Calculating the Potential from the Field — Question 9

Section 24-4 · Physics pages 3868–3869

Question 9 A sinusoidal field has repeated equal-potential points Along x, Ex = E0 cos(πx/L) with E0, L > 0. Set V(0)=0 and consider 0 ≤x ≤2L. x Ex 2L Schematic; labels, not drawing lengths, specify the data. (a) Find V(x) and its largest and smallest values. (b) Find the voltage from L/4 to 3L/4 and explain whether the field vanishes at these endpoints.
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Question 9 - Solution A sinusoidal field has repeated equal-potential points x Ex 2L (a) Method and calculation V (x) = −E0L π sin(πx/L). Vmin = −E0L/π (x = L/2), Vmax = +E0L/π (x = 3L/2). (b) Interpretation and further calculation V (3L/4) −V (L/4) = 0. Ex(L/4) = E0/ √ 2, Ex(3L/4) = −E0/ √ 2. Physical check Cancellation in a definite integral does not require a zero integrand at either endpoint.

Question 9

A sinusoidal field has repeated equal-potential points

Along x, Ex=E0cos⁡(πx/L)E_x=E_0\cos(\pi x/L) with E0,L>0E_0,L>0. Set V(0)=0 and consider 0≤x≤2L0\le x\le2L.

Diagram shown in the original SVG worksheet.

  1. Find V(x) and its largest and smallest values.

  2. Find the voltage from L/4 to 3L/4 and explain whether the field vanishes at these endpoints.

Question 9 - Solution

A sinusoidal field has repeated equal-potential points

Diagram shown in the original SVG worksheet.

(a) Method and calculation

V(x)=−E0Lπsin⁡(πx/L).V(x)=-\frac{E_0L}{\pi}\sin(\pi x/L).Vmin=−E0L/π(x=L/2),Vmax=+E0L/π(x=3L/2).V_{\min}=-E_0L/\pi\ (x=L/2),\quad V_{\max}=+E_0L/\pi\ (x=3L/2).

(b) Interpretation and further calculation

V(3L/4)−V(L/4)=0.V(3L/4)-V(L/4)=0.Ex(L/4)=E0/2,Ex(3L/4)=−E0/2.E_x(L/4)=E_0/\sqrt2,\quad E_x(3L/4)=-E_0/\sqrt2.

Physical check

Cancellation in a definite integral does not require a zero integrand at either endpoint.