Potential Due to a Continuous Charge Distribution — Question 2

Section 24-8 · Physics pages 3934–3935

Question 2 An angular charge ripple is invisible to every axial potential measurement A ring of radius R in the xy plane has line density λ(ϕ) = λ0[1 + η cos ϕ] with λ0 > 0 and |η| < 1. Use infinity reference. x ring Schematic; labels, not drawing lengths, specify the data. (a) Find its total charge and V at every point on the z axis. (b) Find the electric field at the center by direct vector integration and explain what axial potential data fail to determine.
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Question 2 - Solution An angular charge ripple is invisible to every axial potential measurement x ring (a) Method and calculation Q = ∫2π 0 λ(ϕ)R dϕ = 2πRλ0. V (0, 0, z) = 2πkRλ0 √ R2 + z2 , independent of η. (b) Interpretation and further calculation E(0) = −kλ0 R ∫2π 0 (1 + η cos ϕ)(cos ϕˆx + sin ϕˆy)dϕ. E(0) = −πkλ0η R ˆx. Physical check Values of V on one axis determine its axial derivative but do not determine transverse derivatives offthat axis.

Question 2

An angular charge ripple is invisible to every axial potential measurement

A ring of radius R in the xy plane has line density λ(ϕ)=λ0[1+ηcos⁡ϕ]\lambda(\phi)=\lambda_0[1+\eta\cos\phi] with λ0>0\lambda_0>0 and |η|<1|\eta|<1. Use infinity reference.

Diagram shown in the original SVG worksheet.

  1. Find its total charge and V at every point on the z axis.

  2. Find the electric field at the center by direct vector integration and explain what axial potential data fail to determine.

Question 2 - Solution

An angular charge ripple is invisible to every axial potential measurement

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Q=∫02πλ(ϕ)Rdϕ=2πRλ0.Q=\int_0^{2\pi}\lambda(\phi)R\,d\phi=2\pi R\lambda_0.V(0,0,z)=2πkRλ0R2+z2,V(0,0,z)=\frac{2\pi kR\lambda_0}{\sqrt{R^2+z^2}},independent of η\eta.

(b) Interpretation and further calculation

𝑬(0)=−kλ0R∫02π(1+ηcos⁡ϕ)(cos⁡ϕ𝒙̂+sin⁡ϕ𝒚̂)dϕ.\mathbf E(0)=-\frac{k\lambda_0}{R}\int_0^{2\pi}(1+\eta\cos\phi)(\cos\phi\hat{\mathbf x}+\sin\phi\hat{\mathbf y})d\phi.𝑬(0)=−πkλ0ηR𝒙̂.\mathbf E(0)=-\frac{\pi k\lambda_0\eta}{R}\hat{\mathbf x}.

Physical check

Values of V on one axis determine its axial derivative but do not determine transverse derivatives off that axis.