Electric Potential Problem - Fixed Template

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SUMMARY

The discussion focuses on calculating the distance "x" from an electric charge q1 (-6e) at the origin where the net electric potential is zero, given a second charge q2 (+2e) located at a distance "d". The solution involves equating the electric potentials from both charges, leading to the equation (3d/4) = x, confirming that x is indeed less than d. The final answer is validated by multiple participants, ensuring its correctness for exam preparation.

PREREQUISITES
  • Understanding of electric potential and its mathematical representation.
  • Familiarity with Coulomb's law and the concept of electric charge.
  • Basic algebra skills for manipulating equations.
  • Knowledge of the principles of superposition in electric fields.
NEXT STEPS
  • Study the concept of electric potential and its calculation in electrostatics.
  • Learn about Coulomb's law and its application in electric field problems.
  • Practice solving problems involving multiple charges and net electric potential.
  • Explore the principles of superposition and how they apply to electric fields.
USEFUL FOR

Students preparing for exams in physics, particularly those focusing on electrostatics, as well as educators seeking to clarify concepts related to electric potential and charge interactions.

Zackkkkkk
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Homework Statement


An electric charge q1 (-6e) is located at the origin. A second electric charge q2 (+2e) is located a distance "d" from the origin. In terms of "d", what is the distance "x" from q1 where the net electric potential is zero? Assume for this problem that x<d.2.Relevant equations
I need a solid explanation on how to do this problem properly. I have a strong feeling that this can't be right (my attempt). I need to be able to understand and work problems like this for my upcoming exam. Any help would be appreciated.

The Attempt at a Solution


Vtotal = V1 + V2.
If Vtotal = 0 then V1+V2=0.
V1-V2
kq1/x = - kq2/d-x
q1/x = - q2/d-x
(q1(d-x))/(-q2) = x
3(d-x) = x
3d = 4x
(3d/4) = x[/B]
 
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Zackkkkkk said:

Homework Statement


An electric charge q1 (-6e) is located at the origin. A second electric charge q2 (+2e) is located a distance "d" from the origin. In terms of "d", what is the distance "x" from q1 where the net electric potential is zero? Assume for this problem that x<d.2.Relevant equations
I need a solid explanation on how to do this problem properly. I have a strong feeling that this can't be right (my attempt). I need to be able to understand and work problems like this for my upcoming exam. Any help would be appreciated.

The Attempt at a Solution


Vtotal = V1 + V2.
If Vtotal = 0 then V1+V2=0.
V1-V2 [/B]##\quad \quad ## Looks like a missing "=" sign, V1 = -V2
kq1/x = - kq2/(d-x)
q1/x = - q2/(d-x)
(q1(d-x))/(-q2) = x
3(d-x) = x
3d = 4x
(3d/4) = x
Other than what appears to be a typo above, and the lack of some needed sets of parentheses, that looks good.

I'm pretty sure that the final answer is correct.
 
SammyS said:
Other than what appears to be a typo above, and the lack of some needed sets of parentheses, that looks good.

I'm pretty sure that the final answer is correct.
Thank you for your input, friend.
 

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