Math Problem Statement

solve

Compare the following configurations of charges (charges at distances a from the origin). Which charge distribution has the lowest potential V at the origin (pt. O)?

Solution

The problem asks you to compare the electric potentials at the origin (point O) for several configurations of point charges located at a distance aa from the origin. You are tasked with determining which charge distribution results in the lowest potential at the origin.

Key Concept:

The electric potential VV at the origin due to a point charge is given by:

V=kqrV = \frac{kq}{r}

where kk is Coulomb's constant, qq is the charge, and rr is the distance from the charge to the point where potential is being calculated. For multiple charges, the potentials due to each charge sum linearly, since potential is a scalar quantity.

Each configuration has charges at equal distances from the origin, so r=ar = a for all charges. The total potential at the origin for each configuration is the sum of the individual potentials:

Vtotal=kqiaV_{\text{total}} = k \sum \frac{q_i}{a}

We will now compare the charge distributions:

Configurations:

  • (a): Two positive charges (+2q and +q) and two negative charges (-2q and -q)
    • Va=k(+2q)a+k(+q)a+k(2q)a+k(q)a=0V_{\text{a}} = \frac{k(+2q)}{a} + \frac{k(+q)}{a} + \frac{k(-2q)}{a} + \frac{k(-q)}{a} = 0
  • (b): Two positive charges (+q and +q) and two negative charges (-q and -q)
    • Vb=k(+q)a+k(+q)a+k(q)a+k(q)a=0V_{\text{b}} = \frac{k(+q)}{a} + \frac{k(+q)}{a} + \frac{k(-q)}{a} + \frac{k(-q)}{a} = 0
  • (c): Three positive charges (+2q, +q, and +q) and one negative charge (-q)
    • Vc=k(+2q)a+k(+q)a+k(+q)a+k(q)a=+3kqaV_{\text{c}} = \frac{k(+2q)}{a} + \frac{k(+q)}{a} + \frac{k(+q)}{a} + \frac{k(-q)}{a} = \frac{+3kq}{a}
  • (d): Three negative charges (-q, -q, and -q) and one positive charge (+2q)
    • Vd=k(q)a+k(q)a+k(q)a+k(+2q)a=qkaV_{\text{d}} = \frac{k(-q)}{a} + \frac{k(-q)}{a} + \frac{k(-q)}{a} + \frac{k(+2q)}{a} = \frac{-qk}{a}
  • (e): Two positive charges (+q and +q) and two negative charges (-q and -q)
    • Ve=k(+q)a+k(+q)a+k(q)a+k(q)a=0V_{\text{e}} = \frac{k(+q)}{a} + \frac{k(+q)}{a} + \frac{k(-q)}{a} + \frac{k(-q)}{a} = 0

Conclusion:

  • Configurations (a), (b), and (e) have zero potential at the origin because the contributions from the positive and negative charges cancel out.
  • Configuration (c) has the highest potential due to the excess of positive charges.
  • Configuration (d) has the lowest potential because there is a net negative charge contributing to a negative potential at the origin.

Thus, configuration (d) has the lowest potential at the origin.

Let me know if you'd like more details or have questions!

Here are 5 questions that follow from this topic:

  1. How would the potential at the origin change if all charges were twice as far from the origin?
  2. How does the electric field at the origin compare for each configuration?
  3. What is the physical significance of the potential being zero at the origin in certain configurations?
  4. How would adding a charge at the origin affect the system's overall potential?
  5. What role does symmetry play in determining the potential at the origin in these configurations?

Tip: In charge configurations, always check if symmetry helps simplify calculations by canceling contributions from opposing charges.

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Math Problem Analysis

Mathematical Concepts

Electric Potential
Point Charge
Electrostatics

Formulas

V = kq/r
V_total = k ∑(q_i/a)

Theorems

Superposition of Electric Potentials
Coulomb's Law

Suitable Grade Level

College/University - Introductory Physics