Potential Due to a Point Charge — Question 8

Section 24-5 · Physics pages 3886–3887

Question 8 A chord through a spherical equipotential passes through higher potential A positive point charge Q sits at the center of a sphere of radius R. Endpoints A and B on the sphere subtend angle 2θ, where 0 < θ < π/2. A straight guide connects them. Q A B Schematic; labels, not drawing lengths, specify the data. (a) Find the nearest distance of the chord to Q and the maximum potential along the chord. (b) Find the minimum initial energy required energetically for a positive charge q to traverse the chord from A, and compare with an arc at r=R.
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Question 8 - Solution A chord through a spherical equipotential passes through higher potential Q A B (a) Method and calculation rmin = R cos θ. Vmax = kQ R cos θ, VA = VB = kQ R . (b) Interpretation and further calculation Kthreshold = q(Vmax −VA) = kQq R (sec θ −1). Circular arc: ∆V = 0 everywhere, ∆K = 0. Finite-time chord passage requires energy above the smooth-barrier threshold. Physical check Equal endpoint voltage hides the chord barrier; the guide supplies transverse constraint forces but no work.

Question 8

A chord through a spherical equipotential passes through higher potential

A positive point charge Q sits at the center of a sphere of radius R. Endpoints A and B on the sphere subtend angle 2θ2\theta, where 0<θ<π/20<\theta<\pi/2. A straight guide connects them.

Diagram shown in the original SVG worksheet.

  1. Find the nearest distance of the chord to Q and the maximum potential along the chord.

  2. Find the minimum initial energy required energetically for a positive charge q to traverse the chord from A, and compare with an arc at r=R.

Question 8 - Solution

A chord through a spherical equipotential passes through higher potential

Diagram shown in the original SVG worksheet.

(a) Method and calculation

rmin=Rcos⁡θ.r_{\min}=R\cos\theta.Vmax=kQRcos⁡θ,VA=VB=kQR.V_{\max}=\frac{kQ}{R\cos\theta},\qquad V_A=V_B=\frac{kQ}{R}.

(b) Interpretation and further calculation

Kthreshold=q(Vmax−VA)=kQqR(sec⁡θ−1).K_{\mathrm{threshold}}=q(V_{\max}-V_A)=\frac{kQq}{R}(\sec\theta-1).Circular arc:ΔV=0 everywhere,ΔK=0.\text{Circular arc:}\quad\Delta V=0\text{ everywhere},\quad\Delta K=0.Finite-time chord passage requires energy above the smooth-barrier threshold.

Physical check

Equal endpoint voltage hides the chord barrier; the guide supplies transverse constraint forces but no work.