Electric Field Lines — Question 9

Question 9

An oblique counting surface must be projected perpendicular to the field

The same narrow bundle of representative field lines crosses two small patches. The first has area $A_1=2.00$ cm$^2$ normal to the local field. The second has area $A_2=3.00$ cm$^2$ with its normal at $60^\circ$ to the field. The magnitude on a patch is proportional to the line count divided by its perpendicular projected area.

1. Find the ratio $E_2/E_1$.

2. Find the second patch area required for equal field strengths at that angle. Explain the limit as the patch becomes parallel to the field.

Question 9 - Solution

An oblique counting surface must be projected perpendicular to the field

(a) Method and calculation

$$A_{2,\perp}=A_2\cos60^\circ=1.50\,\mathrm{cm^2}.$$$$E_2/E_1=A_{1,\perp}/A_{2,\perp}=2.00/1.50=4/3.$$

(b) Interpretation and further calculation

$$E_2=E_1\Rightarrow A_2\cos60^\circ=A_1\Rightarrow A_2=4.00\,\mathrm{cm^2}.$$$$\theta\to90^\circ:\quad A_\perp\to0.$$A finite field then sends vanishing line count through a fixed nearly parallel patch; keeping the count fixed would require growing patch area.

Physical check

This local density argument is not a count of how many decorative strokes happen to appear in the schematic.