Question 4.12: Given the potential V = 10/r^2 sinθ cosφ, (a) Find the elect...

(a)

D=\varepsilon_{o}E

But

E=-\nabla V=-\left[\frac{\partial V}{\partial r}a_{r}+ \frac{1}{r}\frac{\partial V}{\partial \theta}a_{\theta}+\frac{1}{r\sin\theta}\frac{\partial V}{\partial \phi}a_{\phi} \right]

=\frac{20}{r^{3}}\sin\theta\cos\phi a_{r}-\frac{10}{r^{3}}\cos\theta\cos\phi a_{\theta}+\frac{10}{r^{3}}\sin\phi a_{\phi}

At (2, \pi/2, 0)

D=\varepsilon_{o}E\left(r=2,\theta=\pi/2,\phi=0\right)=\varepsilon_{o}\left(\frac{20}{8}a_{r}-0a_{\theta}+0a_{\phi}\right)

=2.5\varepsilon_{o}a_{r}C/m^{2}=22.1a_{r}pC/m^{2}

(b) The work done can be found in two ways, using either E or V.

Method 1:

W=-Q\int_{L}E\cdot dl

or

-\frac{W}{Q}=\int_{L}E\cdot dl

and because the electrostatic field is conservative, the path of integration is immaterial. Hence the work done in moving Q from A(1, 30°, 120°) to B(4, 90°, 60°) is the same as that in moving Q from A to A^{'}, from A^{'} to B^{'}, and from B^{'} to B, where

\begin{matrix} A\left(1,30^{\circ},120^{\circ}\right) & & B\left(4,90^{\circ},60^{\circ}\right) \\ \downarrow dl=dra_{r} & dl=rd\theta a_{\theta} & \uparrow dl=r\sin\theta d\phi a_{\phi} \\A^{'}\left(4,30^{\circ},120^{\circ}\right) & \rightarrow & B^{'}\left(4,90^{\circ},120^{\circ}\right) \end{matrix}

That is, instead of being moved directly from A to B, Q is moved from A\rightarrow A^{'},A^{'}\rightarrow B^{'},B^{'}\rightarrow B, so that only one variable is changed at a time. This makes the line integral much easier to evaluate. Thus

\frac{-W}{Q}=-\frac{1}{Q}\left(W_{AA^{'}}+W_{A^{'}B^{'}}+W_{B^{'}B}\right)

=\left(\int_{AA^{'}}+\int_{A^{'}B^{'}}+\int_{B^{'}B}\right)E\cdot dl

=\int_{r=1}^{4}\frac{20\sin\theta\cos\phi}{r^{3}}dr|_{\theta=30^{\circ},\phi=120^{\circ}}+\int_{\theta=30^{\circ}}^{90^{\circ}}\frac{-10\cos\theta\cos\phi}{r^{3}}rd\theta|_{r=4,\phi=120^{\circ}}+\int_{\phi=120^{\circ}}^{60^{\circ}}\frac{10\sin\phi}{r^{3}}r\sin\theta d\phi|_{r=4,\theta=90^{\circ}}

=20\left(\frac{1}{2} \right)\left(\frac{-1}{2} \right) \left[-\frac{1}{2r^{2}}|_{r=1}^{4} \right]- \frac{10}{16}\frac{\left(-1\right) }{2}\sin\theta|_{30^{\circ}}^{90^{\circ}}+\frac{10}{16}\left(1\right)\left[-\cos\phi|_{120^{\circ}}^{60^{\circ}}\right]

-\frac{W}{Q}=\frac{-75}{32}+\frac{5}{32}-\frac{10}{16}

or

W=\frac{45}{16}Q=28.125\mu J

Method 2:
Since V is known, this method is much easier.

W=-Q\int_{A}^{B}E\cdot dl=QV_{AB}=Q\left(V_{B}-V_{A}\right)

=10\left(\frac{10}{16}\sin 90^{\circ}\cos 60^{\circ} -\frac{10}{1}\sin 30^{\circ}\cos120^{\circ} \right)\cdot 10^{-6}

=10\left(\frac{10}{32} -\frac{-5}{2} \right)10^{-6}=28.125\mu J  as obtained before