Electric Potential Energy of a System of Point Charges — Question 3

Section 24-10 · Physics pages 3976–3977

Question 3 Removing one corner charge extracts three specific pair contributions Four equal positive point charges q occupy square vertices A,B,C,D of side a. Other charges remain fixed when one is moved; use zero interaction energy at infinite separation. A B C D a Schematic; labels, not drawing lengths, specify the data. (a) Find the full initial interaction energy. (b) Find the quasistatic work removing A to infinity and the energy of the three charges left behind.
Show solutionHide solution
Question 3 - Solution Removing one corner charge extracts three specific pair contributions A B C D a (a) Method and calculation Ui = kq2 (4 a + 2 √ 2a ) = kq2 a (4 + √ 2). qVothers(A) = kq2 a ( 2 + 1 √ 2 ) . (b) Interpretation and further calculation Wext = ∆U = −kq2 a ( 2 + 1 √ 2 ) . Uf = Ui + Wext = kq2 a ( 2 + 1 √ 2 ) . Physical check Removing one charge eliminates its three pairs; using one half of q times the other-source potential would undercount the removed energy.

Question 3

Removing one corner charge extracts three specific pair contributions

Four equal positive point charges q occupy square vertices A,B,C,D of side a. Other charges remain fixed when one is moved; use zero interaction energy at infinite separation.

Diagram shown in the original SVG worksheet.

  1. Find the full initial interaction energy.

  2. Find the quasistatic work removing A to infinity and the energy of the three charges left behind.

Question 3 - Solution

Removing one corner charge extracts three specific pair contributions

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Ui=kq2(4a+22a)=kq2a(4+2).U_i=kq^2\left(\frac4a+\frac2{\sqrt2a}\right)=\frac{kq^2}{a}(4+\sqrt2).qVothers(A)=kq2a(2+12).qV_{\mathrm{others}}(A)=\frac{kq^2}{a}\left(2+\frac1{\sqrt2}\right).

(b) Interpretation and further calculation

Wext=ΔU=−kq2a(2+12).W_{\mathrm{ext}}=\Delta U=-\frac{kq^2}{a}\left(2+\frac1{\sqrt2}\right).Uf=Ui+Wext=kq2a(2+12).U_f=U_i+W_{\mathrm{ext}}=\frac{kq^2}{a}\left(2+\frac1{\sqrt2}\right).

Physical check

Removing one charge eliminates its three pairs; using one half of q times the other-source potential would undercount the removed energy.