Question

Question: To show: The flux of F across a sphere\(S\)with center the origin is independent of the radius of\(S\).

Short answer

The flux is \(4\pi c\) and hence independent of \(R\)

Step-by-step solution

Concept of Flux and Formula Used.

Any impact that appears to pass or travel through a surface or substance is referred to as flux. Flux is a concept found in applied mathematics and vector calculus that has numerous physics applications.

Given:

The inverse square field is given by \({\rm{F}}({\rm{r}}) = \frac{{c{\rm{r}}}}{{|{\rm{r}}{|^3}}}.......(1)\)

The radius is given by \({\rm{r}} = x{\rm{i}} + y{\rm{j}} + z{\rm{k}}\)

Formula used:

Calculate the Flux.

The parameterisation for the sphere is

\({\bf{r}}(\theta ,\phi ) = R\sin \phi \cos \theta {\bf{i}} + R\sin \phi \sin \theta {\bf{j}} + R\cos \phi {\bf{k}}\)

Where \(\theta \in (0,2\pi )\) and \(f \in (0,\pi )\)

\(\begin{array}{l}{{\bf{r}}_\theta } = - R\sin f\sin \theta {\bf{i}} + R\sin f\cos \theta {\bf{j}}\\{{\bf{r}}_f} = R\cos f\cos \theta {\bf{i}} + R\cos f\sin \theta {\bf{j}} - R\sin f{\bf{k}}\\{{\bf{r}}_\theta } \times {{\bf{r}}_f} = \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{ - R\sin f\sin \theta }&{R\sin f\cos \theta }&0\\{R\cos f\cos \theta }&{R\cos f\sin \theta }&{ - R\sin f}\end{array}} \right|\\{{\bf{r}}_\theta } \times {{\bf{r}}_f} = {\bf{i}}\left| {\begin{array}{*{20}{c}}{R\sin f\cos \theta }&0\\{R\cos f\sin \theta }&{ - R\sin f}\end{array}} \right| - {\bf{j}}\left| {\begin{array}{*{20}{c}}{ - R\sin f\sin \theta }&0\\{R\cos f\cos \theta }&{ - R\sin f}\end{array}} \right| + {\bf{k}}\left| {\begin{array}{*{20}{c}}{ - R\sin f\sin \theta }&{R\sin f\cos \theta }\\{R\cos f\cos \theta }&{R\cos f\sin \theta }\end{array}} \right|\\{{\bf{r}}_\theta } \times {{\bf{r}}_f} = - {R^2}{\sin ^2}f\cos \theta {\bf{i}} - {R^2}{\sin ^2}f\sin \theta {\bf{j}} - {R^2}\sin f\cos f{\bf{k}}\end{array}\)

Note that\({{\bf{r}}_\theta } \times {{\bf{r}}_f}\)is pointing towards the origin, therefore

\(\begin{array}{l}d{\bf{S}} = - \left( {{{\bf{r}}_\theta } \times {{\bf{r}}_f}} \right)dA\\ = \left( {{R^2}{{\sin }^2}f\cos \theta {\bf{i}} + {R^2}{{\sin }^2}f\sin \theta {\bf{j}} + {R^2}\sin f\cos f{\bf{k}}} \right)dAd{\bf{S}}\\ = R\sin f(R\sin f\cos \theta {\bf{i}} + R\sin f\sin \theta {\bf{j}} + R\cos f{\bf{k}})dAd{\bf{S}}\\ = (R\sin f){\bf{r}}dA\end{array}\)

The inverse square field is given by\[{\rm{F}}({\rm{r}}) = \frac{{c{\rm{r}}}}{{|{\rm{r}}{|^3}}}........(1)\]

The radius is given by\[{\rm{r}} = x{\rm{i}} + y{\rm{j}} + z{\rm{k}}\]

Therefore,

The flux is \(4\pi c\) and hence independent of \(R\)