Potential Due to an Electric Dipole β€” Question 4

Section 24-7 Β· Physics pages 3918–3919

Question 4 Voltage measurements along two directions reconstruct a planar dipole An ideal dipole has unknown moment p = pxΛ†x + pyΛ†y. At radius R, measured potentials on the positive x and y axes are Vx = 3V0 and Vy = 4V0, where V0>0. x y p Schematic; labels, not drawing lengths, specify the data. (a) Find the two moment components, its magnitude and its orientation. (b) Predict the voltage at radius 2R along (Λ†x + Λ†y)/ √ 2.
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Question 4 - Solution Voltage measurements along two directions reconstruct a planar dipole x y p (a) Method and calculation px = 3V0R2/k, py = 4V0R2/k. p = 5V0R2/k, ΞΈp = tanβˆ’1(4/3) = 53.1301β—¦. (b) Interpretation and further calculation p Β· Λ†r = 7V0R2 k √ 2 . V (2R) = k 4R2 7V0R2 k √ 2 = 7 4 √ 2V0. Physical check A third independent directional measurement would be needed if a nonzero out-of-plane component were allowed.

Question 4

Voltage measurements along two directions reconstruct a planar dipole

An ideal dipole has unknown moment 𝒑=px𝒙̂+pyπ’šΜ‚\mathbf p=p_x\hat{\mathbf x}+p_y\hat{\mathbf y}. At radius R, measured potentials on the positive x and y axes are Vx=3V0V_x=3V_0 and Vy=4V0V_y=4V_0, where V0>0.

Diagram shown in the original SVG worksheet.

  1. Find the two moment components, its magnitude and its orientation.

  2. Predict the voltage at radius 2R along (𝒙̂+π’šΜ‚)/2(\hat{\mathbf x}+\hat{\mathbf y})/\sqrt2.

Question 4 - Solution

Voltage measurements along two directions reconstruct a planar dipole

Diagram shown in the original SVG worksheet.

(a) Method and calculation

px=3V0R2/k,py=4V0R2/k.p_x=3V_0R^2/k,\quad p_y=4V_0R^2/k.p=5V0R2/k,ΞΈp=tan⁡βˆ’1(4/3)=53.1301∘.p=5V_0R^2/k,\quad\theta_p=\tan^{-1}(4/3)=53.1301^\circ.

(b) Interpretation and further calculation

𝒑⋅𝒓̂=7V0R2k2.\mathbf p\cdot\hat{\mathbf r}=\frac{7V_0R^2}{k\sqrt2}.V(2R)=k4R27V0R2k2=742V0.V(2R)=\frac{k}{4R^2}\frac{7V_0R^2}{k\sqrt2}=\frac7{4\sqrt2}V_0.

Physical check

A third independent directional measurement would be needed if a nonzero out-of-plane component were allowed.