A point charge sits at the centre of a sphere of radius

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Vector calculus and field geometry · Vector calculus and field geometry

Mathematical Methods for Physics, Ch Line, surface and volume integrals

Problem

A point charge q=3.0 nCq = 3.0\,\mathrm{nC} sits at the centre of a sphere of radius R=0.050 mR = 0.050\,\mathrm{m}. Someone computes the flux of the charge's field through the sphere and writes: "The area element on this closed sphere is dS=−R2sin⁡θ dθ dϕ r^\mathrm{d}\mathbf{S} = -R^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi\,\hat{\mathbf{r}}, because the charge sits inside and the element should point towards it. Then E⋅dS=q4πϵ0R2r^⋅(−R2sin⁡θ dθ dϕ r^)=−q4πϵ0sin⁡θ dθ dϕ,\mathbf{E}\cdot\mathrm{d}\mathbf{S} = \frac{q}{4\pi\epsilon_0R^2}\hat{\mathbf{r}}\cdot\left(-R^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi\,\hat{\mathbf{r}}\right) = -\frac{q}{4\pi\epsilon_0}\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi, and integrating θ\theta from 00 to π\pi and ϕ\phi from 00 to 2π2\pi gives Φ=−q/ϵ0=−340 V m\Phi = -q/\epsilon_0 = -340\,\mathrm{V\,m}." ((a)) Find the error in this working. ((b)) Give the correct value of the flux. ((c)) State one check that confirms the corrected sign and the radius-independence of the result.

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