Electromagnetism – Physics Hut

Relation between one Coulomb and stat Coulomb

One coulomb is equal to $3\times 10^{9}$ Coulomb.
$1C=3\times 10^{9} statC$
Watch the following video for full proof:

Quantum Mechanics IIT Bombay Assignment-1 Problem-4

Consider a particle in a two dimensional box, whose ground state energy is $\epsilon_{\circ}$ . Then the energy $E=13\epsilon_{\circ}$ is two fold degenerate.
A. True
B. False
Answer:
The correct choice is A.
Watch the following video for full solution:

Quantum Mechanics IIT Bombay Assignment-1 Problem-3

For a particle in one dimensional box, the ground state energy is given by $E_0$ . The allowed energies of such particle are:
A. $2E_0$
B. $4E_0$
C. $6E_0$
D. $9E_0$
Answer:
Correct choices are B. and D.
Watch the following video for full solution:

Quantum Mechanics solved problem-1

The energy eigenvalue and corresponding eigenfunction for a particle in one-dimensional potential $V(x)$ are:

$E=0, \quad \Psi(x)=\frac{A}{x^2+a^2}$

Where $A$ is a positive constant. The form of potential $V(x)$ is:

A. $\frac{\hbar^2(3x^2-a^2)}{m(x^2+a^2)^2}$

B. $\frac{\hbar^2(x^2-a^2)}{2m(x^2+a^2)^2}$

C. $\frac{\hbar^2(3x^2-a^2)}{m(x^2+a^2)^3}$

D. $\frac{\hbar^2(x^2-a^2)}{2m(x^2+a^2)^3}$

The correct Choice is A.

See the full solution from following video tutorial:

Compton Effect Solved Problem

Consider a photon, that is scattered from an electron at rest. If the photon scatters at an angle of $60^{\circ}$ with a wavelength ${\lambda}'$ , that is twice its incident wavelength ${\lambda}$ , then the angle at which electron recoils will be \underline{\hspace{3cm}} degrees. (Answer should be an integer)}
Answer:
The correct Answer is $60^{\circ}$
Watch following video for full solution:

Physical significance of gradient of a scalar field

We know that all scalar fields posses isosurfaces. “An isosurface is the locus of all points in space, which posses same value of scalar field”. Consider a field $\phi(x,y,z)$ . Let $S_{1}$ and $S_{2}$ represent two closely separated isosurfaces of the field $\phi(x,y,z)$ . Let the value of scalar field at all points on isosurface $S_{1}$ is $\phi$ and the value of filed at all points on isosurface $S_{2}$ is $\phi+d\phi$ as shown in Fig. 1

**Fig.1: Isosurfaces of a scalar field**

Consider any point $A(x,y,z)$ on isosurface $S_{1}$ and any point $P(x+dx,y+dy,z+dz)$ on isosurface $S_{2}$ . Let $\overrightarrow{r~}$ and $\overrightarrow{r~}+\overrightarrow{dr~}$ be the position vectors of the points A and P respectively. Therefore, by triangle law of vectors, the displacement $\overrightarrow{AP~}$ is given as follows:

$\overrightarrow{AP~}=\overrightarrow{OP~}-\overrightarrow{OA~}$

$=\overrightarrow{r~}+\overrightarrow{dr~}-\overrightarrow{r~}$

(1) $\begin{equation*} \implies \overrightarrow{AP~}=\overrightarrow{dr~}=dx\hat{i}+dy\hat{j}+dz\hat{k} \end{equation*}$

The magnitude of this displacement is given as follows:

(2) $\begin{equation*} AP=|\overrightarrow{dr~}|=dr=\sqrt{dx^{2}+dy^{2}+dz^{2}} \end{equation*}$

The gradient of scalar field is given according to the following relation:

(3) $\begin{equation*} \overrightarrow{\nabla}\phi= \frac{\partial \phi}{\partial x}\hat{i} + \frac{\partial \phi}{\partial y}\hat{j} + \frac{\partial \phi}{\partial z}\hat{k} \end{equation*}$

Since $\phi$ is a scalar field (function), therefore, the net change in the value of scalar field $\phi$ in going from point A on isosurface $S_{1}$ to some arbitrary point P on the isosurface $S_{2}$ is given by:

(4) $\begin{equation*} \text{ Change in $\phi$ along displacement } \overrightarrow{AP~}=\phi+d\phi-\phi=d\phi \end{equation*}$

From calculus, we know that $\frac{\partial \phi}{\partial x}$ represents the rate at which scalar $\phi$ changes along X-axis, keeping the y, z as constants. Therefore, change in $\phi$ due to net displacement $dx$ (X-component of the displacement $\overrightarrow{AP~}$ ) along X-axis in going from A to P is given by:

$Change~in~\phi~due~ to~ displacement~dx~along~X-axis=\frac{\partial \phi}{\partial x} dx$

Similarly, we can write:

$Change~in~\phi~due~ to~ displacement~dy~along~Y-axis=\frac{\partial \phi}{\partial y}dy$

$Change~in~\phi~due~ to~ displacement~dz~along~Z-axis=\frac{\partial \phi}{\partial z} dz$

Since $\overrightarrow{dr~}= dx\hat{i}+dy\hat{j}+dz\hat{k}$ , therefore, It can be easily understood that:
Change in $\phi$ along displacement $\overrightarrow{AP~}~(or~ \overrightarrow{dr~})=Change~in~\phi~due~ to~ displacement~dx$

$+~Change~in~\phi~due~ to~ displacement~dy$

$+~Change~in~\phi~due~ to~ displacement~dz$

$=\frac{\partial \phi}{\partial x} dx+\frac{\partial \phi}{\partial y} dy+\frac{\partial \phi}{\partial z} dz$

(5) $\begin{equation*} \implies Change~in~\phi~along~displacement~\overrightarrow{AP~}=\frac{\partial \phi}{\partial x} dx+\frac{\partial \phi}{\partial y} dy+\frac{\partial \phi}{\partial z} dz \end{equation*}$

Comparing equations (4) and ( 5), we get the following relation:

(6) $\begin{equation*} d\phi=\frac{\partial \phi}{\partial x} dx+\frac{\partial \phi}{\partial y} dy+\frac{\partial \phi}{\partial z} dz \end{equation*}$

Important Note:
Consider two vectors $\overrightarrow{A~}$ and $\overrightarrow{B~}$ such that

$\overrightarrow{A~}=A_{x}\hat{i}+A_{x}\hat{j}+A_{x}\hat{k}$

and

$\overrightarrow{B~}=B_{x}\hat{i}+B_{x}\hat{j}+B_{x}\hat{k}$

and $\theta$ is the angle between the vectors. Therefore, dot product of two vectors is given by following expressions:

$\overrightarrow{A~}\cdot\overrightarrow{B~}=ABcos\theta$

and

$\overrightarrow{A~}\cdot\overrightarrow{B~}= A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}$

Equation (6) can also be written as follows:

$d\phi=\left(\frac{\partial \phi}{\partial x}\hat{i}+\frac{\partial \phi}{\partial y}\hat{j}+\frac{\partial \phi}{\partial z}\hat{i}\right)\cdot \left(dx\hat{i}+dy\hat{j}+dz\hat{k}\right) ~~~[\text{ Using Important Note}]$

(7) $\begin{equation*} d\phi=\overrightarrow{\nabla~}\phi \cdot \overrightarrow{dr~} \end{equation*}$

[Using equations (1) and ( 6)]
Let $\theta$ is the angle between the vectors $\overrightarrow{\nabla~}\phi$ and $\overrightarrow{dr~}$ , then above equation can be simplified as follows:

(8) $\begin{equation*} \begin{array}{ll} ~~~~~~~~~~d\phi& =|\overrightarrow{\nabla~}\phi | \cdot |\overrightarrow{dr~}|cos\theta \nonumber \\ &=|\overrightarrow{\nabla~}\phi |dr~ cos \theta \end{array} \end{equation*}$

(9) $\begin{equation*} \begin{array}{ll} \implies \frac{d\phi}{dr} &=|\overrightarrow{\nabla~} \phi |~ cos \theta \\ \end{array} \end{equation*}$

The left side of the equation (9) represents rate of change of $\phi$ with respect to position of a point along the path AP (see Fig. 1). It is clear that this rate is always less than or equal to the magnitude of gradient of scalar field, that is $|\overrightarrow{\nabla~} \phi |$ (because value of $cos\theta$ is always less than or equal to 1). Therefore, it is clear that derivative of a scalar quantity / function / field with respect to position is not always equal to gradient magnitude. This equality comes only under one condition that the value of $cos\theta$ must be equal to 1. When such condition is met, then the rate $\frac{d\phi}{dr}$ is maximum and it is written as $\left(\frac{d\phi}{dr}\right)_{max}$ . That is

$\text{When}~~~~~cos\theta=1$

$\text{Then}~~~~~\frac{d\phi}{dr}=maximum=\left(\frac{d\phi}{dr}\right)_{max}~~~\text(say)$

Under this condition, equation (9) will become:

(10) $\begin{equation*} \left(\frac{d\phi}{dr}\right)_{max}=|\overrightarrow{\nabla~} \phi | \nonumber \end{equation*}$

(11) $\begin{equation*} \boxed{ |\overrightarrow{\nabla~} \phi |=\left(\frac{d\phi}{dr}\right)_{max}} \end{equation*}$

From equation (11), we can write the physical significance of gradient of a scalar field as follows:
“The magnitude of gradient of scalar field at a point is equal to the maximum rate of change of field with respect to the position.”
The only task remaining is to find the direction of gradient; as the equation (11) only gives its magnitude. To accomplish this, we turn our attention to the Fig. 1. Note that the direction of $\overrightarrow{dr~}$ is along the path AP. P is arbitrary point on the isosurface $S_{2}$ . Therefore, AP represents all possible displacements from point A on isosurface $S_{1}$ to all points on the isosurface $S_{2}$ . A few paths from $S_{1}$ to $S_{2}$ are shown in Fig. 1, like paths AB, AC and AD.
It is clear that along all these paths, the starting point is on isosurface $S_{1}$ and end point is on isosurface $S_{2}$ . Therefore, change in the field $\phi$ along all these paths is essentially $d\phi$ , however, the path lengths or magnitude of displacements ( $dr$ ) along these paths are not same. It means that, for the rate expression $\frac{d\phi}{dr}$ the value of numerator is same, but the value of denominator is different along all these paths. Therefore, the rate $\frac{d\phi}{dr}$ will be maximum along the shortest path (because denominator $dr$ will be minimum). But it is clear that the shortest path is AB and the displacement $\overrightarrow{AB~}$ is along the normal to the isosurface $S_{1}$ .
Let $\hat{n}$ is unit vector along the normal to point A on isosurface $S_{1}$ . Therefore

$\overrightarrow{AB~} =(AB)\hat{n}$

Moreover along path AB,

$\frac{d\phi}{dr}=maximum=\left(\frac{d\phi}{dr}\right)_{max}$

and

$\overrightarrow{dr~}=\overrightarrow{AB~}=(AB)\hat{n}$

It is also clear that along path AB we have $cos\theta=1$ , that is $\theta=0^{\circ}$ . It means that along the shortest path, the direction of displacement vector and gradient vector are same. Since the direction of shortest displacement is along the normal vector ( $\hat{n}$ ), therefore, the direction of gradient vector is also along the normal vector to isosurface defined by scalar field. Hence we can write:

$~~~~~~\overrightarrow{\nabla~} \phi=|\overrightarrow{\nabla~} \phi |\hat{n}$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\left(\frac{d\phi}{dr}\right)_{max}\hat{n}$

[Using equation (11)]

(12) $\begin{equation*} \implies \boxed{ \overrightarrow{\nabla~} \phi =\left(\frac{d\phi}{dr}\right)_{max}\hat{n}} \end{equation*}$

Thus complete physical significance of gradient of a scalar field can be stated as follows:
“Gradient of a scalar field at a point represents the maximum rate of change of scalar field with respect to position and it is always directed along the normal to the field surface at that point.”
Due to the reason that gradient of a scalar field is always directed along the normal direction, it is also called as directional derivative of a scalar field.

Gradient of a Scalar Field

Let $\phi (x, y, z)$ is a scalar field, which is a function of space variables $x, y, z$ . Then the gradient of scalar field is defined as operation of $\overrightarrow{\nabla}$ on the scalar field.
That is:
$Grad\phi (x, y, z)$ = $\overrightarrow{\nabla} \phi (x, y, z)$ $\equiv \overrightarrow\nabla} \phi$
Here the operator $\overrightarrow{\nabla}$ is called Del or Nabla vector. It is given by the following expression:

(1) $\begin{equation*} \overrightarrow{\nabla}= \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \end{equation*}$

Please note that $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along X, Y and Z axes respectively in cartesian system of cordinates.
Thus using the description given in Eq: (1), the expression for gradient of a scalar field can also be written as:
$Grad\phi (x, y, z)$ = $\overrightarrow{\nabla} \phi (x, y, z)$ $\equiv \overrightarrow\nabla} \phi$ = $\hat{i}\frac{\partial\phi}{\partial x} + \hat{j}\frac{\partial\phi}{\partial y} + \hat{k}\frac{\partial\phi}{\partial z}$

$\implies \boxed{{\overrightarrow\nabla} \phi $ = $\hat{i}\frac{\partial\phi}{\partial x} + \hat{j}\frac{\partial\phi}{\partial y} + \hat{k}\frac{\partial\phi}{\partial z}}$

Example: 1 If $\overrightarrow{r}$ is the position vector of a point (x, y, z) in space, then find the gradient of $\frac{1}{r}$ .
Solution: Since $\overrightarrow{r}$ is the position vector of a point (x, y, z) in space, therefore, it is given as follows:

$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

The magnitude of this vector is given as follows:

(2) $\begin{equation*} r =\sqrt{x^{2}+y^{2}+z^{2}} \end{equation*}$

Thus

(3) $\begin{equation*} \frac{1}{r}=(x^{2}+y^{2}+z^{2})^{-\frac{1}{2}}$ \end{equation*}$

The gradient of $\frac{1}{r}$ is given by:

(4) $\begin{equation*} \overrightarrow\nabla}\left(\frac{1}{r}\right) = \hat{i}\frac{\partial}{\partial x}\left(\frac{1}{r}\right) + \hat{j}\frac{\partial}{\partial y}\left(\frac{1}{r}\right) + \hat{k}\frac{\partial}{\partial z}}\left(\frac{1}{r}\right) \end{equation*}$

Let us now evaluate each term on the right hand side of equation (4) one by one and then solve the problem.
The first term can be calculated using equation (3) as follows:

(5) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial x}\left(\frac{1}{r}\right) &=-\frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}(2x)\\ &=-\frac{x}{(\sqrt{x^{2}+y^{2}+z^{2}})^{3}}\\ &=\frac{-x}{r^{3}} \end{array} \end{equation*}$

(using equation (2))
Similarly, we can find that

(6) $\begin{equation*} \frac{\partial}{\partial y}\left(\frac{1}{r}\right) = \frac{-y}{r^{3}} \end{equation*}$

and

(7) $\begin{equation*} \frac{\partial}{\partial z}\left(\frac{1}{r}\right) = \frac{-z}{r^{3}} \end{equation*}$

Put the values from equations (5), (6) and (7) in equation (4), we get

(8) $\begin{equation*} \overrightarrow{\nabla}\left(\frac{1}{r}\right) = \frac{-(x\hat{i}+y\hat{j}+z\hat{k})}{r^{3}} \nonumber \end{equation*}$

$\implies\boxed{\overrightarrow\nabla\left(\frac{1}{r}\right) = -\frac{\overrightarrow{r}}{r^{3}}}$

Example:2 If $\overrightarrow{r~}$ and $\overrightarrow{r~}'$ be the position vectors of points $(x, y, z)$ and $(x', y', z')$ in space, then find the gradient of $\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}$ .

Solution: Since $\overrightarrow{r~}$ is the position vector of a point (x, y, z) in space, therefore, it is given as follows:

$\overrightarrow{r~}=x\hat{i}+y\hat{j}+z\hat{k}$

Similarly

$\overrightarrow{r~}'={x'}\hat{i}+{y'}\hat{j}+{z'}\hat{k}$

Therefore

$\overrightarrow{r~}-\overrightarrow{r~}'={(x-x')}\hat{i}+{(y-y')}\hat{j}+{(z-z')}\hat{k}$

The magnitude of this difference or displacement vector is given by:

(9) $\begin{equation*} |\overrightarrow{r~}-\overrightarrow{r~}'|=\sqrt{{(x-x')}^2+{(y-y')}^2+{(z-z')}^2} \end{equation*}$

(10) $\begin{equation*} \implies \frac{1}{|\overrightarrow{r~}-\overrightarrow{r~}'|}=[{{(x-x')}^2+{(y-y')}^2+{(z-z')}^2}]^{-\tfrac{1}{2}} \end{equation*}$

The gradient of $\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}$ is given by:

(11) $\begin{equation*} \overrightarrow{\nabla}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = \hat{i}\frac{\partial}{\partial x}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) + \hat{j}\frac{\partial}{\partial y}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) + \hat{k}\frac{\partial}{\partial z}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) \end{equation*}$

Let us now evaluate each term on the right hand side of equation (11) one by one and then solve the problem.
The first term can be calculated using equation (10) as follows:

(12) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial x}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) &=-\frac{1}{2}\left({(x-x')}^2+{(y-y')}^2+{(z-z')}^2\right)^{-\frac{3}{2}}[2(x-x')]\\ &=\frac{-(x-x')}{(\sqrt{(x-x')^2+(y-y')^2+(z-z')^2})^{3}}\\ &=\frac{-(x-x')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{array} \end{equation*}$

(using equation (9))
Similarly, we can find that

(13) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial y}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) &=\frac{-(y-y')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{array} \end{equation*}$

and

(14) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial z}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) &=\frac{-(z-z')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{array} \end{equation*}$

Put the values from equations (12), (13) and (14) in equation (11), we get

(15) $\begin{equation*} \overrightarrow{\nabla}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = \frac{-[(x-x')\hat{i}+(y-y')\hat{j}+(z-z')\hat{k}]}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \nonumber \end{equation*}$

(16) $\begin{equation*} \implies\boxed{\overrightarrow\nabla\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = -\frac{(\overrightarrow{r~}-\overrightarrow{r~}')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}}} \end{equation*}$

Example: 3 If $\overrightarrow{r~}$ and $\overrightarrow{r~}'$ be the position vectors of points $(x, y, z)$ and $(x', y', z')$ in space, then find the gradient of $\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}$ with respect to the point $(x', y', z')$ . That is, find expression for $\overrightarrow\nabla'\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)$

Solution: Since $\overrightarrow{r~}$ is the position vector of a point (x, y, z) in space, therefore, it is given as follows:

$\overrightarrow{r~}=x\hat{i}+y\hat{j}+z\hat{k}$

Similarly

$\overrightarrow{r~}'={x'}\hat{i}+{y'}\hat{j}+{z'}\hat{k}$

Therefore

$\overrightarrow{r~}-\overrightarrow{r~}'={(x-x')}\hat{i}+{(y-y')}\hat{j}+{(z-z')}\hat{k}$

The magnitude of this difference or displacement vector is given by:

(17) $\begin{equation*} |\overrightarrow{r~}-\overrightarrow{r~}'|=\sqrt{{(x-x')}^2+{(y-y')}^2+{(z-z')}^2} \end{equation*}$

(18) $\begin{equation*} \implies \frac{1}{|\overrightarrow{r~}-\overrightarrow{r~}'|}=[{{(x-x')}^2+{(y-y')}^2+{(z-z')}^2}]^{-\tfrac{1}{2}} \end{equation*}$

The gradient of $\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}$ with respect to the point $(x', y', z')$ is given by:

(19) $\begin{equation*} \overrightarrow{\nabla'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = \hat{i}\frac{\partial}{\partial x'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) + \hat{j}\frac{\partial}{\partial y'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) + \hat{k}\frac{\partial}{\partial z'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) \end{equation*}$

Let us now evaluate each term on the right hand side of equation (19) one by one and then solve the problem.
The first term can be calculated using equation (18) as follows:

(20) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial x'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) &=-\frac{1}{2}\left({(x-x')}^2+{(y-y')}^2+{(z-z')}^2\right)^{-\frac{3}{2}}[2(x-x')(-1)]\\ &=\frac{(x-x')}{(\sqrt{(x-x')^2+(y-y')^2+(z-z')^2})^{3}}\\ &=\frac{(x-x')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{array} \end{equation*}$

(using equation (17))
Similarly, we can find that

(21) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial y'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) &=\frac{(y-y')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{array} \end{equation*}$

and

(22) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial z'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) &=\frac{(z-z')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{array} \end{equation*}$

Put the values from equations (20), (21) and (22) in equation (19), we get

(23) $\begin{equation*} \overrightarrow{\nabla'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = \frac{[(x-x')\hat{i}+(y-y')\hat{j}+(z-z')\hat{k}]}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \nonumber \end{equation*}$

(24) $\begin{equation*} \implies\boxed{\overrightarrow\nabla'\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = \frac{(\overrightarrow{r~}-\overrightarrow{r~}')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}}} \end{equation*}$

Example: 4 If $\overrightarrow{r~}$ and $\overrightarrow{r~}'$ be the position vectors of points $(x, y, z)$ and $(x', y', z')$ in space, then find the gradient of $\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}$ with respect to the points $(x, y, z)$ and $(x', y', z')$ . That is, find expression for $\overrightarrow\nabla\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)$ and $\overrightarrow\nabla'\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)$ . Hence show that these two gradients are equal and opposite. That is $\overrightarrow\nabla'\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)=-\overrightarrow\nabla\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)$
Solution: We have already proved in Example 2 and Example 3 that

(25) $\begin{equation*} \overrightarrow\nabla\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = -\frac{(\overrightarrow{r~}-\overrightarrow{r~}')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{equation*}$

and

(26) $\begin{equation*} \overrightarrow\nabla'\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = \frac{(\overrightarrow{r~}-\overrightarrow{r~}')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{equation*}$

Comparing equations (25) and (26), we get the desired result. That is

$\boxed{\overrightarrow\nabla'\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)=-\overrightarrow\nabla\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)}$

Note: This relation is very useful in the derivation and solution of many complex problems of electromagnetism.

If you are interested to know more about scalar and vector fields, then watch video lecture by me on this topic at the following link: https://youtu.be/H4eWdwaGWYI.

Another video, that may be of interest to the reader of this post is ‘Transverse nature of electromagnetic waves’. The link to this video lecture by me is given here: https://youtu.be/k7399IiFEzo