Q7.49P (a) Referring to Prob. 5.52(a) a... [FREE SOLUTION]

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Introduction to Electrodynamics

David J. Griffiths

Physics

4th edition Edition 路 613 pages

ISBN: 9780321856562

Chapter 7: Q7.49P (page 348)

(a) Referring to Prob. 5.52(a) and Eq. 7.18, show that
$E = - \frac{鈭� A}{鈭� t}$ (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 $蟽$ . It spins about a fixed axis at an angular velocity $蝇 (t)$ 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 $B$ . Refer to Ex. 5.11.]

Short Answer

Expert verified

(a) The required equation is $\frac{鈭� A}{鈭� t} = 鈭� E$ proved and also represents by the Faraday鈥檚 law.

(b) The electrical field inside the sphere is $鈭� \frac{渭_{0} R 蟽 \overset{路}{蝇}}{3} r \sin 胃 \hat{蠒}$ and outside the sphere is $\frac{蟽 R^{2}}{r^{2}} (鈭� \frac{渭_{0} R^{2} \overset{路}{蝇}}{3 r} r \sin 胃 \hat{蠒} + \frac{1}{蔚_{0}} \hat{r})$ .

Step by step solution

Write the given data from the question.

Th radius of the spherical shell is $R$ .

The uniform surface charge is $蟽$ .

The angular velocity is $蝇 (t)$ .

Determine the formula to show the equation 7.66, calculate the electric fired inside and outside the sphere.

$A = \frac{1}{4 蟺} 鈭� \frac{B 脳 \hat{r}}{r^{2}} d 蟿$ 鈥︹€� (1)

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

$E = - \frac{1}{4 蟺} \frac{鈭�}{鈭� t} [鈭� \frac{B 脳 \hat{r}}{r^{2}} d 蟿]$ 鈥︹€� (2)

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

$E_{o u t} = \frac{1}{4 蟺蔚_{0}} \frac{Q}{r^{2}} \hat{r}$ 鈥︹€� (3)

The expression for uniform surface charge is given as follows.

$蟽 = \frac{Q}{4 蟺 R^{2}}$ 鈥︹€� (4)

Show the expression E=-∂A∂t.

(a)

Calculate the value of $\frac{鈭� A}{d t}$ .

$\begin{array}{l} \frac{鈭� A}{鈭� t} = \frac{鈭�}{鈭� t} [\frac{1}{4 蟺} 鈭� \frac{B 脳 \hat{r}}{r^{2}} d 蟿] \\ \frac{鈭� A}{鈭� t} = \frac{1}{4 蟺} \frac{鈭�}{鈭� t} 鈭� \frac{B 脳 \hat{r}}{r^{2}} d 蟿 \end{array}$

Substitute $鈭� E$ for $\frac{1}{4 蟺} \frac{鈭�}{鈭� t} 鈭� \frac{B 脳 \hat{r}}{r^{2}} d 蟿$ into above equation.

$\frac{鈭� A}{鈭� t} = 鈭� E$

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

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

$\begin{array}{l} (鈭� 脳 \frac{鈭� A}{鈭� t}) = 鈭� 鈭� 脳 E \\ 鈭� 脳 E = 鈭� (鈭� 脳 \frac{鈭� A}{鈭� t}) \\ 鈭� 脳 E = 鈭� \frac{鈭�}{鈭� t} (鈭� 脳 A) \end{array}$

Substitute $B$ for $鈭� 脳 A$ into above equation.

$鈭� 脳 E = 鈭� \frac{鈭� B}{鈭� t}$

Hence the required equation $\frac{鈭� A}{鈭� t} = 鈭� E$ is proved and also represents by the Faraday鈥檚 law.

Calculate the electric fired inside and outside the sphere.

(b)

The Coulombs field inside the sphere is zero.

Recall the equation (4)

$\begin{array}{l} 蟽 = \frac{Q}{4 蟺 R^{2}} \\ Q = 蟽 (4 蟺 R^{2}) \end{array}$

Calculate the electrical field outside the sphere.

Substitute $蟽 (4 蟺 R^{2})$ for $Q$ into equation (3).

$\begin{array}{l} E = \frac{1}{4 蟺蔚_{0}} \frac{蟽 (4 蟺 R^{2})}{r^{2}} \hat{r} \\ E = \frac{蟽 R^{2}}{蔚_{0} r^{2}} \hat{r} \end{array}$

The Faraday鈥檚 equation.

$E = 鈭� \frac{鈭� A}{鈭� t}$ 鈥︹€� (6)

The vector potential can be written as,

$\begin{array}{l} A (r) = {\begin{cases} \frac{渭_{0} R 蟽}{3} (蝇脳 r) 鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌� r < R \\ \frac{渭_{0} R^{4} 蟽}{3 r^{3}} (蝇脳 r) 鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌� r > R \end{cases} \\ A (r) = {\begin{cases} \frac{渭_{0} R 蟽}{3} 蝇 r \sin 胃 \hat{蠒} 鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌� r < R \\ \frac{渭_{0} R^{4} 蟽}{3 r^{3}} 蝇 r \sin 胃 \hat{蠒} 鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌� r > R \end{cases} \end{array}$

Calculate the electric field inside the sphere.

Substitute $\frac{渭_{0} R 蟽}{3} (蝇脳 r)$ for $A$ into equation (6).

$\begin{array}{l} E_{i n s i d e} = 鈭� \frac{鈭�}{鈭� t} (\frac{渭_{0} R 蟽}{3} 蝇 r \sin 胃 \hat{蠒}) \\ E_{i n s i d e} = 鈭� \frac{渭_{0} R 蟽}{3} r \sin 胃 \hat{蠒} \frac{鈭� 蝇}{鈭� t} \end{array}$

Substitute $\overset{路}{蝇}$ for $\frac{鈭� 蝇}{鈭� t}$ into above equation.

$\begin{array}{l} E_{i n s i d e} = 鈭� \frac{渭_{0} R 蟽}{3} r \sin 胃 \hat{蠒} 脳 \overset{路}{蝇} \\ E_{i n s i d e} = 鈭� \frac{渭_{0} R 蟽 \overset{路}{蝇}}{3} r \sin 胃 \hat{蠒} \end{array}$

Hence, the electrical field inside the sphere is $鈭� \frac{渭_{0} R 蟽 \overset{路}{蝇}}{3} r \sin 胃 \hat{蠒}$ .

Calculate the electric field outside the sphere.

Substitute $\frac{渭_{0} R^{4} 蟽}{3 r^{3}} 蝇 r \sin 胃 \hat{蠒}$ for $A$ into equation (6).

$\begin{array}{l} {E^{'}}_{o u t s i d e} = 鈭� \frac{鈭�}{鈭� t} (\frac{渭_{0} R^{4} 蟽}{3 r^{3}} 蝇 r \sin 胃 \hat{蠒}) \\ {E^{'}}_{o u t s i d e} = \frac{渭_{0} R^{4} 蟽}{3 r^{3}} r \sin 胃 \hat{蠒} \frac{鈭� 蝇}{鈭� t} \end{array}$

Substitute $\overset{路}{蝇}$ for $\frac{鈭� 蝇}{鈭� t}$ into above equation.

$\begin{array}{l} {E^{'}}_{o u t s i d e} = 鈭� \frac{渭_{0} R^{4} 蟽}{3 r^{3}} r \sin 胃 \hat{蠒} 脳 \overset{路}{蝇} \\ {E^{'}}_{o u t s i d e} = 鈭� \frac{渭_{0} R^{4} 蟽 \overset{路}{蝇}}{3 r^{3}} r \sin 胃 \hat{蠒} \end{array}$

The total electric field outside the sphere is given by,

$E = {E^{'}}_{o u t s i d e} + E_{o u t}$

Substitute $鈭� \frac{渭_{0} R^{4} 蟽 \overset{路}{蝇}}{3 r^{3}} r \sin 胃 \hat{蠒}$ for ${E^{'}}_{o u t s i d e}$ and $\frac{蟽 R^{2}}{蔚_{0} r^{2}} \hat{r}$ for $E$ into above equation.

$\begin{array}{l} E = 鈭� \frac{渭_{0} R^{4} 蟽 \overset{路}{蝇}}{3 r^{3}} r \sin 胃 \hat{蠒} + \frac{蟽 R^{2}}{蔚_{0} r^{2}} \hat{r} \\ E = \frac{蟽 R^{2}}{r^{2}} (鈭� \frac{渭_{0} R^{2} \overset{路}{蝇}}{3 r} r \sin 胃 \hat{蠒} + \frac{1}{蔚_{0}} \hat{r}) \end{array}$

Hence, electric field outside the sphere is $\frac{蟽 R^{2}}{r^{2}} (鈭� \frac{渭_{0} R^{2} \overset{路}{蝇}}{3 r} r \sin 胃 \hat{蠒} + \frac{1}{蔚_{0}} \hat{r})$ .