(a) Referring to Prob. 5.52(a) and Eq. 7.18, show that

$E=-\frac{\partial A}{\partial t}$ (7.66) for Faraday-induced electric fields. Check this result by taking the divergence and curl of both sides.

(b) A spherical shell of radius$R$ carries a uniform surface charge $\sigma$. It spins about a fixed axis at an angular velocity $\omega \left(t\right)$that changes slowly with time. Find the electric field inside and outside the sphere. [Hint: There are two contributions here: the Coulomb field due to the charge, and the Faraday field due to the changing $\mathit{B}$. Refer to Ex. 5.11.]

Th radius of the spherical shell is $R$.

The uniform surface charge is $\sigma$.

The angular velocity is $\omega \left(t\right)$.

$A=\frac{1}{4\pi }\int \frac{B\times \widehat{r}}{r^2}d\tau$ …… (1)

The expression for the electric field by using the equation 7.18 and analogous to Bio-savart law is given as follows

$E=-\frac{1}{4\pi }\frac{\partial }{\partial t}[\int \frac{B\times \widehat{r}}{r^2}d\tau ]$ …… (2)

The expression to calculate the coulomb field outside the sphere is given as follows.

$E_{out}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}\widehat{r}$ …… (3)

The expression for uniform surface charge is given as follows.

$\sigma =\frac{Q}{4\pi R^2}$ …… (4)

(a)

Calculate the value of $\frac{\partial A}{dt}$.

$\begin{array}{l}\frac{\partial A}{\partial t}=\frac{\partial }{\partial t}[\frac{1}{4\pi }\int \frac{B\times \widehat{r}}{r^2}d\tau ]\\ \frac{\partial A}{\partial t}=\frac{1}{4\pi }\frac{\partial }{\partial t}\int \frac{B\times \widehat{r}}{r^2}d\tau \end{array}$

Substitute$-E$for $\frac{1}{4\pi }\frac{\partial }{\partial t}\int \frac{B\times \widehat{r}}{r^2}d\tau$into above equation.

$\frac{\partial A}{\partial t}=-E$

Hence, the equation$\frac{\partial A}{\partial t}=-E$is proved.

Take the divergence and curl of the both the sides of the above equation.

$\begin{array}{l}(\nabla \times \frac{\partial A}{\partial t})=-\nabla \times E\\ \nabla \times E=-(\nabla \times \frac{\partial A}{\partial t})\\ \nabla \times E=-\frac{\partial }{\partial t}(\nabla \times A)\end{array}$

Substitute$B$ for$\nabla \times A$ into above equation.

$\nabla \times E=-\frac{\partial B}{\partial t}$

Hence the required equation $\frac{\partial A}{\partial t}=-E$is proved and also represents by the Faraday’s law.

(b)

The Coulombs field inside the sphere is zero.

Recall the equation (4)

$\begin{array}{l}\sigma =\frac{Q}{4\pi R^2}\\ Q=\sigma \left(4\pi R^2\right)\end{array}$

Calculate the electrical field outside the sphere.

Substitute$\sigma \left(4\pi R^2\right)$for $Q$into equation (3).

$\begin{array}{l}E=\frac{1}{4\pi {\epsilon }_0}\frac{\sigma \left(4\pi R^2\right)}{r^2}\widehat{r}\\ E=\frac{\sigma R^2}{{\epsilon }_0{r}^2}\widehat{r}\end{array}$

The Faraday’s equation.

$E=-\frac{\partial A}{\partial t}$ …… (6)

The vector potential can be written as,

$\begin{array}{l}A\left(r\right)=\{\begin{array}{l}\frac{{\mu }_{0}R\sigma }{3}(\omega \times r)r<R\\ \frac{{\mu }_0{R}^4\sigma }{3r^3}(\omega \times r)r>R\end{array}\\ A\left(r\right)=\{\begin{array}{l}\frac{{\mu }_{0}R\sigma }{3}\omega r\sin \theta \widehat{\varphi }r<R\\ \frac{{\mu }_0{R}^4\sigma }{3r^3}\omega r\sin \theta \widehat{\varphi }r>R\end{array}\end{array}$

Calculate the electric field inside the sphere.

Substitute $\frac{{\mu }_{0}R\sigma }{3}(\omega \times r)$for$A$into equation (6).

$\begin{array}{l}{E}_{inside}=-\frac{\partial }{\partial t}\left(\frac{{\mu }_{0}R\sigma }{3}\omega r\sin \theta \widehat{\varphi }\right)\\ E_{inside}=-\frac{{\mu }_{0}R\sigma }{3}r\sin \theta \widehat{\varphi }\frac{\partial \omega }{\partial t}\end{array}$

Substitute $\stackrel{·}{\omega }$for$\frac{\partial \omega }{\partial t}$into above equation.

$\begin{array}{l}{E}_{inside}=-\frac{{\mu }_{0}R\sigma }{3}r\sin \theta \widehat{\varphi }\times \stackrel{·}{\omega }\\ E_{inside}=-\frac{{\mu }_{0}R\sigma \stackrel{·}{\omega }}{3}r\sin \theta \widehat{\varphi }\end{array}$

Hence, the electrical field inside the sphere is $-\frac{{\mu }_{0}R\sigma \stackrel{·}{\omega }}{3}r\sin \theta \widehat{\varphi }$.

Calculate the electric field outside the sphere.

Substitute$\frac{{\mu }_0{R}^4\sigma }{3r^3}\omega r\sin \theta \widehat{\varphi }$ for$A$into equation (6).

$\begin{array}{l}{E^{\text{'}}}_{outside}=-\frac{\partial }{\partial t}\left(\frac{{\mu }_0{R}^4\sigma }{3r^3}\omega r\sin \theta \widehat{\varphi }\right)\\ {E^{\text{'}}}_{outside}=\frac{{\mu }_0{R}^4\sigma }{3r^3}r\sin \theta \widehat{\varphi }\frac{\partial \omega }{\partial t}\end{array}$

Substitute$\stackrel{·}{\omega }$ for $\frac{\partial \omega }{\partial t}$into above equation.

$\begin{array}{l}{E^{\text{'}}}_{outside}=-\frac{{\mu }_0{R}^4\sigma }{3r^3}r\sin \theta \widehat{\varphi }\times \stackrel{·}{\omega }\\ {E^{\text{'}}}_{outside}=-\frac{{\mu }_0{R}^4\sigma \stackrel{·}{\omega }}{3r^3}r\sin \theta \widehat{\varphi }\end{array}$

The total electric field outside the sphere is given by,

$E={E^{\text{'}}}_{outside}+E_{out}$

Substitute $-\frac{{\mu }_0{R}^4\sigma \stackrel{·}{\omega }}{3r^3}r\sin \theta \widehat{\varphi }$for${E^{\text{'}}}_{outside}$ and $\frac{\sigma R^2}{{\epsilon }_0{r}^2}\widehat{r}$for$E$ into above equation.

$\begin{array}{l}E=-\frac{{\mu }_0{R}^4\sigma \stackrel{·}{\omega }}{3r^3}r\sin \theta \widehat{\varphi }+\frac{\sigma R^2}{{\epsilon }_0{r}^2}\widehat{r}\\ E=\frac{\sigma R^2}{r^2}(-\frac{{\mu }_0{R}^2\stackrel{·}{\omega }}{3r}r\sin \theta \widehat{\varphi }+\frac{1}{{\epsilon }_0}\widehat{r})\end{array}$

Hence, electric field outside the sphere is$\frac{\sigma R^2}{r^2}(-\frac{{\mu }_0{R}^2\stackrel{·}{\omega }}{3r}r\sin \theta \widehat{\varphi }+\frac{1}{{\epsilon }_0}\widehat{r})$.