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Chapter 1: Vector Analysis Problem 1.1: Using the definitions in Eqs. 1.1 and 1.4, and appropriate diagrams, show that the dot product and cross product are distributive, a) when the three vectors are co-planar. b) in the general case.  Solution   Problem 1.2 Is the cross product associative? $$(\vec{A}\times \vec{B}) \times \vec{C} \overset{?}{=} \vec{A}\times (\vec{B} \times \vec{C})$$ If so, prove it; if not, provide a counterexample (the simpler the better). Solution   Problem 1.3 Find the angle between the body diagonals of a cube.  Solution Problem 1.4 Use the cross product to find the components of the unit vector $\hat{n}$ perpendicular to the shaded plane in Fig. 1.11. Solution   Problem 1.5 Prove the BAC-CAB rule by writing out both sides in component form. Solution Problem 1.6 Prove that $$[\vec{A}\times (\vec{B}\times \vec{C})]+[\vec{B}\times (\vec{C}\times \vec{A})]+[\vec{C}\times (\vec{A}\times \vec{B})] = 0$$ Under what conditions does $\vec{A}\times (\vec{B}\times \v
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Chapter 1 Vector Analysis: Problem 1.3

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Problem 1.3 Find the angle between the body diagonals of a cube.     If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.
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Chapter 1 Vector Analysis: Problem 1.5

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Problem 1.5 Prove the BAC-CAB rule by writing out both sides in component form.   If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.
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Chapter 1 Vector Analysis: Problem 1.7

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      If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.
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Chapter 1 Vector Analysis: Problem 1.6

Problem 1.6 Prove that $$[\vec{A}\times (\vec{B}\times \vec{C})]+[\vec{B}\times (\vec{C}\times \vec{A})]+[\vec{C}\times (\vec{A}\times \vec{B})] = 0$$ Under what conditions does $\vec{A}\times (\vec{B}\times \vec{C}) = (\vec{A}\times \vec{B})\times \vec{C}$? Solution: Note That: $$\vec{A}\times (\vec{B}\times \vec{C}) = (\vec{A}\cdot \vec{C})\vec{B} - (\vec{A}\cdot \vec{B})\vec{C}$$ Consider, $$[\vec{A}\times (\vec{B}\times \vec{C})]+[\vec{B}\times (\vec{C}\times \vec{A})]+[\vec{C}\times (\vec{A}\times \vec{B})]$$ $$ = [(\vec{A}\cdot \vec{C})\vec{B}-(\vec{A}\cdot \vec{B})\vec{C}] + [(\vec{B}\cdot \vec{A})\vec{C}-(\vec{B}\cdot \vec{C})\vec{A}]+ [(\vec{C}\cdot \vec{B})\vec{A}-(\vec{C}\cdot \vec{A})\vec{B}]$$ Using the commutative property of dot product i.e $\vec{A}\cdot \vec{B} = \vec{B}\cdot \vec{A}$   $$[\vec{A}\times (\vec{B}\times \vec{C})]+[\vec{B}\times (\vec{C}\times \vec{A})]+[\vec{C}\times (\vec{A}\times \vec{B})] = \vec{0}$$   Say $\vec{A}\times (\vec{B}\times \vec{C}) = (\vec
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Chapter 1 Vector Analysis: Problem 1.2

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Problem 1.2 Is the cross product associative? $$(\vec{A}\times \vec{B}) \times \vec{C} \overset{?}{=} \vec{A}\times (\vec{B} \times \vec{C})$$ If so, prove it; if not, provide a counterexample (the simpler the better). Solution: No, cross product is not associative. Counter Example Consider Three Non-Zero Vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$. Just to keep the example simple, let $\vec{C}$ be perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ $$\vec{A}\times \vec{B} = |A||B|\sin\theta \hat{n}$$ where $\theta$ is the angle between them and $\theta<90^{\circ}$ and $\hat{n}$ is the direction perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ i.e the direction of $\vec{C}$. Hence, the angle between the cross product of $\vec{A}$ and $\vec{B}$ and that of $\vec{C}$ is $0$. $$(\vec{A}\times \vec{B})\times \vec{C} = |A||B|\sin\theta\hat{n} \times \vec{C} = \vec{0}$$ Now Consider the plane containing $\vec{B}$ and $\vec{C}$. $$\vec{A}\times (\vec{B} \times \vec{C}
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Chapter 1 Vector Analysis: Problem 1.1

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Problem 1.1: Using the definitions in Eqs. 1.1 and 1.4, and appropriate diagrams, show that the dot product and cross product are distributive, a) when the three vectors are co-planar. b) in the general case. Solution:  Eq$^n$ 1.1 $$\vec{A}\cdot \vec{B} = |A||B| \cos\theta$$where $\theta$ is the angle between the vectors $\vec{A}$ and $\vec{B}$ Eq$^n$ 1.4 $$\vec{A}\times \vec{B} = |A||B| \sin\theta \hat{n}$$where $\theta$ is the angle between the vectors $\vec{A}$ and $\vec{B}$, and $\hat{n}$ is a unit vector normal to both $\vec{A}$ and $\vec{B}$. We have to show that: $$\vec{A}\cdot (\vec{B}+\vec{C}) = \vec{A}\cdot \vec{B}+\vec{A}\cdot \vec{C}$$ $$\vec{A}\times (\vec{B}+\vec{C}) = \vec{A}\times \vec{B}+\vec{A}\times \vec{C}$$ $a) $ When $\vec{A}$, $\vec{B}$ and $\vec{C}$ are co-planar.  $\theta_1$ and $\theta_2$ are the angles the vector $\vec{B}$ and $\vec{C}$ makes with the $\vec{A}$ respectively. To get $\vec{B}+\vec{C}$ put the tail of $\vec{C}$ on the head of $\vec{B}$.   Let $
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List of Questions and link to solutions

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Chapter 1: Vector Analysis Problem 1.1: Using the definitions in Eqs. 1.1 and 1.4, and appropriate diagrams, show that the dot product and cross product are distributive, a) when the three vectors are co-planar. b) in the general case.  Solution   Problem 1.2 Is the cross product associative? $$(\vec{A}\times \vec{B}) \times \vec{C} \overset{?}{=} \vec{A}\times (\vec{B} \times \vec{C})$$ If so, prove it; if not, provide a counterexample (the simpler the better). Solution   Problem 1.3 Find the angle between the body diagonals of a cube.  Solution Problem 1.4 Use the cross product to find the components of the unit vector $\hat{n}$ perpendicular to the shaded plane in Fig. 1.11. Solution   Problem 1.5 Prove the BAC-CAB rule by writing out both sides in component form. Solution Problem 1.6 Prove that $$[\vec{A}\times (\vec{B}\times \vec{C})]+[\vec{B}\times (\vec{C}\times \vec{A})]+[\vec{C}\times (\vec{A}\times \vec{B})] = 0$$ Under what conditions does $\vec{A}\times (\vec{B}\times \v
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Chapter 2 Electrostatics: Problem 2.38

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Problem 2.38: A metal sphere of radius $R$, carrying charge $q$, is surrounded by a thick concentric metal shell (inner radius $a$, outer radius $b$, as in Fig. 2.48). The shell carries no net charge. (a) Find the surface charge density $\sigma$ at $R$, at $a$, and at $b$. (b) Find the potential at the center, using infinity as the reference point. (c) Now the outer surface is touched to a grounding wire, which drains off charge and lowers its potential to zero (same as at infinity). How do your answers to (a) and (b) change?                       Solution: Create a spherical guassian surface with radius $r$ such that $a\le r\le b$.                       Since electric field inside a conductor is zero.  Hence the inner part of the shell with inner radius a will get induced with a charge $-q$. Now a make the shell electrically neutral, positive charge $+q$ will get induced on the outer surface of shell with outer radius $b$.    $a) $  $$\sigma_b = \frac{+q}{4\pi b^2}$$  $$\sigma_a=
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Chapter 2 Electrostatics: Problem 2.14

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Problem 2.14 Find the electric field inside a sphere that carries a charge density proportional to the distance from the origin, $ρ = kr$, for some constant $k$. [Hint: This charge density is not uniform, and you must integrate to get the enclosed charge.] Solution: Given, $\rho = kr$  Consider a spherical Gaussian surface of radius $r_0$ such that $r_0\le R$ where $R$ is the radius of the sphere.                        By Gauss Law, $$\oint \vec{E}\cdot \vec{da} = \frac{Q_{enc}}{\epsilon_0}$$ Since the charge distribution is radially symmetrical, the direction of the electric field will be radially outwards. $$\therefore E\times 4\pi {r_0}^2 = \frac{Q_{enc}}{\epsilon_0}$$ $$Q_{enc} = \int_{vol.}\rho \times d\tau$$ $$Q_{enc} = \int_{vol.}kr \times r^2 sin\theta dr d\theta d\phi$$ $$Q_{enc} = k\int_0^{r_0} r^3 dr \int_0^{\pi} sin\theta d\theta \int_0^{2\pi}d\phi$$ $$Q_{enc} = k\times \frac{r^4}{4} \times 2 \times 2\pi$$  $$Q_{enc} = k\pi {r_0}^4$$ $$\therefore E \times 4\pi r_0^2 =
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Chapter 2 Electrostatics: Problem 2.13

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Problem 2.13: Find the electric field a distance s from an infinitely long straight wire that carries a uniform line charge $\lambda$. Compare Eq. 2.9. Solution:                     Consider a Gaussian cylindrical surface of radius $s$ and length $h$.  Let the wire passes through the axis of the cylinder.  By Guass Law,  $$\oint \vec{E}\cdot \vec{da} = \frac{Q_{enc}}{\epsilon_0}$$  $\because $ Electric field through the plane surfaces of the cylinder is zero.  Also Electric field is directed outward and normal to the curved surface. $$\therefore E\times 2\pi s h= \frac{\lambda\times h}{\epsilon_0}$$  $$\therefore E= \frac{1}{2\pi \epsilon_0}\frac{\lambda}{s}$$   $$\therefore \vec{E}= \frac{1}{4\pi \epsilon_0}\frac{2\lambda}{s}\hat{s}$$      If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If
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Chapter 2 Electrostatics: Problem 2.12

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Problem 2.12 Use Gauss’s law to find the electric field inside a uniformly charged solid sphere (charge density $\rho$). Compare your answer to Prob. 2.8. Solution:                     Consider a gaussian spherical surface of radius $r$ where $r\ge R$  According to Gauss Law, $$\oint \vec{E}\cdot \vec{da} = \frac{Q_{enc}}{\epsilon_{0}} = \frac{\rho\times \frac{4}{3}\pi R^3}{\epsilon_0}$$  Since the charge distribution is symmetrical, $\vec{E}$ and $\vec{da}$ point in the same direction $\hat{r}$  $$E\times 4\pi r^2 = \frac{\rho\times \frac{4}{3}\pi R^3}{\epsilon_0}$$  $$\vec{E} = \frac{\rho\times R^3}{ 3 \epsilon_0 r^2}\hat{r}$$  Now, For $r< R, Q_{enc} = \rho \times \frac{4}{3}\pi r^3$  $$E\times 4\pi r^2 = \frac{\rho \times \frac{4}{3}\pi r^3}{\epsilon_0}$$  $$E = \frac{\rho \times r}{3\epsilon_0}\hat{r}$$ Hence, For $r< R, \vec{E}=\frac{\rho \times r}{3\epsilon_0}\hat{r}$  For $r\ge R, \vec{E}=\frac{\rho\times R^3}{ 3 \epsilon_0 r^2}\hat{r}$  If you have any doubt regardin
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Chapter 2 Electrostatics: Problem 2.11

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Problem 2.11: Use Gauss’s law to find the electric field inside and outside a spherical shell of radius R that carries a uniform surface charge density σ. Compare your answer to Prob. 2.7. Solution:                      Consider a gaussian spherical surface of radius $r$ where $r\ge R$. According to Gauss Law,  $$\oint \vec{E}\cdot \vec{da} = \frac{Q_{enc}}{\epsilon_{0}} = \frac{\sigma\times 4\pi R^2}{\epsilon_0}$$  Since the charge distribution is symmetrical, $\vec{E}$ and $\vec{da}$ point in the same direction and the direction is $\hat{r}$  $$E\times 4\pi r^2 = {\sigma\times 4\pi R^2 \over \epsilon_0}$$  $$\vec{E} = {\sigma\times R^2 \over \epsilon_0\times r^2}\hat{r}$$  Now, For $r<R, Q_{enc} = 0$ $$\therefore \oint \vec{E}\cdot \vec{da} = \frac{Q_{enc}}{\epsilon_{0}} = 0$$ Since, $\vec{da}\ne 0$ and both $\vec{E}$ and $\vec{da}$ points in the same direction. $$\therefore \vec{E} = 0$$ Hence , For $r<R, \vec{E}=0$ and For $r\ge R$, $\vec{E} = {\sigma\times R^2 \over \epsil
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Chapter 1 Vector Analysis: Problem 2.10

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Problem 2.10: A charge $q$ sits at the back corner of a cube, as shown in Fig. 2.17. What is the flux of E through the shaded side?                   Solution:  Surround the charge by 8 such cubes such that 4 cubes are above and 4 cubes are below and all are geometrically equivalent.                   Consider, the bigger cube having all the 8 small cubes. By guass law, the total flux would be $$\oint \vec{E}\cdot \vec{da} = {Q_{enc}\over \epsilon} = {q\over \epsilon}$$  Let $\phi$ be the flux through the required surface (The dark one in the diagram).  There would be $4\times 6 $ such faces and all are geometrically equivalent. Hence, the flux through them would be same and summation of them would be the total flux. Hence,  $$\oint \vec{E}\cdot \vec{da} = {24 \phi} = {q\over \epsilon_0}$$  $$\therefore \phi = {q\over 24\epsilon_0}$$   If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This enco
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Chapter 1 Vector Analysis: Problem 1.56

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Problem 1.56: Compute the line integral of $$v = 6 \hat{x} + yz^2 \hat{y} + (3y + z) \hat{z}$$ along the triangular path shown in Fig. 1.49. Check your answer using Stokes’ theorem.   Solution:  $$\vec{v} = 6\hat{i} + yz^2\hat{j} + (3y+z)\hat{k}$$  $$\vec{dl} = dx\hat{i} + dy\hat{j} + dz \hat{k}$$  Break the line integral into 3 parts where $L_1$ denotes the line integral where $y$ changes from $0$ to $1$, $x=0, z=0$  $$\int_{L_1}\vec{v}\cdot \vec{dl} =\int_{L_1} (6\hat{i} + 3y\hat{k})\cdot (dy\hat{j}) = 0$$ $L_2$ denotes the line integral of the line $z = -2y+2, x=0$ $dz = -2dy$  $$\int_{L_2}\vec{v}\cdot \vec{dl} =\int_{L_2} (6\hat{i} + yz^2\hat{j} + (3y+z)\hat{k})\cdot (dy\hat{j} + dz\hat{k})$$  $$\int_{L_2}\vec{v}\cdot \vec{dl} =\int_{L_2} (yz^2dy + (3y+z)dz$$  $$\int_{L_2}\vec{v}\cdot \vec{dl} =\int_{L_2} (y(-2y+2)^2dy + (3y+(-2y+2))(-2dy)$$ $$\int_{L_2}\vec{v}\cdot \vec{dl} =\int_1^0 (4y^3 - 8y^2+2y-4)dy$$  $$\int_{L_2}\vec{v}\cdot \vec{dl} =\big[y^4 - \frac{8y^3}{3}+y^2-4y
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Chapter 1 Vector Analysis: Problem 1.55

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Problem 1.55: Check Stokes’ theorem using the function $v = ay\vec{x} + bx \vec{y}$ ($a$ and $b$ are constants) and the circular path of radius $R$, centered at the origin in the $xy$ plane.  Solution: Stokes Theorem States That for a vector $\vec{v}$, $$\int_S (\vec{\nabla} \times \vec{v})\cdot d\vec{a} = \oint_P \vec{v}\cdot d\vec{l}$$  Given function is $$\vec{v} = ay\hat{x} + bx\hat{y} $$ where $a$ and $b$ are constants.  The path given is a circular path of radius $R$ in the $xy$ plane.                In Cartesian coordinate System $$\vec{da} = dxdy\hat{z}$$  and  $$\vec{dl} = dx\hat{x} + dy\hat{y}$$ $$ (\vec{\nabla} \times \vec{v})= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ ay & bx & 0 \end{vmatrix}$$ $$\qquad \quad \space = \hat{x}(\frac{\partial}{\partial{y}} (0)- \frac{\partial}{\partial{z}} (x)) - \hat{y}(\frac{\partial}{\partial{x}} (0) - \frac{\partial}{\
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Chapter 1 Vector Analysis: Problem 1.54

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Problem 1.54: Check the divergence theorem for the function $$v = r^2\cos\theta\hat{r} + r^2\cos \phi \hat{θ} − r^2 \cos\theta \sin \phi \hat{\phi}$$ , using as your volume one octant of the sphere of radius $R$ (Fig. 1.48). Make sure you include the entire surface.      Solution:  Divergence Theorem States That for a verctor $\vec{v}$, $$\int_V (\vec{\triangledown}\cdot \vec{v})d\tau = \oint_S \vec{v}\cdot d\vec{a}$$ Let's consider the octant where $r$ varies from $0$ to $R$, $\theta$ varies from $0$ to $\pi\over 2$ and $\phi$ varies from $0$ to $\pi\over 2$  Given $$\vec{v} = r^2cos\theta \hat{r} + r^2 cos\phi\hat{\theta} - r^2 cos\theta sin\phi \hat{\phi} $$ $$\vec{\triangledown}\cdot \vec{v} = \frac{1}{r^2 sin\theta}\big[\frac{\partial}{\partial r}(r^2sin\theta \cdot r^2 cos\theta) + \frac{\partial}{\partial \theta}(rsin\theta \cdot r^2 cos\phi) + \frac{\partial}{\partial \phi} (r\cdot (-r^2 cos\theta sin\phi))\big]$$ $$\vec{\triangledown}\cdot \vec{v} = \frac{1}{r^2 s
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