Gauss' Law and Coulomb's Law — Question 10

Section 23-4 · Physics pages 3710–3711

Question 10 A centered charge is not generally normal to an ellipsoidal surface A positive point charge q is at the center of the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1. A local normal is parallel to (x/a2, y/b2, z/c2). At P = (a/ √ 2, b/ √ 2, 0) take a = 2b. ˆr ˆn q Equatorial section of the ellipsoid; the two direction arrows are schematic. (a) Calculate the cosine of the angle between the radial Coulomb field and the outward normal at P. (b) Find the flux through the whole ellipsoid, and state why Gauss law does not justify replacing the field by one constant scalar over its surface.
Show solutionHide solution
Question 10 - Solution A centered charge is not generally normal to an ellipsoidal surface ˆr ˆn q (a) Method and calculation ˆr = (2, 1, 0) √ 5 , ˆn = (1, 2, 0) √ 5 . cos γ = ˆr · ˆn = 4 5. (b) Interpretation and further calculation Φellipsoid = q/ϵ0. dΦ = kq r2 cos γ dA. Distance r and projection factor vary across the surface. The enclosed-charge result is exact even though the usual spherical shortcut is unavailable. Physical check For a sphere with a = b = c, the normal becomes radial everywhere and the angle reduces to zero.

Question 10

A centered charge is not generally normal to an ellipsoidal surface

A positive point charge q is at the center of the ellipsoid x2/a2+y2/b2+z2/c2=1x^2/a^2+y^2/b^2+z^2/c^2=1. A local normal is parallel to (x/a2,y/b2,z/c2)(x/a^2,y/b^2,z/c^2). At P=(a/2,b/2,0)P=(a/\sqrt2,b/\sqrt2,0) take a=2ba=2b.

Diagram shown in the original SVG worksheet.

  1. Calculate the cosine of the angle between the radial Coulomb field and the outward normal at P.

  2. Find the flux through the whole ellipsoid, and state why Gauss law does not justify replacing the field by one constant scalar over its surface.

Question 10 - Solution

A centered charge is not generally normal to an ellipsoidal surface

Diagram shown in the original SVG worksheet.

(a) Method and calculation

𝒓̂=(2,1,0)5,𝒏̂=(1,2,0)5.\hat{\mathbf r}=\frac{(2,1,0)}{\sqrt5},\quad\hat{\mathbf n}=\frac{(1,2,0)}{\sqrt5}.cos⁡γ=𝒓̂⋅𝒏̂=45.\cos\gamma=\hat{\mathbf r}\cdot\hat{\mathbf n}=\frac45.

(b) Interpretation and further calculation

Φellipsoid=q/ϵ0.\Phi_{\mathrm{ellipsoid}}=q/\epsilon_0.dΦ=kqr2cos⁡γdA.d\Phi=\frac{kq}{r^2}\cos\gamma\,dA.Distance r and projection factor vary across the surface. The enclosed-charge result is exact even though the usual spherical shortcut is unavailable.

Physical check

For a sphere with a=b=ca=b=c, the normal becomes radial everywhere and the angle reduces to zero.