Electric Potential Energy of a System of Point Charges — Question 7

Section 24-10 · Physics pages 3984–3985

Question 7 Three equal repelling charges expand self-similarly from an equilateral triangle Three particles each of charge q>0 and mass m start at rest at the vertices of an equilateral triangle of side a and are released in isolation. A B C a Schematic; labels, not drawing lengths, specify the data. (a) Use symmetry and energy conservation to find each speed when the side length has grown to b>a. (b) Find each limiting speed and explain why the particles remain at the vertices of an expanding equilateral triangle.
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Question 7 - Solution Three equal repelling charges expand self-similarly from an equilateral triangle A B C a (a) Method and calculation U(a) = 3kq2/a, U(b) = 3kq2/b. 3 ( 1 2mv2) = 3kq2(1/a −1/b) ⇒v = √ 2kq2 m (1 a −1 b ) . (b) Interpretation and further calculation v∞= √ 2kq2/(ma). Fi ∥(ri −rcenter), |F1| = |F2| = |F3|. Identical radial evolution preserves all threefold symmetries. Physical check The displayed v is each particle speed relative to the fixed center of mass; the side-length rate is ˙b = √ 3v.

Question 7

Three equal repelling charges expand self-similarly from an equilateral triangle

Three particles each of charge q>0 and mass m start at rest at the vertices of an equilateral triangle of side a and are released in isolation.

Diagram shown in the original SVG worksheet.

  1. Use symmetry and energy conservation to find each speed when the side length has grown to b>a.

  2. Find each limiting speed and explain why the particles remain at the vertices of an expanding equilateral triangle.

Question 7 - Solution

Three equal repelling charges expand self-similarly from an equilateral triangle

Diagram shown in the original SVG worksheet.

(a) Method and calculation

U(a)=3kq2/a,U(b)=3kq2/b.U(a)=3kq^2/a,\quad U(b)=3kq^2/b.3(12mv2)=3kq2(1/a−1/b)⇒v=2kq2m(1a−1b).3\left(\tfrac12mv^2\right)=3kq^2(1/a-1/b)\Rightarrow v=\sqrt{\frac{2kq^2}{m}\left(\frac1a-\frac1b\right)}.

(b) Interpretation and further calculation

v∞=2kq2/(ma).v_\infty=\sqrt{2kq^2/(ma)}.𝑭i∥(𝒓i−𝒓center),|𝑭1|=|𝑭2|=|𝑭3|.\mathbf F_i\parallel(\mathbf r_i-\mathbf r_{\mathrm{center}}),\quad |\mathbf F_1|=|\mathbf F_2|=|\mathbf F_3|.Identical radial evolution preserves all threefold symmetries.

Physical check

The displayed v is each particle speed relative to the fixed center of mass; the side-length rate is ḃ=3v\dot b=\sqrt3v.