Electric Potential Energy of a System of Point Charges — Question 8

Section 24-10 · Physics pages 3986–3987

Question 8 A constrained middle charge minimizes a three-body energy along its track Two positive charges q are fixed at x=0 and x=L. A third positive charge q of mass m moves on a frictionless straight guide within 0 < x < L. Include all three pair energies. x q q at x q Schematic; labels, not drawing lengths, specify the data. (a) Find U(x), its equilibrium position and minimum value. (b) Find the curvature at equilibrium and the small-oscillation angular frequency along the guide.
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Question 8 - Solution A constrained middle charge minimizes a three-body energy along its track x q q at x q (a) Method and calculation U(x) = kq2 ( 1 L + 1 x + 1 L −x ) . U′ = kq2[−x−2 + (L −x)−2] = 0 ⇒x = L/2, Umin = 5kq2/L. (b) Interpretation and further calculation U′′ = kq2 [ 2 x3 + 2 (L −x)3 ] ⇒U′′(L/2) = 32kq2/L3. ω = √ 32kq2 mL3 . Physical check The fixed endpoint pair adds a constant to U, affecting the minimum value but not the equilibrium or frequency.

Question 8

A constrained middle charge minimizes a three-body energy along its track

Two positive charges q are fixed at x=0 and x=L. A third positive charge q of mass m moves on a frictionless straight guide within 0<x<L0<x<L. Include all three pair energies.

Diagram shown in the original SVG worksheet.

  1. Find U(x), its equilibrium position and minimum value.

  2. Find the curvature at equilibrium and the small-oscillation angular frequency along the guide.

Question 8 - Solution

A constrained middle charge minimizes a three-body energy along its track

Diagram shown in the original SVG worksheet.

(a) Method and calculation

U(x)=kq2(1L+1x+1L−x).U(x)=kq^2\left(\frac1L+\frac1x+\frac1{L-x}\right).U′=kq2[−x−2+(L−x)−2]=0⇒x=L/2,Umin=5kq2/L.U\prime=kq^2[-x^{-2}+(L-x)^{-2}]=0\Rightarrow x=L/2,\quad U_{\min}=5kq^2/L.

(b) Interpretation and further calculation

U′′=kq2[2x3+2(L−x)3]⇒U′′(L/2)=32kq2/L3.U\prime\prime=kq^2\left[\frac2{x^3}+\frac2{(L-x)^3}\right]\Rightarrow U\prime\prime(L/2)=32kq^2/L^3.ω=32kq2mL3.\omega=\sqrt{\frac{32kq^2}{mL^3}}.

Physical check

The fixed endpoint pair adds a constant to U, affecting the minimum value but not the equilibrium or frequency.