\(\mathrm{I} ∝ \frac{1}{\lambda^{4}}\)
According to Rayleigh's scattering equation, violet colour which has the shortest wavelength gets much scattered during day time. The next scattered colour is blue. As our eyes are more sensitive to blue colour than violet colour the sky appears blue during day time.

Biot-Savart's law states that, the magnitude of magnetic field \(d\vec { B } \) at a point P at a distance of r from the small elemental length taken on a conductor carrying current varies
(i) directly as the strength of the current I
(ii) directly as the magnitude of the length of element \(\vec { dl } \)
(iii) directly as the sine of the angle θ between \(\vec { dl } \) and \(\hat { r } \).
(iv) inversely as the square of the distance r between the point P and length of element \(\vec { dl } \).
\(d\vec { B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { I\vec { dl } \times \hat { r } }{ { r }^{ 2 } } \)

The thumb, index finger and middle finger of right hand are stretched out in mutually perpendicular directions. If the index finger points the direction of the magnetic field and the thumb indicates the direction of motion of the conductor, then the middle finger will indicate the direction of the induced current.

An electromagnetic waves are the waves that are radiated by an accelerated charge which propagates through space as coupled electric and magnetic fields, oscillating perpendicular to each other and to the direction of propagation of the wave.

(i) Two equal and opposite charges separated by a small distance constitute an electric dipole.
(ii) The magnitude of the electric dipole moment is equal to the product of the magnitude of one of the charges and the distance between them, \(|\vec{p}|=2 q a\).
(iii) The electric dipole moment vector lies along the line joining two charges and is directed from -q to +q.

It there are more than two charges, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
Consider a system of n charges namely q₁, q₂, q₃ ...q_n. The force on q₁ exerted by the charge q₂ is \(\overrightarrow{F_{12}}=k \frac{q_{1} q_{2}}{r_{21}^{2}} \hat{r}_{21}\) .
The force on q₁ exerted by the charge q₃ is \(\overrightarrow{F_{13}}=k \frac{q_{1} q_{3}}{r_{31}^{2}} \hat{r}_{31}\)
By continuing this, the total force acting on the charge q₁ due to all other charges is given by
\( \vec{F}_{1}^{\text { tot }}=\overrightarrow{F_{12}}+\overrightarrow{F_{13}}+\overrightarrow{F_{14}}+\ldots+\vec{F}_{1 n} \)
\(\vec{F}_{1} ^{\text { tot }}=k\left\{\frac{q_{1} q_{2}}{r_{21}^{2}} \hat{r}_{21}+\frac{q_{1} q_{3}}{r_{31}^{2}} \hat{r}_{31}+\frac{q_{1} q_{4}}{r_{41}^{2}} \hat{r}_{41}+\ldots+\frac{q_{1} q_{n}}{r_{n 1}^{2}} \hat{r}_{n 1}\right\}\) 

Applying Kirchoff’s rule to the point P in the circuit,
The arrows pointing towards P are positive and away from P are negative.
Therefore, 0.2A - 0.4A + 0.6A - 0.5A + 0.7A - I = 0
0.7 A - I = 0
1.5A - 0.9A – I = 0
0.6A - I = 0
I = 0.6 A

(i) When the spectrum obtained from the Sun is examined, it consists of large number of dark lines . These dark lines in the solar spectrum are called Fraunhofer lines.
(ii) The Absorption spectra for various materials are compared with the Fraunhofer lines in the solar spectrum, which helps in identifying elements present in the Sun's atmosphere.

(i) The photons of light of frequency (v) and wavelength (λ) will have energy, given by 
\(E=h v=\frac{h c}{\lambda}\) 
(ii) The energy of a photon is determined by the frequency of the radiation and not by its intensity and the intensity has no relation with the energy of the individual photons in the beam.
(iii) The photons travel with the velocity of light and its momentum is given by 
\(p=\frac{h}{\lambda}=\frac{h v}{c}\) 
(iv) Since photons are electrically neutral, they are unaffected by electric and magnetic fields.
(v) When a photon interacts with matter (photon-electron collision), the total energy, total linear momentum, and angular momentum are conserved. Since photons may be absorbed (or) a new photon may be produced in such interactions, the number of photons may not be conserved.

(i) Photo cells are used as switches and sensors.
(ii) Automatic switch on and off of street lights.
(iii) They are used for reproduction of sound in motion pictures
(iv) They are used as timers to measure the speed of athletes during a race.
(v) In photography, they are used to measure the intensity of the given light and to calculate the exact time of exposure.

(i) An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy acquired by the electron is given by
\(\cfrac { 1 }{ 2 } { mv }^{ 2 }=ev\) 
(ii) Therefore, the speed v of the electron is
\(v=\sqrt { \cfrac { 2ev }{ m } } \) 
Hence, the de Broglie wavelength of the matter waves associated with electron is
\(\lambda =\cfrac { h }{ mv } =\cfrac { h }{ \sqrt { 2mev } } \) 
(iii) Substituting the known values in the above equation, we get
\(\lambda =\cfrac { 6.26\times { 10 }^{ -34 } }{ \sqrt { 2V\times 1.6\times { 10 }^{ -19 }\times 9.11\times { 10 }^{ -31 } } } \) 
= \(\cfrac { 12.27\times { 10 }^{ -10 } }{ \sqrt { V } } m\) 
\(\lambda =\cfrac { 12.27 }{ \sqrt { V } } \overset { o }{ A } \)
(iv) Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as
\(\lambda =\cfrac { h }{ \sqrt { 2mK } } \)

Capacitance of a parallel plate capacitor:
(i) Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d as shown in Figure.

(ii) The electric field between two infinite parallel plates is uniform and is given by \(E=\frac { \sigma }{ { \varepsilon }_{ o} } \) where σ is the surface charge density on the plates \(\left( \sigma =\frac { Q }{ A } \right) \).
iii) If the separation distance d is very much smaller than the size of the plate (d² < < A), then the above result is used even for finite-sized parallel plate capacitor.
The electric field between the plates is 
\(E=\frac { Q }{ A{ \varepsilon }_{ 0 } } ...(1)\)
iv) Since the electric field is uniform, the electric potential between the plates having separation d is given by
\(V=Ed=\frac { Qd }{ A{ \varepsilon }_{ 0 } } \quad \quad \quad ...(2)\)
Therefore the capacitance of the capacitor is given by
\(C=\frac { Q }{ V } =\frac { Q }{ \left( \frac { Qd }{ A{ \varepsilon }_{ 0 } } \right) } =\frac { { \varepsilon }_{ 0 }A }{ d } \quad \quad ....(3)\)
(v) From equation (3), it is evident that capacitance is directly proportional to the area of cross section and is inversely proportional to the distance between the plates.

\(\frac { P }{ Q } =\frac { R }{ S } \)
\(S=\frac { Q }{ P } \times R\)
\(S=\frac { 1000 }{ 100 } \times 40S=400\Omega \)

Since the resistances are connected in parallel, therefore, the equivalent resistance in the circuit is
\(\frac { 1 }{ { R }_{ p } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } =\frac { 1 }{ 4 } +\frac { 1 }{ 6 } \)
\(\frac { 1 }{ { R }_{ p } } =\frac { 5 }{ 12 } \Omega \quad or\quad { R }_{ p }=\frac { 12 }{ 5 } \Omega \)
The resistors are connected in parallel, the potential difference (voltage) across them is the same.
\({ I }_{ 1 }=\frac { V }{ { R }_{ 1 } } =\frac { 24V }{ 4\Omega } =6A\)
\({ I }_{ 2 }=\frac { V }{ { R }_{ 2 } } =\frac { 24 }{ 6 } =4A\)
The current I is the sum of the currents in the two branches. Then,
I = I₁ + I₂ = 6 A + 4 A = 10 A

U =\(\vec{p_m}.\vec {B}\)
U = -(0.5\(\hat{i}\) + 0.4\(\hat{j}\)).(0.2\(\hat{j}\)) 
U = 0.1 J

Convex Mirror is diverging in nature and for all positions of objects, convex mirror forms virtual and erect image.

Dipole moment, \(\vec{p}_m=n\times I\times\vec{A}\)
\(\vec{p}_m\) = 50 x 3 x 3.14 x 25 x 10^-4 ≈ 1.2 A m²

Magnetic filed at the centre of a circular
loop, B = \(\frac{μ_oI}{2\pi R}\)
From the figure, R =\(\frac{2r}{\pi}\)
\(\therefore B'=\frac{μ_oI}{2\pi \times\frac{2r}{\pi}}=\frac{μ_oI}{4r}\)
\(B'=\frac { { \mu }_{ ° }I }{ 4r } \bigotimes \)

\(Q_{midpoint}=\frac{Q}{\sqrt{1^2+1^2}}=\frac{Q}{\sqrt2}\)

\(\text {emf } e=8 \mathrm{~V}  \)
\(d I=I_1-I_0=2-(-2)=4 \mathrm{~A}  \)
\(\text {dt }=0.05 \mathrm{~s}  \)
\(L=\frac{-e}{d I / d t}=\frac{-8}{4 / 0.05}  \)
\(=\frac{-8 \times 0.05}{4}=\frac{-0.40}{4}  \)
=-0.1 H
-ve sign indicates that self-induced emf always opposes the current w.r.t. time.

\( { E } ={ E }_{ 0 }\hat { k } \) and \({ B } ={ B }_{ 0 }\hat { j } \)

B_o  = 3 x 10^-6T
Amplitude of electric field, E_o = B_oc
E_o = 3 x 10^-6 x 3 x 10⁸ = 900V m ^-1  

\(U=\frac{1}{4\piε_0}\frac{q_1q_2}{r_{12}}\)
\(i) U=\frac{1}{4\piε_0}\frac{Q(-Q)}{r}=\frac{1}{4\piε_0}[\frac{-Q^2}{r}]\)
\(ii) U=\frac{1}{4\piε_0}\frac{(-Q)(-Q)}{r}=\frac{1}{4\piε_0}[\frac{Q^2}{r}]\)
\(iii) U=\frac{1}{4\piε_0}\frac{Q(2Q)}{r}=\frac{1}{4\piε_0}[\frac{2Q^2}{r}]\)
\(iv) U=\frac{1}{4\piε_0}\frac{Q(-2Q)}{2r}=\frac{1}{4\piε_0}[\frac{-Q^2}{r}]\)
From the values, 1 = 4 < 2 < 3

The charge + q will be stable between B₁ and B₂ with respect to the displacement.

Yellow - 4
Violet - 7
Orange - 10³
Silver - Tolerance - 10%

Resistance, \(R=\frac{V}{I}=\frac{4}{2}=2 \ ohm\)

(i) Let us consider a thin lens made up of a medium of refractive index n₂ is placed in a medium of refractive index n₁. Let R₁ and R₂ be the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole.

(ii) Consider a point object 'O' on the principal axis. A paraxial ray from 'O' which falls very close to P, after refraction at the surface (1) forms image at 1'.
(iii) Before it does so, it is again refracted by the surface (2). Therefore the final image is formed at I.
(iv) The general equation for the refraction at a single spherical surface is given from Equation,
 \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ v} =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ { R } } \)
For the refracting surface (1), the light goes from n₁ to n₂
\(\cfrac { { n }_{ 2 } }{ v' } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ { R }_{ 1 } } \)   .....(1)
For the refracting surface (2), the light goes from n₂ to n₁
\(\frac{n_{1}}{v}+\frac{n_{2}}{v^{\prime}}=\frac{\left(n_{1}-n_{2}\right)}{R_{2}}\)    ......(2)
For surface (2) I' acts as virtual object.
Adding the above two equations (1) and (2)
\(\cfrac { { n }_{ 1 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
on further simplifying and rearranging,
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 }-{ n }_{ 1 } }{ { n }_{ 1 } } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 } }{ n_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ....(3)
If the object is at infinity, the image is formed at the focus of the lens. Thus, for u = \(\infty\), v = f. Then the equation becomes.
\(\cfrac { 1 }{ f } -\cfrac { 1 }{ \infty } =\left( \cfrac { { n }_{ 2 } }{ { { n }_{ 1 } } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ f } =\left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ......(4)
If the lens is kept in air, then we can take n₂ = n and n₁ = 1. So the equation (4) becomes,
\(\\ \cfrac { 1 }{ f } =\left( n-1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)   ..(5)
The above equation is called the lens maker's formula.
Significance:
It tells the lens manufacturers what curvature is needed to make a lens of desired focal length with a material of particular refractive index to make a lens of desired focal length. This formula holds good also for a concave lens.

Cyclotron:
Device used to accelerate the charged particles to gain large kinetic energy.
Principle:
When a charged particle moves perpendicular to the magnetic field, it experiences magnetic Lorentz force.
Construction:
(i) The particles are allowed to move in between two semi-circular metal containers called Dees (hollow D - shaped objects).
(ii) The uniform magnetic field is controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees.
(iii) Source is kept between two Dees.
(vi) Dees are connected to high frequency alternating potential difference.
Working:
(i) The ion ejected from source is positively charged.
(ii) It is accelerated towards negative potential Dees
(iii) This ion undergoes a circular path.
(iv) At this time, the polarities of the Dees are reversed, so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle q is provided by Lorentz force.

\(\frac { m{ v }^{ 2 } }{ r } \) = qvB
⇒ r = \(\frac { m }{ qB } \)v ........(1)
⇒ r ∝ v
(v) If radius of the circular paths, increases, velocity also increases particles undergo spiral path with increasing radius.
(vi) When the frequency f at which the positive ion ciculates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_osc. This is called Resonance condition.
From equation, f = \(\frac { qB }{ 2\pi m } \) we have
f_osc = \(\frac { qB }{ 2\pi m } \),
The time period of oscillation is
T = \(\frac { 2\pi m }{ qB } \)
The kinetic energy of the charged particle is,
KE = \(\frac { 1 }{ 2 } mv^{ 2 }=\frac { { q }^{ 2 }B^{ 2 }{ r }^{ 2 } }{ 2m } \)   ........(2)
Limitations:
(i) The speed of ion is limited.
(ii) Electron cannot be accelerated.
(iii) Uncharged particles cannot be accelerated.

Let YY' be an infinitely long straight conductor carry current I. In order to calculate magnetic field at a point P which is at a distance a from the wire, let us consider a small line element dl (segment AB).
According to Biot Savart law, the magnetic field at a point P due to current element Idl is,
\({ d \vec B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { { Idl sin \theta} }{ { r }^{ 2 } }\hat n \).
To apply trigonometry, draw a perpendicular AC to the line BP as shown in Figure.
In triangle ΔABC, \(\sin \theta=\frac{\mathrm{AC}}{\mathrm{AB}}\)
∴ AC = AB sinθ
\(\text { But, } A B  =d l \Rightarrow A C=d l \sin \theta\)
Let dΦ be the angle subtended between AP and BP
ie., \(\angle \mathrm{APB}=\angle \mathrm{APC}=d \phi\)
In a triangle \(\triangle \mathrm{APC}, \sin (d \phi) \simeq A C / A P\)
Since, dΦ is very small, \(\sin (d \phi) \simeq d \phi\)
But,  \(\mathrm{AP}  =r \Rightarrow A C=r d \phi \)
\(\therefore \mathrm{AC} =d l \sin \theta=r d \phi \)
\(\therefore d \vec{B}  =\frac{\mu_0}{4 \pi} \frac{I}{r^2}(r d \phi) \hat{n}=\frac{\mu_0}{4 \pi} \frac{I d \phi}{r} \hat{n}\)
Let Φ be the angle between AP and OP
\(\text {In a } \triangle \mathrm{OPA}, \cos \phi =\frac{\mathrm{OP}}{\mathrm{AP}}=\frac{\mathrm{a}}{\mathrm{r}}  \)
\(r  =\frac{a}{\cos \phi}  \)
\(\text {Now, } d \vec{B}  =\frac{\mu_0}{4 \pi} \frac{I}{a / \cos \phi} d \phi . \hat{n}  \)
\(d \vec{B}  =\frac{\mu_0 I}{4 \pi a} \cos \phi d \phi \hat{n}\)
The total magnetic field at P due to the conductor YY' is
\(\vec { B } = \int _{- \Phi _{ 1 } }^{ { \Phi }_{ 2} }d\vec B =\int _{ -\Phi _{ 1 } }^{ { \Phi }_{ 2 } }\frac { { \mu }_{ 0 }I }{ 4\pi a }{ cos\phi d\phi } \hat { n }\)
\(=\frac { { \mu }_{ 0 }I }{ 4\pi a }[{ sin\phi ]^{\phi_2} _{\phi_-1}} \hat { n }\)
\( \vec{B}=\frac { { \mu }_{ 0 }I }{ 4\pi a } (sin{ \Phi }_{ 1 }+sin{ \Phi }_{ 2 })\hat { n } \)
For infinitely long conductor, Φ₁ =  Φ₂ = 90^o
\(\therefore \vec{B}=\frac { { \mu }_{ 0 }I }{ 4\pi a } \times 2\hat{n}\Rightarrow\vec { B } =\frac { { \mu }_{ 0 }I }{ 2\pi a } \hat { n } \)
                                                                                                                 

Principle:
The principle of transformer is the mutual induction between two coils. That is, when an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil.

Construction:
(i) In the simple construction of transformers, there are two coils of high mutual inductance wound over the same transformer core.
(ii) The core is generally laminated and is made up of a good magnetic material like silicon steel. Coils are electrically insulated but magnetically linked via transformer core.
(iii) The coil across which alternating voltage is applied is called primary coil P and the coil from which output power is drawn out is called secondary coil S. The assembled core and coils are kept in a container which is filled with suitable medium for better insulation and cooling purpose.
Working:
(i) If the primary coil is connected to a source of alternating voltage, an alternating magnetic flux is set up in the laminated core.
(ii) If there is no magnetic flux leakage, then whole of magnetic flux linked with primary coil is also linked with secondary coil.
(iii) This means that rate at which magnetic flux changes through each turn is same for both primary and secondary coils.
(iv) As a result of flux change, emf is induced in both primary and secondary coils. The emf induced in the primary coil v_p or back of ε_p is given by,
v_p = ε_p = -N_p \(\frac{dФ_B}{dt}\)             ........(1)
(vi) The frequency of alternating magnetic flux in the core is same as the frequency of the applied voltage. Therefore, induced emf in secondary will also have same frequency as that of applied voltage. The emf induced in the secondary coil ε_s is given by ,
ε_s = -N_s\(\frac{dФ_B}{dt}\)  
Where N_p and N_s are the number of turns in the primary and secondary coil respectively. If the secondary circuit is open, then ε_s = ሀ_s  where u is the voltage  ሀ_s across secondary coil
ሀ_s = ε_s = -N_s \(\frac{dФ_B}{dt}\)               ........(2)
From equations (1) and (2),
\(\frac{v_s}{v_p}=\frac{N_s}{N_p}\) = K                    ........(3)
(vii) This constant K is known as voltage transformation ratio. For an ideal transformer, input power v_pi_p= Output power v_si_s 
where i_p and i_s are the currents in the primary and secondary coil respectively.
(ix) Therefore,
\(\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}\)             ........(4)
Equation (4) is written in terms of amplitude of corresponding quantities
\(\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}\) = K
i) If N_s> N_p (or K > 1), ∴ V_s > V_p and I_s <  I_p This is the case of step-up transformer in which voltage is increased and the corresponding current is decreased.
ii) If N_s< N_p (or K < 1), ∴ V_s < V_p and I_s >  I_p This is step-down transformer where voltage is decreased and the current is increased.
Efficiency of a transformer:
The efficiency η of a transformer is defined as the ratio of the useful output power to the input power. Thus,
\(η=\frac{Outpur \ power}{Input \ power}\times100% \) %          ....(5)

Case (i): Electric field due to an electric dipole at points on the axial Iine:
Consider an electric dipole placed on the x-axis as shown in Figure. A point C is located at a distance of r from the midpoint O (of the dipole) along the axial line.

The electric field at a point C due to +q is
\({ \vec { E } }_{ + }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { (r-a) }^{ 2 } } \) along BC
Since the electric dipole moment vector \(\vec { p } \) is from -q to +q and is directed along BC, the above equation is rewritten as
\({ \vec { E } }_{ + }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { (r-a) }^{ 2 } } \hat { p } \)  ....(1)
When \(\vec { p } \) is the electric dipole moment unit vector from -q to +q. The electric field at a point C due to -q is
\({ \vec { E } }_{ - }=-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { (r+a) }^{ 2 } } \hat { p } \)              ....(2)
Since +q is located closer to the point C than -q, \({ \vec { E } }_{ + }\) is stronger than \({ \vec { E } }_{ - }\). Therefore, the length of the \({ \vec { E } }_{ + }\) vector is drawn larger than that of \({ \vec { E } }_{ - }\) vector.
The total electric field at point C is calculated using the superposition principle of the electric field.
\({ \vec { E } }_{ tot }={ \vec { E } }_{ + }+{ \vec { E } }_{ - }\)
\(=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { (r-a) }^{ 2 } } \hat { p } -\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { (r+a) }^{ 2 } } \hat { p } \)   
\({ \vec { E } }_{ tot }=\frac { q }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { 1 }{ { (r-a) }^{ 2 } } -\frac { 1 }{ { (r+a) }^{ 2 } } \right) \hat { p } \)    ....(3)
\({ \vec { E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } q\left( \frac { 4ra }{( { r }^{ 2 }-{ a }^{ 2 })^2 } \right) \hat { p } \)  ...(4)
Note that the total electric field is along \({ \vec { E } }_{ + }\), since +q is closer to C than -q.
If the point C is very far away from the dipole then (r >> a). Under this limit the term \(({ r }^{ 2 }-{ a }^{ 2 })\approx { r }^{ 2 }\).
Substituting this into equation (4), we get
\({ \vec { E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { 4aq }{ { r }^{ 3 } } \right) \hat { p } (r>>a)\)
\(since\quad 2aq\hat { p } =\vec { p } \)
\({ \vec { E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { 2\vec { p } }{ { r }^{ 3 } } (r>>a)\)        ...(5)
The direction \({ \vec { E } }_{ tot }\) is shown in Figure.

NOTE: If the point C is chosen on the left side of the dipole, the total electric field is still in the direction of \(\vec { p } \). 
Case (ii) Electic field due to an electric dipole at a point on the equatorial plane:

Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane. Since the point C is equidistant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of \({ \vec { E } }_{ + }\) is along BC and the direction of \({ \vec { E } }_{ - }\) is along CA. \({ \vec { E } }_{ + }\) and \({ \vec { E } }_{ - }\)  are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. The perpendicular components \(|{ \vec { E } }_{ + }|\) sinθ and \(|{ \vec { E } }_{ -}|\) sinθ are equal in magnitude and oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of \({ \vec { E } }_{ + }\) and \({ \vec { E } }_{ - }\) and its direction is along \(-\hat{p}\) as shown in the Figure.
\({ \vec { E } }_{ tot }=-|{ \vec { E } }_{ + }|cos\theta \hat { p } -|{ \vec { E } }_{ - }|cos\theta \hat { p } \)   ...(6) 
The magnitudes \({ \vec { E } }_{ + }\) and \({ \vec { E } }_{ - }\) are the same and given by,
\(|{ \vec { E } }_{ + }|=|{ \vec { E } }_{ - }|=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ (r^2+a^2) } \)  ...(7)
By substituting equation (7) into equation (6), we get
\({ \vec { E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { 2qcos\theta }{ ({ r }^{ 2 }+{ a }^{ 2 }) } \hat { p } ....(8)\)
\(=-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { 2qa }{ ({ r }^{ 2 }+{ a }^{ 2 })^{ \frac { 3 }{ 2 } } } \hat { p } \)
Since  \(cos \theta =\frac { a }{ \sqrt { { r }^{ 2 }+{ a }^{ 2 } } } \)
\({ \vec { E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { \vec { p } }{ ({ r }^{ 2 }+{ a }^{ 2 })^{ \frac { 3 }{ 2 } } } \)
Since \(\vec { p } \) = 2qa\(\hat { p } \)              ...(9)
At very large distances (r >> a), the equation (9) becomes
\({ \vec { E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { \vec { p } }{ { r }^{ 3 } } (r>>a)\)          ...(10) 
Negative sign shows that direction of Electric field is opposite to the direction of dipole moment vector.

Wheatstone's bridge:
i) An important application of Kirchhoff's rule is Wheatstone's bridge. It is used to compare Resistances and also helps in determining the unknown resistance in electrical network. The bridge consists of four resistances P, Q, R and S connected as shown in Figure.
ii) A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is I_G and its resistance is G.
Applying Kirchhoff's current rule to junction B
I₁ - I_G - I₃ = 0    ..(1)
Applying Kirchhoff's current rule to junction D,
I₂ + I_G - I₄ = 0      ...(2)

Applying Kirchhoff's voltage rule to loop ABDA,
I₁P + I_GG - I₂R = 0      ...(3)
Applying Kirchhoff's voltage rule to loop ABCDA,
I₁P + I₃Q - I₄S - I₂R = 0   ...(4)
(iii) When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B and D, no current flows through galvanometer (I_G = 0). Substituting I_G = 0 in equation (1), (2) and (3), we get
I₁ = I₃  ..(5)
I₂ = I₄  ..(6)
I₁P = I₂R    ..(7)
Substituting the equation (7) in equation (4),
I₃Q = I₄S  .....(8)
Dividing equation (7) by equation (8), we get
\(\cfrac { P }{ Q } =\cfrac { R }{ S } \)   .....(9)
(iv) This is the bridge balance condition. Only under this condition, galvanometer shows null deflection.