The Electric Field β€” Question 4

Section 22-1 Β· Physics pages 3478–3479

Question 4 A compensating source must cancel unwanted components as well as set magnitude Two independent fixed sources produce EA = (30, βˆ’10, 20) N/C and EB = (βˆ’5, 25, βˆ’8) N/C at P. A third adjustable source must make the total field point along +z with magnitude 50 N/C. Component triples refer to Cartesian axes. P A B C Schematic; labels, not drawing lengths, specify the data. (a) Find the required third-source field. (b) If that source operates at 90 percent of its required strength with direction unchanged, find the actual field and the residual transverse magnitude.
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Question 4 - Solution A compensating source must cancel unwanted components as well as set magnitude P A B C (a) Method and calculation EA + EB = (25, 15, 12) N/C. EC = (0, 0, 50) βˆ’(25, 15, 12) = (βˆ’25, βˆ’15, 38) N/C. (b) Interpretation and further calculation Eactual = (25, 15, 12) + .9(βˆ’25, βˆ’15, 38) = (2.5, 1.5, 46.2) N/C. EβŠ₯= √ 2.52 + 1.52 = √ 8.5 β‰ˆ2.91548 N/C. A strength error in the compensator introduces direction error as well as longitudinal error. Physical check Vector superposition applies component by component; adding field magnitudes would not solve the compensation problem.

Question 4

A compensating source must cancel unwanted components as well as set magnitude

Two independent fixed sources produce 𝑬A=(30,βˆ’10,20)\mathbf E_A=(30,-10,20) N/C and 𝑬B=(βˆ’5,25,βˆ’8)\mathbf E_B=(-5,25,-8) N/C at P. A third adjustable source must make the total field point along +z+z with magnitude 50 N/C. Component triples refer to Cartesian axes.

Diagram shown in the original SVG worksheet.

  1. Find the required third-source field.

  2. If that source operates at 90 percent of its required strength with direction unchanged, find the actual field and the residual transverse magnitude.

Question 4 - Solution

A compensating source must cancel unwanted components as well as set magnitude

Diagram shown in the original SVG worksheet.

(a) Method and calculation

𝑬A+𝑬B=(25,15,12)N/C.\mathbf E_A+\mathbf E_B=(25,15,12)\,\mathrm{N/C}.𝑬C=(0,0,50)βˆ’(25,15,12)=(βˆ’25,βˆ’15,38)N/C.\mathbf E_C=(0,0,50)-(25,15,12)=(-25,-15,38)\,\mathrm{N/C}.

(b) Interpretation and further calculation

𝑬actual=(25,15,12)+.9(βˆ’25,βˆ’15,38)=(2.5,1.5,46.2)N/C.\mathbf E_{\mathrm{actual}}=(25,15,12)+.9(-25,-15,38)=(2.5,1.5,46.2)\,\mathrm{N/C}.EβŸ‚=2.52+1.52=8.5β‰ˆ2.91548N/C.E_\perp=\sqrt{2.5^2+1.5^2}=\sqrt{8.5}\approx2.91548\,\mathrm{N/C}.A strength error in the compensator introduces direction error as well as longitudinal error.

Physical check

Vector superposition applies component by component; adding field magnitudes would not solve the compensation problem.