Potential Due to a Group of Point Charges — Question 2

Section 24-6 · Physics pages 3894–3895

Question 2 Unequal opposite sources produce two axial zero-potential points Charges +Q at x=0 and −2Q at x=d have Q,d>0. Use zero potential at infinity and exclude the source points. x −d +Q d/3 −2Q Schematic; labels, not drawing lengths, specify the data. (a) Find all finite points on the x axis where V=0. (b) Find the direction and magnitude of the field at each zero.
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Question 2 - Solution Unequal opposite sources produce two axial zero-potential points x −d +Q d/3 −2Q (a) Method and calculation 1 |x| = 2 |x −d| ⇒|x −d| = 2|x|. (x −d)2 = 4x2 ⇒x = d/3 or x = −d. (b) Interpretation and further calculation E(d/3) = ( 9 + 9 2 ) kQ d2 ˆx = 27kQ 2d2 ˆx. E(−d) = ( −1 + 1 2 ) kQ d2 ˆx = −kQ 2d2 ˆx. Physical check Both roots satisfy the unsquared distance equation; neither has a vanishing field.

Question 2

Unequal opposite sources produce two axial zero-potential points

Charges +Q at x=0 and −2Q-2Q at x=d have Q,d>0. Use zero potential at infinity and exclude the source points.

Diagram shown in the original SVG worksheet.

  1. Find all finite points on the x axis where V=0.

  2. Find the direction and magnitude of the field at each zero.

Question 2 - Solution

Unequal opposite sources produce two axial zero-potential points

Diagram shown in the original SVG worksheet.

(a) Method and calculation

1|x|=2|x−d|⇒|x−d|=2|x|.\frac1{|x|}=\frac2{|x-d|}\Rightarrow|x-d|=2|x|.(x−d)2=4x2⇒x=d/3orx=−d.(x-d)^2=4x^2\Rightarrow x=d/3\ \text{or}\ x=-d.

(b) Interpretation and further calculation

𝑬(d/3)=(9+92)kQd2𝒙̂=27kQ2d2𝒙̂.\mathbf E(d/3)=\left(9+\frac92\right)\frac{kQ}{d^2}\hat{\mathbf x}=\frac{27kQ}{2d^2}\hat{\mathbf x}.𝑬(−d)=(−1+12)kQd2𝒙̂=−kQ2d2𝒙̂.\mathbf E(-d)=\left(-1+\frac12\right)\frac{kQ}{d^2}\hat{\mathbf x}=-\frac{kQ}{2d^2}\hat{\mathbf x}.

Physical check

Both roots satisfy the unsquared distance equation; neither has a vanishing field.