at working problems without numbers and am not very good at physics, so any help would be greatly appreciated!

Thanks in advance :smile:
 
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You're on the right track.

So you have

[tex] \frac{\sqrt{q}}{x} = \frac{\sqrt{2q}}{3d-x}[/tex]

and you want to solve that for x. That's difficult when the "x" terms are in the denominator. Try to think of how to get the x's into the numerator, or how to get rid of the fractions altogether.
 
That worked! Thanks a bunch! :)
 

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