Math Problem Statement

Point charges 50nC and xnC located at (3,4,0), (4,2,-3) and (2,-6,5) while line x=3, y=-4 carries uniform charge 55nC/m. Here x=23 and y=3. If V=0V at O(0,0,0) find V at A(y,y,5).

Solution

The problem involves multiple point charges and a line charge distributed uniformly, and asks for the potential VV at a certain point. Let's break down the problem.

Problem Breakdown:

  1. Point charges:

    • Charge 1: 50nC50 \, \text{nC} located at (3,4,0)(3, 4, 0)
    • Charge 2: xnCx \, \text{nC} located at (4,2,3)(4, 2, -3)
    • Charge 3: (2,6,5)(2, -6, 5), the charge is unspecified but it appears to be part of the setup.
  2. Line charge:

    • A line along x=3x = 3, y=4y = -4, with a uniform charge density of 55nC/m55 \, \text{nC/m}.
  3. Potential given:

    • V=0VV = 0 \, \text{V} at O(0,0,0)O(0, 0, 0).
  4. We are asked to find:

    • The potential at point AA at (y,y,5)(y, y, 5).

Potential Due to Point Charges

The potential VV at a point due to a point charge qq located at a distance rr from the point is given by the formula: V=kqrV = \frac{kq}{r} where k=9×109N\cdotpm2/C2k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 is Coulomb's constant, and rr is the distance between the charge and the point of interest.

We will need to compute the distance between the given point charges and the point AA, and sum up the contributions from each charge.

Potential Due to the Line Charge

For a line charge, the potential at a point a distance rr from the line is: Vline=λ2πϵ0ln(r0r)V_{\text{line}} = \frac{\lambda}{2\pi \epsilon_0} \ln\left(\frac{r_0}{r}\right) where:

  • λ=55nC/m\lambda = 55 \, \text{nC/m} is the linear charge density,
  • ϵ0=8.85×1012F/m\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m},
  • rr is the perpendicular distance from the point to the line.

Let me calculate the potential at the given point A(y,y,5)A(y, y, 5) considering the contributions of the point charges and the line charge.Here are the results for the contributions to the potential at point A(3,3,5)A(3, 3, 5):

  1. Potential due to the first point charge (50 nC at (3,4,0)(3, 4, 0)): V1=88.25VV_1 = 88.25 \, \text{V}

  2. Distance from point AA to the line charge located along x=3x = 3 and y=4y = -4 is 7m7 \, \text{m}.

  3. Potential due to the line charge with uniform charge density 55nC/m55 \, \text{nC/m}: Vline=1924.70VV_{\text{line}} = -1924.70 \, \text{V}

Next, we would need the unknown charges and then sum the contributions to find the total potential at point AA. Let me know if you'd like to proceed with assumptions for the missing charges or if you'd like to explore any part in more detail.


Here are some relative questions you might explore:

  1. How would the potential change if the second point charge were given?
  2. Can we assume symmetry or make approximations to estimate the missing charge?
  3. How does the distance from the line charge affect the potential in general?
  4. What is the significance of V=0V = 0 at O(0,0,0)O(0,0,0) in terms of reference potential?
  5. How would we calculate the potential at a different point from these same charges?

Tip: Always remember to consider both point charges and continuous charge distributions (like line or surface charges) when calculating potentials.

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Math Problem Analysis

Mathematical Concepts

Electrostatics
Electric Potential
Point Charges
Line Charge

Formulas

Potential due to a point charge: V = kq/r
Potential due to a line charge: V_line = (lambda / (2π * epsilon_0)) ln(r0 / r)

Theorems

Coulomb's Law
Superposition Principle

Suitable Grade Level

Undergraduate Physics