Math Problem Statement

A rod of length L lies along the x axis with its left end at the origin. It has a nonuniform charge density 𝜆 = 𝛼x, where 𝛼 is a positive constant.

Calculate the electric potential at point B , which lies on the perpendicular bisector of the rod a distance b above the x axis. (Use the following as necessary: 𝛼, ke, Lb, and d.)

Solution

To calculate the electric potential VBV_B at point BB, which lies a distance bb above the perpendicular bisector of the rod, we need to integrate the contributions of the infinitesimal charges along the length of the rod.

Given that the charge density is non-uniform, λ=αx\lambda = \alpha x, where α\alpha is a constant and xx is the distance along the rod from the origin, we need to find the potential due to each infinitesimal charge dqdq.

Step-by-step breakdown:

  1. Set up the infinitesimal charge element dqdq: The linear charge density λ(x)\lambda(x) varies as: λ(x)=αx\lambda(x) = \alpha x The infinitesimal charge dqdq over an infinitesimal length dxdx is given by: dq=λ(x)dx=αxdxdq = \lambda(x) dx = \alpha x dx

  2. Determine the distance from each charge element to point BB: Let the position of point BB be (0,b)(0, b). The distance from an infinitesimal charge element at position xx on the rod to the point BB is: d=x2+b2d = \sqrt{x^2 + b^2}

  3. Write the infinitesimal potential dVdV: The potential at point BB due to an infinitesimal charge dqdq is: dV=kedqd=keαxdxx2+b2dV = \frac{k_e dq}{d} = \frac{k_e \alpha x dx}{\sqrt{x^2 + b^2}} where kek_e is the Coulomb constant.

  4. Integrate to find the total potential VBV_B: The rod extends from x=0x = 0 to x=Lx = L, so we integrate the expression for dVdV from 0 to LL: VB=0Lkeαxdxx2+b2V_B = \int_0^L \frac{k_e \alpha x dx}{\sqrt{x^2 + b^2}}

  5. Solve the integral:

    This is a standard integral, and the result is: VB=keα[x2+b2]0LV_B = k_e \alpha \left[ \sqrt{x^2 + b^2} \right]_0^L

  6. Evaluate the limits: VB=keα(L2+b202+b2)V_B = k_e \alpha \left( \sqrt{L^2 + b^2} - \sqrt{0^2 + b^2} \right) Simplifying: VB=keα(L2+b2b)V_B = k_e \alpha \left( \sqrt{L^2 + b^2} - b \right)

  7. Substitute α\alpha in terms of total charge (if needed): If we are given the total charge QQ on the rod, we could express α\alpha in terms of QQ. However, for this problem, α\alpha is left as a constant, so the final expression for the potential at point BB is:

VB=keα(L2+b2b)V_B = k_e \alpha \left( \sqrt{L^2 + b^2} - b \right)

Final Answer:

VB=keα(L2+b2b)V_B = k_e \alpha \left( \sqrt{L^2 + b^2} - b \right)

Let me know if you want more details or have any questions. Here are five related questions:

  1. How does the potential change if the charge density was constant?
  2. How would the electric field be calculated from the electric potential in this scenario?
  3. How does the potential behave as the distance bb increases significantly?
  4. Can you calculate the total charge on the rod if α\alpha is known?
  5. How would the potential change if the rod were bent into a semicircle?

Tip: When solving for potentials due to non-uniform charge distributions, always consider symmetry and the behavior of charge density carefully.

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Math Problem Analysis

Mathematical Concepts

Electrostatics
Calculus
Integration
Non-uniform Charge Distribution

Formulas

dq = 𝜆(x) dx = 𝛼 x dx
V_B = ∫(k_e * α * x dx) / √(x^2 + b^2) from 0 to L
V_B = k_e * α * (√(L^2 + b^2) - b)

Theorems

Superposition Principle in Electrostatics
Coulomb's Law

Suitable Grade Level

Undergraduate Physics (Introductory level)