Electric Field Question: Finding Zero Points for Point Charges

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SUMMARY

The discussion focuses on determining the zero points of the electric field created by a positive and a negative point charge. The user initially calculated a distance of 1.03m for the zero point but was corrected to 1.82m to the left of the negative charge, as per the textbook. The user also speculated about the existence of four zero points, including locations at infinity and near the charges, but was guided to understand that only specific locations yield a net electric field of zero. The key takeaway is the importance of accurately applying the electric field equation, E = k_e * q / r^2, to find these points.

PREREQUISITES
  • Understanding of electric fields and point charges
  • Familiarity with the equation E = k_e * q / r^2
  • Basic knowledge of vector addition in physics
  • Concept of electric field direction and magnitude
NEXT STEPS
  • Study the concept of electric field lines and their behavior around point charges
  • Learn about superposition of electric fields from multiple charges
  • Explore the concept of equipotential surfaces in electrostatics
  • Investigate the effects of distance on electric field strength and direction
USEFUL FOR

Students and educators in physics, particularly those studying electrostatics, as well as anyone seeking to understand the behavior of electric fields generated by point charges.

altegron
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Homework Statement



http://img184.imageshack.us/img184/8125/chapter23number13qh8.png

Homework Equations



[tex]E = {k}_{e} \frac{q} {r^2}[/tex]

The Attempt at a Solution



I figured that the electric field vector from the negative point charge would have to cancel out the electric field vector from the positive point charge so I said that 0 = E from the negative + E from the positive. I substituted values in and made the distance for the positive one r and the distance for the negative one (1-r). When I solved this for r, I get an answer like 1.03m, but the book says the answer is 1.82m to the left of the negative charge.

Even if I did the math wrong, what I am really wondering about it is the concept here. To me it seems like there should be 4 points where the field is 0. 1. At infinity, which the book mentions. 2. Somewhere far to the left of the negative charge, which I guess is the answer they put in the back. 3. Just slightly to the left of the negative charge. 4. And just slightly right of the negative charge.

My thinking on 3 and 4 is that E will get bigger and bigger as you get closer to the point (r->0, so E goes to infinity, right?). So, E will be large and positive close to the (+) charge, but as you go left to the (-) charge it will decrease to 0, and then become more negative as it reaches the (-) charge. Once it passes this, it will get bigger (from -infinity) until it equalizes with the (+) charge. Then, when you are rather far away it will be 0 again.
 
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altegron said:
3. Just slightly to the left of the negative charge. 4. And just slightly right of the negative charge.

My thinking on 3 and 4 is that E will get bigger and bigger as you get closer to the point (r->0, so E goes to infinity, right?). So, E will be large and positive close to the (+) charge, but as you go left to the (-) charge it will decrease to 0, and then become more negative as it reaches the (-) charge. Once it passes this, it will get bigger (from -infinity) until it equalizes with the (+) charge. Then, when you are rather far away it will be 0 again.

Re 3: How can the field add up to zero, both "far to the left"(not infinity) and "slightly to the left" of the negative charge? Only one can be true.

And as for 4, what are the directions of the electric fields at a point between the charges (along the line joining them)?
 
neutrino said:
Re 3: How can the field add up to zero, both "far to the left"(not infinity) and "slightly to the left" of the negative charge? Only one can be true.

And as for 4, what are the directions of the electric fields at a point between the charges (along the line joining them)?

Ohhhh I get it now! Thanks for the help!
 

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