Simple Point Charge Potential Problem

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SUMMARY

The discussion centers on calculating the electric potential at a point above the center of a charge distribution formed by two point charges, Charge A and Charge B, located at (-d/2, 0) and (d/2, 0) respectively. The potential is calculated using the formula V(r) = (1/4πε₀)Σ(qᵢ/rᵢ). The confusion arises from the expectation of a non-zero potential despite the charges being of opposite signs, which leads to potential cancellation. The key takeaway is that while the charges are equal in magnitude and opposite in sign, the potential at the point (0, z) does not cancel out to zero due to the distance factor in the potential calculation.

PREREQUISITES
  • Understanding of electric potential and point charges
  • Familiarity with the formula V(r) = (1/4πε₀)Σ(qᵢ/rᵢ)
  • Knowledge of coordinate systems in physics
  • Basic grasp of vector addition and cancellation principles
NEXT STEPS
  • Study the concept of electric potential due to multiple point charges
  • Learn about the superposition principle in electrostatics
  • Explore the implications of charge sign and magnitude on electric potential
  • Investigate the derivation and application of the potential formula V(r) = (1/4πε₀)Σ(qᵢ/rᵢ)
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding electric potential in systems with multiple point charges.

dgreenheck
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Homework Statement


There are two point charges aligned on the X-axis. Charge A is a distance -d/2 from the origin and Charge B is a distance d/2 from the origin. What is the potential at a distance z above the center of the charge distribution?

To further clarify
Charge A location at (-d/2,0)
Charge B location at (d/2,0)
Point of analysis at (0,z)

Homework Equations


V(r) = \frac{1}{4\pi\epsilon\stackrel{}{0}}\sum\frac{q\stackrel{}{i}}{r\stackrel{}{i}}

The Attempt at a Solution


I'm confused because I'm getting 0 and the answer should be non-zero. There is no direction information carried in r, correct? Because it's just supposed to be the distance to point (0,z) which is \sqrt{z\stackrel{2}{}+(\frac{d}{2})\stackrel{2}{}}. And since the charges are of opposite sign, the sum will cause them to cancel out. Obviously my understanding of the summation is wrong here so if someone could point me in the right direction that would be very helpful. Thanks
 
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dgreenheck said:
1. Homework Statement
There are two point charges aligned on the X-axis. Charge A is a distance -d/2 from the origin and Charge B is a distance d/2 from the origin. What is the potential at a distance z above the center of the charge distribution?

To further clarify
Charge A location at (-d/2,0)
Charge B location at (d/2,0)
Point of analysis at (0,z)

Homework Equations


V(r) = \frac{1}{4\pi\epsilon\stackrel{}{0}}\sum\frac{q\stackrel{}{i}}{r\stackrel{}{i}}

The Attempt at a Solution


I'm confused because I'm getting 0 and the answer should be non-zero. There is no direction information carried in r, correct? Because it's just supposed to be the distance to point (0,z) which is \sqrt{z\stackrel{2}{}+(\frac{d}{2})\stackrel{2}{}}. And since the charges are of opposite sign, the sum will cause them to cancel out. Obviously my understanding of the summation is wrong here so if someone could point me in the right direction that would be very helpful. Thanks

I "bolded" a couple of parts of your original post.

Are these charges equal in magnitude? Are these charges of the same sign or opposite sign?
 
The charges are of opposite sign and of equal magnitude.
 

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