Potential Due to a Group of Point Charges — Question 6

Section 24-6 · Physics pages 3902–3903

Question 6 A remote cluster exposes net charge before its internal arrangement Charges +2Q at x=a and -Q at x=-a are viewed at x=r>a. Let Q,a be positive and r much larger than a. Use infinity reference. x −Q +2Q P Schematic; labels, not drawing lengths, specify the data. (a) Find the exact axial potential and expand through order a/r. (b) Estimate the fractional correction to the net-charge approximation and find the leading criterion for it to be below one percent.
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Question 6 - Solution A remote cluster exposes net charge before its internal arrangement x −Q +2Q P (a) Method and calculation V (r) = kQ ( 2 r −a − 1 r + a ) = kQ r + 3a r2 −a2 . V (r) = kQ r [ 1 + 3a r + O(a2/r2) ] . (b) Interpretation and further calculation Vmono = kQ/r, V −Vmono Vmono ≃3a/r. 3a/r < .0100 ⇒r > 300a to leading order. Physical check The exact relative correction is (3a/r + a2/r2)/(1 −a2/r2); the leading criterion is an approximation, not a strict bound.

Question 6

A remote cluster exposes net charge before its internal arrangement

Charges +2Q at x=a and -Q at x=-a are viewed at x=r>a. Let Q,a be positive and r much larger than a. Use infinity reference.

Diagram shown in the original SVG worksheet.

  1. Find the exact axial potential and expand through order a/ra/r.

  2. Estimate the fractional correction to the net-charge approximation and find the leading criterion for it to be below one percent.

Question 6 - Solution

A remote cluster exposes net charge before its internal arrangement

Diagram shown in the original SVG worksheet.

(a) Method and calculation

V(r)=kQ(2r−a−1r+a)=kQr+3ar2−a2.V(r)=kQ\left(\frac2{r-a}-\frac1{r+a}\right)=kQ\frac{r+3a}{r^2-a^2}.V(r)=kQr[1+3ar+O(a2/r2)].V(r)=\frac{kQ}{r}\left[1+\frac{3a}{r}+O(a^2/r^2)\right].

(b) Interpretation and further calculation

Vmono=kQ/r,V−VmonoVmono≃3a/r.V_{\mathrm{mono}}=kQ/r,\quad\frac{V-V_{\mathrm{mono}}}{V_{\mathrm{mono}}}\simeq3a/r.3a/r<.0100⇒r>300ato leading order.3a/r<.0100\Rightarrow r>300a\quad\text{to leading order}.

Physical check

The exact relative correction is (3a/r+a2/r2)/(1−a2/r2)(3a/r+a^2/r^2)/(1-a^2/r^2); the leading criterion is an approximation, not a strict bound.