Equipotential Surfaces — Question 6

Section 24-3 · Physics pages 3842–3843

Question 6 A circular equipotential requires a constraint to guide a moving particle A positive source Q creates V = kQ/r. A positive particle q of mass m moves at constant speed v around the circle r=R, held there by an ideal radial guide. Q q v Schematic; labels, not drawing lengths, specify the data. (a) Find electric work over any circular arc and the electric force. (b) Find the required guide force, specifying inward as positive for that calculation.
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Question 6 - Solution A circular equipotential requires a constraint to guide a moving particle Q q v (a) Method and calculation ∆V = 0 ⇒We = −q∆V = 0. Fe = kQq R2 ˆr outward. (b) Interpretation and further calculation Fguide,in −kQq R2 = mv2 R . Fguide,in = mv2 R + kQq R2 . Physical check An equipotential is a geometric locus, not a free trajectory; the electric force here points away from the required centripetal acceleration.

Question 6

A circular equipotential requires a constraint to guide a moving particle

A positive source Q creates V=kQ/rV=kQ/r. A positive particle q of mass m moves at constant speed v around the circle r=R, held there by an ideal radial guide.

Diagram shown in the original SVG worksheet.

  1. Find electric work over any circular arc and the electric force.

  2. Find the required guide force, specifying inward as positive for that calculation.

Question 6 - Solution

A circular equipotential requires a constraint to guide a moving particle

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ΔV=0⇒We=−qΔV=0.\Delta V=0\Rightarrow W_e=-q\Delta V=0.𝑭e=kQqR2𝒓̂outward.\mathbf F_e=\frac{kQq}{R^2}\hat{\mathbf r}\quad\text{outward}.

(b) Interpretation and further calculation

Fguide,in−kQqR2=mv2R.F_{\mathrm{guide,in}}-\frac{kQq}{R^2}=\frac{mv^2}{R}.Fguide,in=mv2R+kQqR2.F_{\mathrm{guide,in}}=\frac{mv^2}{R}+\frac{kQq}{R^2}.

Physical check

An equipotential is a geometric locus, not a free trajectory; the electric force here points away from the required centripetal acceleration.