Potential Due to a Continuous Charge Distribution — Question 8

Section 24-8 · Physics pages 3946–3947

Question 8 A graded solid sphere has a center voltage sensitive to inner charge weighting A sphere of radius R has volume density ρ(r) = ρ0r/R with ρ0 > 0. Use zero potential at infinity. R Cross-section of concentric spherical surfaces; labels specify radii. (a) Find total charge Q and center potential by summing thin spherical shells. (b) Compare V(0) with its surface value and with a uniform solid sphere having the same Q and R; the uniform sphere has center potential 3kQ/(2R).
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Question 8 - Solution A graded solid sphere has a center voltage sensitive to inner charge weighting R (a) Method and calculation Q = 4π ∫R 0 ρ0 r Rr2dr = πρ0R3. V (0) = k ∫R 0 4πr2ρ(r)dr r = 4πkρ0R2 3 = 4kQ 3R . (b) Interpretation and further calculation V (R) = kQ/R, V (0)/V (R) = 4/3. Vgraded(0) Vuniform(0) = 4/3 3/2 = 8 9. Physical check Moving a larger fraction of fixed total charge outward lowers the center potential because each shell is weighted by inverse radius.

Question 8

A graded solid sphere has a center voltage sensitive to inner charge weighting

A sphere of radius R has volume density ρ(r)=ρ0r/R\rho(r)=\rho_0r/R with ρ0>0\rho_0>0. Use zero potential at infinity.

Diagram shown in the original SVG worksheet.

  1. Find total charge Q and center potential by summing thin spherical shells.

  2. Compare V(0) with its surface value and with a uniform solid sphere having the same Q and R; the uniform sphere has center potential 3kQ/(2R)3kQ/(2R).

Question 8 - Solution

A graded solid sphere has a center voltage sensitive to inner charge weighting

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Q=4π∫0Rρ0rRr2dr=πρ0R3.Q=4\pi\int_0^R\rho_0\frac rR r^2dr=\pi\rho_0R^3.V(0)=k∫0R4πr2ρ(r)drr=4πkρ0R23=4kQ3R.V(0)=k\int_0^R\frac{4\pi r^2\rho(r)dr}{r}=\frac{4\pi k\rho_0R^2}{3}=\frac{4kQ}{3R}.

(b) Interpretation and further calculation

V(R)=kQ/R,V(0)/V(R)=4/3.V(R)=kQ/R,\quad V(0)/V(R)=4/3.Vgraded(0)Vuniform(0)=4/33/2=89.\frac{V_{\mathrm{graded}}(0)}{V_{\mathrm{uniform}}(0)}=\frac{4/3}{3/2}=\frac89.

Physical check

Moving a larger fraction of fixed total charge outward lowers the center potential because each shell is weighted by inverse radius.