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Physics / Chapter 24: Electric Potential / Section 24-9

Calculating the Field from the Potential — Question 8

Section 24-9 · Physics pages 3966–3967

Question 8 A central difference estimates the field with a computable curvature error Along x, V (x) = V0 + Ax + Bx3. An instrument estimates the field at zero by Eh = −[V (h) −V (−h)]/(2h) with h>0. x −h −h/2 0 h/2 h Schematic; labels, not drawing lengths, specify the data. (a) Find the exact field at zero and the finite-spacing error. (b) Combine estimates at h and h/2 to remove the cubic error exactly for this polynomial.
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Question 8 - Solution A central difference estimates the field with a computable curvature error x −h −h/2 0 h/2 h (a) Method and calculation E(0) = −V ′(0) = −A. Eh = −A −Bh2, Eh −E(0) = −Bh2. (b) Interpretation and further calculation Eh/2 = −A −Bh2/4. Eimproved = 4Eh/2 −Eh 3 = −A. Physical check The cancellation is exact for the stated cubic; higher odd powers leave higher-order errors.

Question 8

A central difference estimates the field with a computable curvature error

Along x, V(x)=V0+Ax+Bx3V(x)=V_0+Ax+Bx^3. An instrument estimates the field at zero by Eh=−[V(h)−V(−h)]/(2h)E_h=-[V(h)-V(-h)]/(2h) with h>0.

Diagram shown in the original SVG worksheet.

  1. Find the exact field at zero and the finite-spacing error.

  2. Combine estimates at h and h/2 to remove the cubic error exactly for this polynomial.

Question 8 - Solution

A central difference estimates the field with a computable curvature error

Diagram shown in the original SVG worksheet.

(a) Method and calculation

E(0)=−V′(0)=−A.E(0)=-V\prime(0)=-A.Eh=−A−Bh2,Eh−E(0)=−Bh2.E_h=-A-Bh^2,\quad E_h-E(0)=-Bh^2.

(b) Interpretation and further calculation

Eh/2=−A−Bh2/4.E_{h/2}=-A-Bh^2/4.Eimproved=4Eh/2−Eh3=−A.E_{\mathrm{improved}}=\frac{4E_{h/2}-E_h}{3}=-A.

Physical check

The cancellation is exact for the stated cubic; higher odd powers leave higher-order errors.

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