Electric Field Lines — Question 2

Section 22-2 · Physics pages 3494–3495

Question 2 A saddle field produces hyperbolic lines with a singular direction at the null In a plane, Ex = ax and Ey = −ay with a > 0. A line in the first quadrant passes through (L, L), where L > 0. x y Schematic; labels, not drawing lengths, specify the data. (a) Integrate the field-line equation to find that curve and its arrow orientation. (b) Analyze the field on both coordinate axes and at the origin. Explain why extending the hyperbola through the origin would be incorrect.
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Question 2 - Solution A saddle field produces hyperbolic lines with a singular direction at the null x y (a) Method and calculation dy dx = −y x ⇒dy y = −dx x ⇒xy = C. C = L2, y = L2/x; x > 0, y > 0 : Ex > 0, Ey < 0. (b) Interpretation and further calculation y = 0 : E = (ax, 0), x = 0 : E = (0, −ay). E(0, 0) = 0, xy = L2̸ = 0. Arrows point away from the null along the x axis and toward it along the y axis. No direction is defined at the null itself. Physical check The field-line equation using a ratio is not valid at points where both numerator and denominator vanish.

Question 2

A saddle field produces hyperbolic lines with a singular direction at the null

In a plane, Ex=axE_x=ax and Ey=−ayE_y=-ay with a>0a>0. A line in the first quadrant passes through (L,L)(L,L), where L>0L>0.

Diagram shown in the original SVG worksheet.

  1. Integrate the field-line equation to find that curve and its arrow orientation.

  2. Analyze the field on both coordinate axes and at the origin. Explain why extending the hyperbola through the origin would be incorrect.

Question 2 - Solution

A saddle field produces hyperbolic lines with a singular direction at the null

Diagram shown in the original SVG worksheet.

(a) Method and calculation

dydx=−yx⇒dyy=−dxx⇒xy=C.\frac{dy}{dx}=-\frac yx\Rightarrow\frac{dy}{y}=-\frac{dx}{x}\Rightarrow xy=C.C=L2,y=L2/x;x>0,y>0:Ex>0,Ey<0.C=L^2,\quad y=L^2/x;\quad x>0,y>0:\ E_x>0,E_y<0.

(b) Interpretation and further calculation

y=0:𝑬=(ax,0),x=0:𝑬=(0,−ay).y=0:\ \mathbf E=(ax,0),\qquad x=0:\ \mathbf E=(0,-ay).𝑬(0,0)=𝟎,xy=L2≠0.\mathbf E(0,0)=\mathbf0,\quad xy=L^2\ne0.Arrows point away from the null along the x axis and toward it along the y axis. No direction is defined at the null itself.

Physical check

The field-line equation using a ratio is not valid at points where both numerator and denominator vanish.