The minimum energy needed for an electron to escape from the metal surface is called work function of that metal.
Unit: electron volt (eV).

If the scattering of light is by atoms and molecules which have size a yery less than that of the wavelength \({\lambda}\) of light a << \({\lambda}\) the scattering is called Rayleigh's scattering.
The intensity of Rayleigh's scattering is inversely proportional to fourth power of wavelength
\(\mathrm{I} ∝ \frac{1}{\lambda^{4}}\)

Primary focus:

(i) The primary focus F_I is defined as a point where an object should be placed to give parallel emergent rays to the principal axis after passing through lens.
Secondary focus:

(ii) The secondary focus F₂ is defined as a point where all the parallel rays travelling close to the principal axis converge to form an image on the principal axis after passing through lens.

The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors. \(C=\frac{Q}{V}\)
Its unit is Coulomb per volt or farad (F).

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
\( \vec{F}_{1}^{\text { tot }}=\overrightarrow{F_{12}}+\overrightarrow{F_{13}}+\overrightarrow{F_{14}}+\ldots+\vec{F}_{1 n} \)
\(\vec{F}_{1} ^{\text { tot }}=k\left\{\frac{q_{1} q_{2}}{r_{21}^{2}} \hat{r}_{21}+\frac{q_{1} q_{3}}{r_{31}^{2}} \hat{r}_{31}+\frac{q_{1} q_{4}}{r_{41}^{2}} \hat{r}_{41}+\ldots+\frac{q_{1} q_{n}}{r_{n 1}^{2}} \hat{r}_{n 1}\right\}\) 

R = 10 Ω, I = 5 A, t = 5 minutes = 5 x 60 s
H = I² R t
= 5² x 10 x 5 x 60
= 25 x 10 x 300
= 25 x 3000
= 75000 J (or) 75 kJ

Electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having unit area of cross section.

(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.

\(\text { K.E }=81.9 \times 10^{-15} \mathrm{~J} \)
\(\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}=\frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-3} \times 1836 \times 81.9 \times 10^{-15}}} \)
\(\lambda =\mathbf{4 . 0 0} \times 10^{-14} \mathrm{~m} \)

(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
\(\lambda =\cfrac { h }{ \sqrt { 2mK } } \)

Relation between f and R:
C ⇒ Center of curvature
F ⇒ Principal focus
i ⇒ Angle of incidence

The angles From right angle triangles ∆MCR, and ∆MFP,
\(\tan i=\frac{P M}{P C} \text { and } \tan 2 i=\frac{P M}{P F}\)
As the angles are small, tan i = i and tan 2i = 2i.
\(\mathrm{i}=\frac{\mathrm{PM}}{\mathrm{PC}} \text { and } 2 \mathrm{i}=\frac{\mathrm{PM}}{\mathrm{PF}}\)
Simplifying further,
\(2 \frac{\mathrm{PM}}{\mathrm{PC}}=\frac{\mathrm{PM}}{\mathrm{PF}} ; 2 \mathrm{PF}=\mathrm{PC}, \mathrm{R}=2 \mathrm{f}\)
PF is focal length f and PC is the radius of curvature R.
R = 2f (or) f = R/2

(a) Microwaves:
It is produced by special vacuum tubes such as klystron, magnetron and gunn diode. The frequency range of microwaves is 10⁹ Hz to 10¹¹ Hz. These waves undergo reflection and can be polarised.
Uses:
It is used in radar system for aircraft navigation, speed of the vehicle, microwave oven for cooking and very long distance wireless communication through satellites.
(b) X-rays:
lt is produced when there is sudden stopping of high speed electrons at high-atomic number target, and also by electronic transitions among the innermost orbits of atoms. The frequency range of X-rays is from 10¹⁷ Hz to 10¹⁹ Hz. X-rays have more penetrating power than ultraviolet radiation.
Uses:
X-rays are used extensively in studying structures of inner atomic electron shells and crystal structures. It is used in detecting fractures, diseased organs, formation of bones and stones, observing the progress of healing bones. Further, in a finished metal product, it is used to detect faults, cracks, flaws and holes.
(c) Radio waves:
It is produced by accelerated motion of charges in conducting wires. The frequency range is from few Hz to 10⁹ Hz. It obeys reflection and diffraction.
Uses:
It is uses in radio and television communication systems and also in cellular phones to transmit voice communication in the ultra high frequency band.
(d) Visible light:
Visible light is produced by incandescent bodies and also it is radiated by excited atoms in gases. The frequency range is from 4 x 10¹⁴ Hz to 8 x 10¹⁴ Hz. It obeys the laws of interference, diffraction and can be polarised. It exhibits photo-electric effect also.
Uses:
It can be used to study the structure of molecules, arrangement of electrons in external shells of atoms and it causes sensation of vision.

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.

Since the resistors are connected in series, the effective resistance in the circuit
= 4 Ω + 6 Ω = 10 Ω
The Current I in the circuit =\(\frac { V }{ { R }_{ eq } } =\frac { 24 }{ 10 } =2.4A\)
Voltage across 4Ω resistor
V₁= IR₁ = 2.4A x 4Ω = 9.6V
Voltage across 6 Ω resistor
V₂ = IR₂ = 2.4A x 6Ω = 14.4V

\(\vec { B } =\frac { \beta }{ \sqrt { 3 } } =(\hat { i } +\hat { j } +\hat { k } )T\)
Using an equation,
Lorentz force, \(\vec{F}=Il\hat{j}\times\vec B\)
We can get,
Lorentz force \(F=\sqrt { \frac { 2 }{ { 3 } } } \beta Il\)

Electric field between the plates \(= \frac{σ}{ε_0}\)
Electric force on an electron \(= e\frac{σ}{ε_0}\)
Magnetic force on an electron, F = BIl
But, \(I= \frac{e}{t}\)
∵Electron moves in a straight line. So,
E_F = M_F
\(e\frac{σ}{ε_0}=B(\frac{e}{t})l\)
\(\therefore t = ε_0\frac{lB}{σ}\)

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

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

L = 20 x 10^-3H. C = 50 x 10^-6 F, R= 40Ω
enf V = 10 sin 340 t
\(\therefore V_0=10 \mathrm{~V}, \omega=340  \)
\(X_1=1 \omega^{\prime}=20 \times 10^3 \times 340  \)
\(=6800 \times 10^{-1}=6.8 \Omega  \)
\(X_C=\frac{1}{C .}  \)
\(=\frac{1}{50 \times 10^{-\alpha} \times 340}=\frac{10^{\circ}}{17000}=\frac{10^{\prime}}{17}=58.823 \Omega   \)
\(Z=\sqrt{R^2+\left(X_6-X_1\right)^2}  \)
\(=\sqrt{(40)^2+(58.82-6.8)^2}  \)
\(=\sqrt{(40)^2+(52.02)^2}  \)
\(=65.62 \Omega\)
The peak current in the circuit is,
\(I_0=\frac{V_0}{Z}=\frac{10}{65.62}  \)
\(\cos 0=\frac{R}{Z}=\frac{40}{65.62} \)
\(\text{Power loss in A.C. circuit }=V_{r m} 1_{r \rightarrow \infty} \cos \phi \)
\(=\frac{1}{2} V_{\mathrm{o}} I_{\mathrm{c}} \cos \phi  \)
\(=\frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62}\)
\(\frac{2000}{4305.98}\)
= 0.46 W

\(\Phi = \frac{\pi}{3}-\frac{\pi}{3}=0 \)
Power factor = cosФ = cos 0 = 1

The direction of conventional current is always opposite to the direction of flow of electrons.

If voltage is increased from V to 2 V
Then Q₁ = CV
Q₂ = C(2 V) = 2 CV
∴ Q is doubled and C remains same

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

Brown - 1
Black - 0
Yellow - 10⁴
(∴ R = 10 x 10⁴ Ω = 100 kΩ)

\(P=\frac{V^2}{R}=\frac{240 \times 240}{120}=480 \ W\)

(i) When a photon of energy hv is incident on a metal surface, it is completely absorbed by a single electron and the electron is ejected.
(ii) In this process, a part of the photon energy is used for the ejection of the electrons from the metal surface (photoelectric work function Φ₀) and the remaining energy as the kinetic energy of the ejected electron. From the law of conservation of energy,
\(\\ \\ \\ hv=\phi { _{ 0 }+\cfrac { 1 }{ 2 } { mv }^{ 2 } }\)   ......(1) 
(iii) where m is the mass of the electron and v its velocity.

(iv) If  we reduce the frequency of the incident light is reduced, the speed or kinetic energy of photo electrons is also reduced. At some frequency v₀ of incident radiation, the photo electrons are ejected with almost zero kinetic energy.
Then the equation becomes.
\({ hv }_{ 0 }=\phi _{ 0 }\)    ......(2) 
(v) Where v₀ is the threshold frequency. B rewriting the equation, we get 
\(hv={ hv }_{ o }+\cfrac { 1 }{ 2 } { { mv }^{ 2 } }\)   ......(3) 
The equation is known as einstein's photoelectric equation.
(vi) If the electron does not lose energy by internal collisions, then it is emitted with maximum kinetic energy K_max. Then
\({ K }_{ max }=\cfrac { 1 }{ 2 } { mv }^{ 2 }_{ max }\)    ......(4) 
(vii) where v_maxis the maximum velocity of max the electron ejected. The equation (1) is rearranged as follows:
\({ K }_{ max }=hv-{ \phi }_{ 0 }\) 

A graph between maximum kinetic energy K_max of the photoelectron and frequency v of the incident light is a straight line.

(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.

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
\(\therefore \vec{B}=\frac { { \mu }_{ 0 }I }{ 4\pi a } \times 2\hat{n}\Rightarrow\vec { B } =\frac { { \mu }_{ 0 }I }{ 2\pi a } \hat { n } \)
                                                                                                                 

Construction:
Alternator consists of two major parts, namely stator and rotor. As their names suggest, stator is stationary while rotor rotates inside the stator. In any standard construction of commercial alternators, the armature winding is mounted on stator and the field magnet on rotor.
The construction details of stator, rotor and various other components involved in them are given below.

(a) Stator:
The stationary part which has armature windings mounted in it is called stator. It has two  components, namely stator frame, stator core and armature winding.
Stator core:
Stator core or armature core is made up of iron or steel alloy. It is a hollow cylinder and is laminated to minimize eddy current loss. The slots are cut on inner surface of the core to accommodate armature windings.
Armature winding:
Armature winding is the coil, wound on slots provided in the armature core. 
(b) Rotor
Rotor contains magnetic field windings. The magnetic poles are magnetized by DC source. The ends of field windings are connected to a pair of slip rings, attached to a common shaft about which rotor rotates. Slip rings rotate along with rotor. To maintain connection between the DC source and field windings, two brushes are used which continuously slide over the slip rings.

MaxWell's equations in integral form
i) Gauss law in electricity, \(\oint _s\vec{E} \vec{d} A=\frac{Q_{\text {enclosed }}}{\varepsilon_{o}}\)
ii) Gauss law in magnetism \(\oint _s \vec{B} \cdot \vec{d} A=0\)
iii) Faraday's law \(\oint_l \vec E. \vec {d l}=-\frac{d \phi _B}{d t}\)
iv) Ampere-Maxwell's law \(\oint_l \vec {B}. \vec {d l}=\mu_{o} i_c+\mu_{o} \varepsilon_{o} \frac{d}{d t} \oint_s \vec{E} \cdot {d} \vec A\)

Electrostatic potential at a point due to an electric dipole :
(i) Consider two equal and opposite charges separated by a small distance 2a as shown in Figure. The point P is located at a distance r from the midpoint 'O' of the dipole. Let θ be the angle between the line OP and dipole axis AB.

(ii) Let r₁ be the distance of point P from +q and r₂ be the distance of point P from -q.
Potential at P due to charge +q\(=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }_{ 1 } } \)
Potential at P due to charge -q \(=-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }_{ 2 } } \)
Total potential at the point P,
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } q\left( \frac { 1 }{ { r }_{ 1 } } -\frac { 1 }{ { r }_{ 2 } } \right) \) ....(1)
(iii) Suppose if the point P is far away from the dipole, such that (r >> a), then equation (1) can be expressed in terms of r.
By the cosine law for triangle BOP,
\({ r }_{ 1 }^{ 2 }={ r }^{ 2 }+{ a }^{ 2 }-2ra cos\theta \)
\({ r }_{ 1 }^{ 2 }={ r }^{ 2 }\left( 1+\frac { { a }^{ 2 } }{ { r }^{ 2 } } -\frac { 2a }{ r } cos\theta \right) \)
Since the point P is very far from dipole, then(r >> a). Then term \(\frac{a^{2}}{r^{2}}\) is very small and can be neglected. Therefore 
\({ r }_{ 1 }^{ 2 }={ r }^{ 2 }\left( 1-2a\frac { cos\theta }{ r } \right) \)
\((or){ r }_{ 1 }=r{ \left( 1-\frac { 2a }{ r } cos\theta \right) }^{ \frac { 1 }{ 2 } }\)
\(\frac { 1 }{ { r }_{ 1 } } =\frac { 1 }{ r } { \left( 1-\frac { 2a }{ r } cos\theta \right) }^{ -\frac { 1 }{ 2 } }\)
iv) Since \(\frac{a}{r}\) << 1, we can use binomial theorem and retain the terms up to first order. 
\(\frac { 1 }{ { r }_{ 1 } } =\frac { 1 }{ r } \left( 1+\frac { a }{ r } cos\theta \right) ...(2)\)
Similarly applying the cosine law for triangle AOP,
r₂² = r² + a² - 2ra cos (180-θ)
Since cos(180 - θ) = - cos θ we get
r₂²= r² + a² + 2ra cos θ
Neglecting the term \(\frac { { a }^{ 2 } }{ { r }^{ 2 } } \) because (r >> a)
\({ r }_{ 2 }^{ 2 }={ r }^{ 2 }\left( 1+\frac { 2acos\theta }{ r } \right) \)
\({ r }_{ 2 }=r{ \left( 1+\frac { 2acos\theta }{ r } \right) }^{ \frac { 1 }{ 2 } }\)
Using Binomial theorem, we get
\(\frac { 1 }{ { r }_{ 2 } } =\frac { 1 }{ r } \left( 1-a\frac { cos\theta }{ r } \right) \quad \quad ...(3)\)
Substituting equation (3) and (2) in equation (1),
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } q\left( \frac { 1 }{ r } \left( 1+a\frac { cos\theta }{ r } \right) -\frac { 1 }{ r } \left( 1-a\frac { cos\theta }{ r } \right) \right) \)
\(V=\frac { q }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { 1 }{ r } \left( 1+a\frac { cos\theta }{ r } -1+a\frac { cos\theta }{ r } \right) \right) \)
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { 2aq }{ { r }^{ 2 } } cos\theta \)
v) But the electric dipole moment p = 2qa and we get,
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { pcos\theta }{ { r }^{ 2 } } \right) \)
Now we can write p cos\(\theta =\vec { p } .\hat { r } \) where \(\hat { r } \) is the unit vector from the point O to point P. Hence the electric potential at a point P due to an electric dipole is given by
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { \vec { p } .\hat { r } }{ { r }^{ 2 } } (r>>a)\quad ...(4)\)
Equation (4) is valid for distances very large compared to the size of the dipole. But for a point dipole, the equation (4) is valid for any distance.
Special cases:
Case (i) : If the point P lies on the axial line of the dipole on the side of +q, then θ = 0. Then the electric potential becomes
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { p }{ { r }^{ 2 } } \quad \quad ...(5)\)
Case (ii) : If the point P lies on the axial line of the dipole on the side of -q, then θ = 180°, then
\(V=-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { p }{ { r }^{ 2 } } \quad \quad ...(6)\)
Case (iii) : If the point P lies on the equatorial line of the dipole, then θ = 90°. Hence,
V = 0  .....(7)

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.