ELECTRIC DIPOLE -

A system of two equal and opposite point charges separated by a fixed small distance is called an electric dipole. Many of the atoms/molecules are dipoles. Ex- HCl, H2O etc

$\fn_cm \large {\color{Red} \textbf{NOTE:-}}$

(1) The total charge of the electric dipole is zero. This does not mean that the field of the electric dipole is zero. Because charges are separated by some distance. So the electric field due to them, when added, does not exactly cancel out.

(2) However, at distances much larger than the dipole size, the fields of both charges nearly cancel out. Hence we expect a dipole field to fall off, at a larger distance, faster than 1/r², typical of the field due to a single charge. In fact, a dipole field at larger distances falls off as 1/r³.

Dipole length:- It is the separation between two charges. It is a vector quantity and its direction is -ve to +ve charge.

Dipole moment:- It is the property of the dipole, it gives the strength of an electric dipole. ” The dipole moment of an electric dipole is a vector quantity whose magnitude is equal to the product of either charge and its length, and its direction is along -ve to +ve charge.”

i.e. $\dpi{120} \fn_cm \left [ \vec{p}=q\vec{l} \right ]$

Its S.I unit is (C-m) and its dimension is [M°LTA]

$\fn_cm \large {\color{Red} \textbf{NOTE:-}}$

(1) If the centre of +ve charges and -ve charges lie at the same place. Then their dipole moment is zero. ex- CO2, CH4 etc.

(2) If there are two or more dipoles, then the resultant dipole moment is the vector sum of the dipole moment due to each dipole.

Exercise 1

(1) An HCl molecule has a dipole moment of $\dpi{120} \fn_cm 3.4\times10^{-30}Cm$ . Assuming that equal and opposite charges lie on the two atoms to form a dipole, what is the magnitude of this charge? The separation between the two atoms of HCl is $\dpi{120} \fn_cm 1.0\times10^{-10}m$ . $\dpi{120} \fn_cm (3.4\times10^{-20}C)$

(2) 3 C and -3 C are two equal charges separated by a distance of 4 cm, then calculate the dipole moment of a dipole. (0.12 Cm)

(3) Two charges, one +5μC and another -5μC are placed 1mm apart. Calculate the dipole moment. $\dpi{120} \fn_cm \left ( 5\times 10^{-9}Cm \right )$

(4) A system has two charges $\dpi{120} \fn_cm \pm 2.5\times10^{-7}C$ located at points A(0,0,-15) and B (0,0,+15) cm respectively. Find the total charge and electric dipole moment of the system. $\dpi{120} \fn_cm (0, 7.5\times10^{-8}Cm)$

(5) Calculate the electric dipole moment between an electron and a proton $\dpi{120} \fn_cm 4.3\times10^{-9}m$ apart. $\dpi{120} \fn_cm (6.88\times10^{-28}Cm)$

(6) Find the resultant dipole moment (magnitude and direction) of the system in each case as shown in the figure.

(7) Find the dipole moment of the system as shown in figure.

(8) A thin non-conducting ring of radius R has linear charge density $\dpi{120} \fn_cm \lambda=\lambda_0\cos\theta$ , where $\dpi{120} \fn_cm \theta$ is measured as shown in the figure. Find the dipole moment for the ring. $\dpi{120} \fn_cm (\pi \lambda_0 R^2)$

Electric field due to dipole

On the axial position (End on position)

Consider AB is an electric dipole consisting of charges -q and +q separated by a small distance of 2d in free space.

We have to calculate the electric field strength at point P on the axial line which is distance r from the centre of dipole. From the figure Net electric field at point P due to dipole is

$\dpi{120} \fn_cm \vec{E}=\vec{E}_A+\vec{E}_B$

$\dpi{120} \fn_cm =\frac{kq}{(r+d)^2}(-\hat{i})+\frac{kq}{(r-d)^2}(\hat{i})$

$\dpi{120} \fn_cm =\frac{kq4rd\hat{i}}{(r^2-d^2)^2}$

$\dpi{120} \fn_cm \left [ \vec{E}_{axial}=\frac{2kr\vec{p}}{(r^2-d^2)^2} \right ]$

For small dipole (r>>>d) then we can write $\dpi{120} \fn_cm (r^2-d^2)\approx r^2$

$\dpi{120} \fn_cm \therefore \vec{E}=\frac{2kr\vec{p}}{r^4}$

$\dpi{120} \fn_cm \left [ \vec{E}_{axial}=\frac{2k\vec{p}}{r^3} \right ]\;\;\;\;\;(\vec{E}_{axial}\;is\;always\;along\; \vec{p})$

$\fn_cm \large {\color{Red} \textbf{NOTE:-}}$ Here $\dpi{120} \fn_cm \left [ E\propto \frac{1}{r^3} \right ]$ i.e. we expect a dipole field to fall off, a large distance, faster than $\dpi{120} \fn_cm \frac{1}{r^2}$ , typical of the field due to a single charge.

On the equatorial position (Broad on position)

$\dpi{120} \fn_cm here\;\left | \vec{E}_A \right |=\left | \vec{E}_B \right |=E$

Consider AB is an electric dipole consisting of charges -q and +q separated by a small distance of 2d in free space.

We have to calculate the electric field strength at point P on the equatorial line which is distance r from the centre of dipole. From the figure Net electric field at point P due to dipole is

$\dpi{120} \fn_cm \vec{E}=\vec{E}_A+\vec{E}_B$

From Fig $\dpi{120} \fn_cm E_A\sin\theta\; and\; E_B\sin\theta$ are cancelled from each other but $\dpi{120} \fn_cm E_A\cos\theta\; and\; E_B\cos\theta$ are added.

i.e.

$\dpi{120} \fn_cm E=\vec{E}_A\cos\theta(-\hat{i})+\vec{E}_B\cos\theta(-\hat{i})$

$\dpi{120} \fn_cm =-2E\cos\theta\;\hat{i}$

$\dpi{120} \fn_cm =-2\frac{kq}{r^2+d^2}\frac{d}{\sqrt{r^2+d^2}}$

$\dpi{120} \fn_cm \left [ \vec{E}_{eq}=\frac{-k\vec{p}}{(r^2+d^2)^{3/2}} \right ]$

For small dipole (r>>>d) then we can write $\dpi{120} \fn_cm (r^2+d^2)\approx r^2$

$\dpi{120} \fn_cm \left [ \vec{E}_{eq}=\frac{-k\vec{p}}{r^3} \right ]\;\;\;\;\;(here\; \vec{E}_{eq}\;is\;opposite\;to\; \vec{p})$

$\fn_cm \large {\color{Red} \textbf{NOTE:-}}$ clearly $\dpi{120} \fn_cm \left [ E_{axial}=2 E_{eqatorial} \right ]$

On any point

Consider a dipole AB of dipole moment P along A to B. We have to calculate the electric field strength at P ( any point) which is the distance r from the centre of the dipole.

Let $\dpi{120} \fn_cm \vec{r}$ is the position vector of P w.r.t. O and θ is the angle between $\dpi{120} \fn_cm \vec{p}\;and\;\vec{r}$ .

To find electric field at point P, resolve dipole moment $\dpi{120} \fn_cm \vec{p}$ into two component $\dpi{120} \fn_cm p\cos\theta\;(along\;A_1B_1)$ and $\dpi{120} \fn_cm p\sin\theta\;(along\;A_2B_2)$ as shown in Fig.

For dipole A1B1, point P is on the axial line, electric field due to the dipole of dipole moment $\dpi{120} \fn_cm p\cos\theta$ is

$\dpi{120} \fn_cm E_a=\frac{2kp\cos\theta}{r^3}$

For dipole A2B2, point P is on the equatorial line, electric field due to the dipole of dipole moment $\dpi{120} \fn_cm p\sin\theta$ is

$\dpi{120} \fn_cm E_e=\frac{kp\sin\theta}{r^3}$

$\dpi{120} \fn_cm \therefore$ The net electric field at P due to the dipole of dipole moment p is

$\dpi{120} \fn_cm \vec{E}_{net}=\vec{E}_a+\vec{E}_e$

$\dpi{120} \fn_cm E_{net}=\sqrt{E_a^2+E_e^2}$

$\dpi{120} \fn_cm \left [ E_{net}=\frac{kp\sqrt{3\cos^2\theta+1}}{r^3} \right ]$

If $\dpi{120} \fn_cm \vec{E}_{net}$ makes an angle $\dpi{120} \fn_cm \alpha$ with $\dpi{120} \fn_cm E_a$

then $\dpi{120} \fn_cm \tan\alpha=\frac{E_e}{E_a}$

$\dpi{120} \fn_cm \tan\alpha=\frac{kp\sin\theta}{r^3}\frac{r^3}{2kp\cos\theta}=\frac{\tan\theta}{2}$

$\dpi{120} \fn_cm \left [ \alpha=\tan^{-1}\left ( \frac{\tan\theta}{2} \right ) \right ]$

Exercise 2

(1) Calculate the electric field due to an electric dipole of length 10 cm having charges of 1μC at an equatorial point 12 cm from the centre of the dipole. $\dpi{120} \fn_cm (4.096\times 10^5 N/C))$

(2) Two point charges, each of 5μC but opposite in sign, are placed 4 cm apart. Calculate the electric field intensity at a point distance 4 cm from the midpoint on the axial line of the dipole. $\dpi{120} \fn_cm (10^8 N/C)$

(3) Two charges ±10μC are placed 5mm apart. Determine the electric field at (a) a point P on the axis of the dipole 15 cm away from its centre O on the side of the +ve charge, (b) a point Q, 15 cm away from O on the passing through O and normal to the axis of the dipole. $\dpi{120} \fn_cm (2.66\times10^5N/C, 1.33\times 10^5 N/C)$

(4) Calculate the field due to an electric dipole of length 10 cm and consisting of charges of ±100μC at a point 20 cm from each charge. $\dpi{120} \fn_cm (1.125\times10^7 N/C)$

(5) What is the magnitude of electric intensity due to a dipole of moment $\dpi{120} \fn_cm 2\times10^{-8}C-m$ at a point distance 1m from the centre of the dipole, when the line joining the point to the centre of the dipole makes an angle of 60° with the dipole axis? $\dpi{120} \fn_cm (238.1N/C)$

(6) At what angle θ from the dipole will the electric field due to the dipole be at the right angle to the dipole moment? Also find the magnitude of the electric field, if the distance between the point and the dipole is r. $\dpi{120} \fn_cm (\tan^{-1}\sqrt{2},E=\frac{p}{2\sqrt{2}\pi\epsilon_0r^3})$

(7) Find the magnitude of electric field intensity at point (2,0,0) due to a dipole of dipole moment $\dpi{120} \fn_cm \vec{p}=(\hat{i}+\sqrt{3}\hat{j})$ kept at origin. $\dpi{120} \fn_cm (E=\sqrt{7k}/8)$

(8) Find out the magnitude of electric field intensity at a point (-1,√3,0) due to a dipole of dipole moment $\dpi{120} \fn_cm \vec{p}=\hat{i}+\sqrt{3}\hat{j}$ kapt at origin.

(9) Find the magnitude and direction of the electric field at point P as shown in the figure.

(10) For a given dipole at a point ( away from the centre of the dipole) intensity of the electric field is E. Charges of the dipole are brought closer such that the distance between point charges is half, and the magnitude of charges is also halved. Find the intensity of the field at the same point. $\dpi{120} \fn_cm (E_f=E_i/4)$

(11) Find the force acting on the dipole in each case.

(12) Two point dipoles $\dpi{120} \fn_cm p\hat{k}$ and $\dpi{120} \fn_cm \frac{p}{2}\hat{k}$ are located at (0,0,0) and (1,0,2) respectively. Find the resultant electric field due to the two dipoles at the point (1,0,0). All distances are in meters. $\dpi{120} \fn_cm \left ( \frac{-7p}{32\pi\epsilon_0}\hat{k} \right )$

(13) A dipole having a dipole moment $\dpi{120} \fn_cm \vec{p}$ is placed at the axis of a charged ring having charge Q and radius R, at a distance x from the centre as shown in fig. Find the force experienced by the dipole of dipole moment $\dpi{120} \fn_cm \vec{p}=p\hat{i}$ placed along the axis of uniformly charged ring.

$\dpi{120} \fn_cm \left ( \frac{Qp}{4\pi\epsilon_0}\frac{R^2-2x^2}{(R^2+x^2)^{5/2}} \right )$

(14) An electric dipole of dipole moment $\dpi{120} \fn_cm p$ is kept at a distance $\dpi{120} \fn_cm r$ from an $\dpi{120} \fn_cm \infty$ long charged wire of linear charge density $\dpi{120} \fn_cm \lambda$ . Find the force acting on the dipole. $\dpi{120} \fn_cm \left ( \frac{p\lambda}{2\pi\epsilon_0 r^2} \right )$

Electric Dipole in a uniform electric field

Torque experienced on dipole

Consider an electric dipole AB consisting of charges -q and +q placed in a uniform electric field of strength E and making an angle θ with it.

The force exerted on charge +q and -q by field is given by $\dpi{120} \fn_cm \vec{F_1}=q\vec{E}\;and\;\vec{F_2}=-q\vec{E}$ respectively.

$\dpi{120} \fn_cm \therefore$ The net force on the dipole is zero $\dpi{120} \fn_cm (F_{net}=0)$

i.e. Dipole has no translational motion in a uniform electric field.

Now, the Torque on the dipole due to $\dpi{120} \fn_cm \vec{F}_1$ is given by

$\dpi{120} \fn_cm \vec{\tau}_1=\vec{OB}\times q\vec{E}$

Similarly, torque on dipole due to $\dpi{120} \fn_cm \vec{F}_2$ is given by

$\dpi{120} \fn_cm \vec{\tau}_2=\vec{OA}\times (-q\vec{E})$

$\dpi{120} \fn_cm \therefore$ Net torque on the dipole is

$\dpi{120} \fn_cm \vec{\tau}=\vec{\tau}_1+\vec{\tau}_2$

$\dpi{120} \fn_cm \vec{\tau}=\left [ \vec{OB}\times q\vec{E} \right ]+\left [ \vec{OA}\times (-q\vec{E}) \right ]$

$\dpi{120} \fn_cm =\left ( \vec{OB}\times q\vec{E} \right )+\left ( \vec{AO}\times q\vec{E} \right )$

$\dpi{120} \fn_cm =\left ( \vec{AO}+\vec{OB} \right )\times q\vec{E}$

$\dpi{120} \fn_cm =\vec{AB}\times q\vec{E}$

$\dpi{120} \fn_cm =q\vec{AB}\times \vec{E}$

$\dpi{120} \fn_cm \left [ \vec{\tau}=\vec{P}\times \vec{E} \right ]$

This torque will tend to align the dipole with the field $\dpi{120} \fn_cm \vec{E}$

$\dpi{120} \fn_cm \left [ \tau=PE\sin\theta \right ]$ ( it is not a torque, it is a couple because it is independent from axis of rotation)

Where $\dpi{120} \fn_cm \theta$ is the angle between $\dpi{120} \fn_cm \vec{P}\;and\; \vec{E}$

The direction of the torque is perpendicular to both $\dpi{120} \fn_cm \vec{P}\;and\; \vec{E}$ according to right-hand thumb rule.

If $\dpi{120} \fn_cm \theta=0^0\;or\; \theta=180^0$

$\dpi{120} \fn_cm \tau=PE\sin 0^0\; or\; PE\sin 180^0=0$

$\dpi{120} \fn_cm \left [ \tau_{min}=0 \right ]$

If $\dpi{120} \fn_cm \theta=90^0$

$\dpi{120} \fn_cm \tau=PE\sin 90^0=PE$

$\dpi{120} \fn_cm \left [ \tau_{max}=PE \right ]$

Potential Energy of dipole

Consider an electric dipole of dipole moment $\dpi{120} \fn_cm \vec{p}$ placed in a uniform electric field $\dpi{120} \fn_cm \vec{E}$ with an angle $\dpi{120} \fn_cm \theta$ . (Same as previous Fig)

We know that the torque acting on the dipole is given by

$\dpi{120} \fn_cm \vec{\tau}=\vec{P}\times \vec{E}$

$\dpi{120} \fn_cm \tau=PE\sin\theta$

If the dipole is rotated through a very small angle against this torque, then the small amount of work done is

$\dpi{120} \fn_cm dw=\tau d\theta$

$\dpi{120} \fn_cm dw=PE\sin\theta d\theta$

$\dpi{120} \fn_cm \therefore$ The total work done in rotating the dipole from $\dpi{120} \fn_cm \theta_1\;to\; \theta_2$ is

$\dpi{120} \fn_cm w=PE\int_{\theta_1}^{\theta_2}\sin\theta d\theta$

$\dpi{120} \fn_cm w=PE\left [ -\cos\theta \right ]_{\theta_1}^{\theta_2}$

$\dpi{120} \fn_cm \left [ w=-PE\left ( \cos\theta_2-\cos\theta_1 \right ) \right ]$

This work done is stored in the dipole in the form of potential energy

i.e. $\dpi{120} \fn_cm \left [ U=-PE\left ( \cos\theta_2-\cos\theta_1 \right ) \right ]$

if $\dpi{120} \fn_cm \theta_1=90^0$ (initial position) and $\dpi{120} \fn_cm \theta_2=\theta$ (Final position)

Then $\dpi{120} \fn_cm \left [ U=-PE\cos\theta \right ]$

In vector form $\dpi{120} \fn_cm \left [ U=-\vec{P}.\vec{E} \right ]$

Special Case:-

If $\dpi{120} \fn_cm \theta=0^0$

$\dpi{120} \fn_cm U=-PE\cos 0^0$ ( $\dpi{120} \fn_cm \vec{P}\;and \; \vec{E}$ are in the same direction )

$\dpi{120} \fn_cm \left [ U_{min}=-PE \right ]$ ( minimum P.E in this condition) (Stable Equilibrium)

If $\dpi{120} \fn_cm \theta=90^0$

$\dpi{120} \fn_cm U=-PE \cos 90^0\;\;\;\;\;\Rightarrow \;\;\;\;\;\left [ U=0 \right ]$

if $\dpi{120} \fn_cm \theta=180^0$

$\dpi{120} \fn_cm U=-PE\cos 180^0$ ( $\dpi{120} \fn_cm \vec{P}\;and \; \vec{E}$ are in the opposite direction )

$\dpi{120} \fn_cm \left [ U_{max}=PE \right ]$ ( maximum P.E in this condition) (Unstable Equilibrium)

NOTE:-

If we calculate work done by external agents use $\dpi{120} \fn_cm W_{ext}=\Delta U$

Q: Why do we take it $\dpi{120} \fn_cm \theta=90^0$ as P.E of dipole is zero

Ans: If we hold the dipole perpendicular to the uniform electric field and bring it from infinity into the field, then the work done on charge $\dpi{120} \fn_cm -q$ by the external agent $\dpi{120} \fn_cm (-qV)$ is equal to the work done on charge $\dpi{120} \fn_cm +q$ which is ( $\dpi{120} \fn_cm (+qV)$ . The net workdone on the dipole will be zero and hence P.E is zero

Exercise 3

(1) An electric dipole with dipole moment $\dpi{120} \fn_cm 4\times10^{-9}Cm$ is aligned at $\dpi{120} \fn_cm 30^0$ with the direction of a uniform electric field of magnitude $\dpi{120} \fn_cm 5\times10^4 N/C$ . Calculate the magnitude of the torque acting on the dipole.

(2) An electric dipole is placed at an angle $\dpi{120} \fn_cm 60^0$ with an uniform electric field of magnitude $\dpi{120} \fn_cm 4\times10^5 N/C$ . It experience a torque of $\dpi{120} \fn_cm 8\sqrt{3}N-m$ . If the length of dipole is 4 cm, determine the magnitude of either charge. $\dpi{120} \fn_cm \left ( 10^{-3}C \right )$

(3) An electric dipole consists of two charges of 0.1μC separated by a distance of 2 cm. The dipole is placed in an uniform electric field of magnitude $\dpi{120} \fn_cm 10^5 N/C$ . What maximum torque does the field exert on the dipole? $\dpi{120} \fn_cm (2\times 10^{-4}N.m)$

(4) Two particles, carrying charges -q and +q and having equal masses m each, are fixed at the ends of a light rod of length d to form a dipole. The rod is clamped at an end and is placed in uniform electric field E with the axis of the dipole along the electric field. The rod is slightly titled and then released. Neglect gravity finds the time period of small oscillations. $\dpi{120} \fn_cm \left ( 2\pi \sqrt{\frac{md}{qE}} \right )$

(5) An electric dipole consists of two charges of $\dpi{120} \fn_cm \pm 16\times10^{-19}C$ separated by a distance of $\dpi{120} \fn_cm 3.9\times10^{-12}m$ . The dipole is placed in a uniform electric field of $\dpi{120} \fn_cm 10^5 N/C$ . Calculate

(a) The electric dipole moment
(b) P.E of the dipole in the stable equilibrium position.

(6) Two point charges $\dpi{120} \fn_cm \pm 3.2\times10^{-19}C$ are situated at a distance of $\dpi{120} \fn_cm 2.4\times10^{-10}m$ from each other and constitute an electric dipole. What is the work done in rotating the dipole through $\dpi{120} \fn_cm 180^0$ from the stable equilibrium position in a uniform electric field of $\dpi{120} \fn_cm 4\times10^5 V/m$ ? $\dpi{120} \fn_cm (6.14\times10^{-23}J)$

Electric Dipole in a non-uniform electric field

If an electric dipole is placed in a non-uniform electric field, there is a net force on the dipole in addition to the torque tending to align the dipole with the field.

The net force depends on

(a) The orientation of the dipole w.r.t the electric field.
(b) The dipole moment of the dipole.
(c) How rapidly the field varies in space.

$\dpi{120} \fn_cm {\color{Red} \textbf{Case 1:-}}$ When $\dpi{120} \fn_cm \vec{P}$ is parallel to $\dpi{120} \fn_cm \vec{E}$

i.e. Dipole moves towards the region of a strong electric field i.e. in the direction of increasing electric field.

$\dpi{120} \fn_cm {\color{Red} \textbf{Case 2:-}}$ When $\dpi{120} \fn_cm \vec{P}$ is anti-parallel to $\dpi{120} \fn_cm \vec{E}$

i.e. dipole moves towards the region of weak electric field i.e. in the direction of decreasing electric field.

Force calculation

The net force on the dipole is

$\dpi{120} \fn_cm F=q(E+dE)-qE$

$\dpi{120} \fn_cm F=qdE$

$\dpi{120} \fn_cm F=(qdx)\frac{dE}{dx}$

$\dpi{120} \fn_cm \left [ F=P\left | \frac{dE}{dx} \right | \right ]$ ( usually, this formula is valid when x and E are in the same line)

where $\dpi{120} \fn_cm \frac{dE}{dx}$ is known as the electric field gradient

Special Case:-

A system of two dipoles

Potential energy of a system of two dipoles

Let us take two examples

In case (a)

The electric field at the axial point due to first dipole on the second dipole is

$\dpi{120} \fn_cm E_1=\frac{2kP_1}{r^3}$

Potential energy of the second dipole is

$\dpi{120} \fn_cm U=-\vec{P}_2.\vec{E}_1=-(P_2\hat{i}).\left ( \frac{2kP_1}{r^3}\hat{i} \right )$

$\dpi{120} \fn_cm \left [ U=-\frac{2kP_1P_2}{r^3} \right ]$

In case (b)

The electric field due to the first dipole on the second dipole can be expressed as

$\dpi{120} \fn_cm \vec{E}_1=\frac{2kP_1cos\alpha}{r^3}\hat{i}+\frac{kP_1sin\alpha}{r^3}(-\hat{j})$

and $\dpi{120} \fn_cm \vec{P}_2=P_2\cos\beta\hat{i}+P_2\sin\beta\hat{j}$

The potential energy of interaction is

$\dpi{120} \fn_cm U=\vec{P}_2.\vec{E}_1$

$\dpi{120} \fn_cm =-\left [ \frac{2kP_1\cos\alpha P_2\cos\beta}{r^3}-\frac{kP_1\sin\alpha P_2\sin\beta}{r^3} \right ]$

$\dpi{120} \fn_cm \left [ U=\frac{kP_1P_2}{r^3}\left [ \sin\alpha\sin\beta-2\cos\alpha\cos\beta \right ] \right ]$

Force between two small dipoles placed at a very large distance

If we take the dipoles are very small and placed at a large distance, then we can assume the electric field is nearly non-uniform like $\dpi{120} \fn_cm E=f(x)$ or varies along electric field lines

Generally, two methods can be used to find the force between two dipoles

First method:- Calculate the electric field due to the first dipole at the place of the second dipole and use formula $\dpi{120} \fn_cm F_2=P_2\frac{dE_1}{dr}$ to find the force on the second dipole due to the first dipole,where $\dpi{120} \fn_cm P_2$ is the dipole moment of the second dipole.

Second method:- Calculate the electric field due to the first dipole at the place of the second dipole and use formula $\dpi{120} \fn_cm U=-\vec{P}_2.\vec{E_1}$ to calculate the potential energy between the combination of two dipoles. Finally, use formula $\dpi{120} \fn_cm F=-\frac{dU}{dr}$ to calculate the force on the second dipole due to the first dipole.

Let’s Take an example

Dipole P2 is placed in a non-uniform electric field due to dipole P1

The Electric Field due to P1 at a distance r on the axial line is given by

$\dpi{120} \fn_cm E_1=\frac{2kP_1}{r^3}$

$\dpi{120} \fn_cm \therefore \frac{dE_1}{dr}=\frac{-6kP_1}{r^3}$

$\dpi{120} \fn_cm \therefore$ Force on P2 due to P1 is $\dpi{120} \fn_cm F_2=P_2\left | \frac{dE_1}{dr} \right |$

$\dpi{120} \fn_cm \left [ F_2=\frac{6kP_1P_2}{r^4}\right ]\;\;\;\;\;(Attractive)$

——————————————————————————

We can use another process ( Generally we use this method)

We know that the electric field due to dipole P1 at an axial point at distance r is given by

$\dpi{120} \fn_cm E_1=\frac{2kP_1}{r^3}$

Hence, the Potential energy of two dipole systems is given by

$\dpi{120} \fn_cm U=-\vec{P}_2.\vec{E}_1=-P_2E_1\cos 0^0=-P_2.E_1$

$\dpi{120} \fn_cm U=-\frac{2kP_1P_2}{r^3}$

The force between dipoles is

$\dpi{120} \fn_cm F=-\frac{dU}{dr}$

$\dpi{120} \fn_cm \left [ F=\frac{6kP_1P_2}{r^4}\right ]\;\;\;\;\;(Attractive)$

Dipole in different situations

for situation (5) if the force on both charges of dipole P2 is different then the net force between dipoles is not zero, in this case, we solve this by the long process method as

Electric field on the equatorial line at a distance r from the first dipole is

$\dpi{120} \fn_cm E=\frac{kp}{r^3}$

Now, the force acting on the charge +q in the second dipole is

$\dpi{120} \fn_cm F_1=\frac{qkp}{(r-d)^3}\;\;\;(downward)$

The force acting on the charge -q in the second dipole is

$\dpi{120} \fn_cm F_2=\frac{qkp}{(r-d)^3}\;\;\;(upward)$

so, the net force on the second dipole due to the first dipole is

$\dpi{120} \fn_cm F=F_1-F_2\;\;\;(downward)$

$\dpi{120} \fn_cm F=qkp\left [ \frac{1}{(r-d)^3}-\frac{1}{(r+d)^3} \right ]$

Sole it, we get

$\dpi{120} \fn_cm \left [ \frac{3kp_1p_2}{r^4} \right ]\;\;\;(Perpendicular\;to\;r)$

Exercise 4

(1) The electric dipole moment $\dpi{120} \fn_cm \vec{p}=p\hat{i}$ is kept at a point (x,y) in an electric field $\dpi{120} \fn_cm \vec{E}=4xy^2\hat{i}+4x^2y \hat{j}$ . Find the magnitude of the force acting on the dipole. $\dpi{120} \fn_cm \left ( 4py\sqrt{y^2+4x^2} \right )$

(2) Find the net torque due to the interaction of the system of two dipoles of dipole moment P1 and P2 placed perpendicular to each other with distance r. $\dpi{120} \fn_cm \left ( \frac{3kP_1P_2}{r^3} \right )$