Its working depends on the electron emission from a metal cathode due to irradiation of light or other radiations.

Energy of the incident photon is
E = hv = \(\frac { hc }{ \lambda } \) (in joules)
E = \(\frac { hc }{ \lambda e } \) (in eV)
Substituting the known values, we get
E = \(\frac { 6.626\times { 10 }^{ -34 }\times 3\times 10^{ 8 } }{ 300\times { 10 }^{ -9 }\times 1.6\times { 10 }^{ -19 } } \)
E = 4.14 eV
The work function of silver = 4.7 eV. Since the energy of the incident photon is less than the work function of silver, photoelectrons are not observed in this case.

The ejection of electrons from the metal plate when illuminated by light or any electromagnetic radiation of suitable wavelength (or frequency) is called photoelectric effect.

While electric potential is work done per unit charge the potential energy is the total work done in bringing different charges from infinity to their respective positions.
\(V=\frac { { 1 } }{ 4\pi { \varepsilon }_{ 0 }^{ } } .\frac { { q }_{ 1 } }{ r } and\quad U=\frac { { 1 } }{ 4\pi { \varepsilon }_{ 0 }^{ } } .\frac { { q }_{ 1 }{ q }_{ 2 } }{ r } \)
\(U={ q }_{ 2 }\left[ \frac { { 1 } }{ 4\pi { \varepsilon }_{ 0 }^{ } } .\frac { { q }_{ 1 } }{ r } \right] ={ q }_{ 2 }V\)
∴ Electric potential energy = charge q₂ x potential due to charge q₁ at the location of charge q₂.

When an irregular shaped conductor is given positive charge, the electric field near the sharp end is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge. This reduces the total charge of the conductor near the sharp edge. This is called corona discharge.

Electric field lines are a set of continuous lines which represent the electric field in some region of space visually.

By using superposition principle, the net electric field at point A is
\(\vec { E_{ A } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1A }^{ 2 } } \hat { { r }_{ 1A } } +\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 2A }^{ 2 } } \hat { { r }_{ 2A } } \)
where r_1A and r_2A are the distances of point A from the two charges respectively
\(\vec { E_{ A } } =\frac { 9\times 10^{ 4 }\times 1\times { 10 }^{ -6 } }{ (2\times { 10 }^{ -3 })^{ 2 } } (\hat { j } )+\frac { 9\times 10^{ 9 }\times 1\times { 10 }^{ -6 } }{ (2\times 10^{ -3 })^{ 2 } } (\hat { i } )\)\(\)
= 2.25 x 10⁹\(\hat { j } \) + 2.25 x 10⁹\(\hat { i } \) = 2.25 x 10⁹ (\((\hat { i } +\hat { j } )\)
The magnitude of electric field
\(|\vec { { E }_{ A } } |=\sqrt { (2.25\times { 10 }^{ 9 })^{ 2 }+(2.25\times 10^{ 9 })^{ 2 } } \)
= 2.25 x \(\sqrt { 2 } \) x 10⁹ NC^-1
The direction of \(\vec { { E }_{ A } } \) is given by \(\frac { \vec { { E }_{ A } } }{ |\vec { { E }_{ A } } | } =\frac { 2.25\times 10^{ 9 }(\hat { i } +\hat { j } ) }{ 2.25\times \sqrt { 2 } \times { 10 }^{ 9 } } =\frac { (\hat { i } +\hat { j } ) }{ \sqrt { 2 } } \), which is the unit vector along OA as shown in the figure.

The acceleration experienced by an electron placed at point A is
\(\vec { a_{ A } } =\frac { \vec { F } }{ m } =\frac { q\vec { { E }_{ A } } }{ m } \)
=\(\frac { (-1.6\times 10^{ -19 })\times (2.25\times 10^{ 9 })(\hat { i } +\hat { j } ) }{ 9.1\times { 10 }^{ -31 } } \)
= -3.95 x 10²⁰ \((\hat { i } +\hat { j } )\)Nkg^-1
The electron is accelerated in a direction exactly opposite to \(\vec { { E }_{ A } } \).

\(K .E_{\max } =\mathrm{hv}-\phi  \)
\(=\frac{\mathrm{hc}}{\lambda}-\phi  \)
\(\mathrm{E}  =\frac{6.6 \times 10^{-34} \times 3 \times 10^8-1.235}{500 \times 10^{-9} \times 1.6 \times 10^{-19}}  \)
\(=2.475-1.235  \)
\(\text {K. } \mathrm{E}_{\max } =1.24 \mathrm{eV}\)

\(\mathrm{P}  =\frac{\mathrm{n}}{\mathrm{t}} \frac{\mathrm{hc}}{\lambda}  \)
\(\frac{\mathrm{n}}{\mathrm{t}}  =\frac{\mathrm{P} \lambda}{\mathrm{hc}}  \)
\(\frac{\mathrm{n}}{\mathrm{t}}  =\frac{3.8 \times 10^{26} \times 550 \times 10^{-9}}{6.6 \times 10^{-3} \times 3 \times 10^8}=1 \times 10^{-15}\)

\(\frac{\mathrm{hc}}{\lambda}  =\phi+\mathrm{K} . \mathrm{E}  .....(1)  \)
\(\frac{2 \mathrm{hc}}{\lambda}  =\phi+3 \mathrm{~K} . \mathrm{E}......(2)\)
multiply eqn. (1) by 3, we get
\(\frac{3 \mathrm{hc}}{\lambda}  =3\phi+3 \mathrm{~K} . \mathrm{E}  .....(3)\)
Subtract eqn. (2) from (3), we get
\(\frac{\mathrm{hc}}{\lambda}  =2\phi \)
\(\phi=\frac{ \mathrm{hc}}{2\lambda}  \)

\(\lambda\propto \frac{1}{\sqrt{V}}\)
\(\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{224\times10^3}}{\sqrt{14\times 10^3}}\)
\(=\sqrt{16}=4\)
\(\lambda_{\mathrm{2}}= \frac{\lambda_1}{4}\)

\(\mathrm{E}_{\mathrm{p}}  =\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}}  \)
\(\mathrm{E}_{\mathrm{e}}  =\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda_{\mathrm{e}}^2}  \)
\(\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}}  =\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda_{\mathrm{e}}^2}  \)
\(\lambda_{\mathrm{p}}  \propto \lambda_{\mathrm{e}}^{{ }^2}\)

Q = q₁ + q₂ = 4 x 10^-2C
\(q_{2f}=Q[\frac{r_2}{r_1+r_2}]\)
= 4 x 10^-2 \([\frac{3}{4}]\)
q_2f = 3 x 10^-2 C

Energy density u_E \(=\frac{U}{volume}\)
If A' = 2A     d ' = 2d
Then V ' = 2A  x 2d = 4Ad = 4V
Then volume would be increased. So, energy density will change.

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

\(\vec {E}\) = 10x\(\hat{i},\)  when x = 2 m
\(\vec {E}\) = 10 x 2 x \(\hat{i}\) = 20\(\hat{i}\)
Since, \(E=\frac{-dV}{dx}\therefore V=+20 V\)

Force ∝ charge 
After the separation, the magnitude of charge will be increased. So the force will be more than before.

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

Davisson - Germer experiment
(i) The filament F is heated by a low tension (L . T) battery. Electrons are emitted from the hot filament by thermionic emission.
(ii) They are then accelerated due to the potential diference between the filament and the anode aluminum cylinder by a high tension (H.T) battery.
(iii) Electron beam is collimated by using two thin aluminum diaphragms and is allowed to strike a single crystal of Nickel.
(iv) The electrons scattered by Niatoms in diflerent directions are received by the electron detector which measures the intensity of scattered electron beam.
(v) The detector is capable of rotation in the plane of the paper, so that the angle (\(\theta\)) between the incident beam and the scattered beam can be changed at our will.
(vi) The intensity of the scattered electron beam is measured as a function of the angle \(\theta\).

(i) Figure shows the variation of intensity of the scattered electrons with the angle \(\theta\) for the accelerating voltage of 54 V.
(ii) For a given accelerating voltage V, the scattered wave shows a peak or maximum at an angle of 50^o to the incident electron beam.
(iii) This peak in intensity is attributed to the constructive interference of electrons diffracted from various atomic layers of the target material.


(iv) From the known value of interplanar spacing of Nickel, the wavelength of the electron wave has been experimentally calculated as 1.65\(\overset { o }{ A }\).
(v) The wavelength can also be calculated from de Broglie relation for V = 54 V from equation as
\(\lambda =\cfrac { 12.27 }{ \sqrt { V } } \overset { o }{ A } =\cfrac { 12.27 }{ \sqrt { 54 } } \) 
\(\lambda =1.67\overset { o }{ A } \)
(vi) This value agrees very well with the experimentally observed wavelength of 1.65 \(\overset { o }{ A }\). Thus this experiment directly verifies de Broglie's hypothesis of the wave nature of moving particles.

Principle:
(i) The wave nature of the electron is used in the construction of microscope called electron microscope.
(ii)The resolving power of a microscope is inversely proportional to the wavelength of the radiation used for illuminating the object under study.
(iii) Higher magnification as well as higher resolving power can be obtained by employing the waves of shorter wavelengths.
(iv) De Broglie's wavelength of electron is very much less than (a few thousand less) that of the visible light being used in optical microscopes.
(v) As a result, the microscopes employing de Broglie waves of electrons have very much higher resolving power than optical microscope.
(vi)  Electron microscopes giving magnification more than 2,00,000 times are common in research laboratories.

Working:
(i) The construction and working of an electron microscope is similar to that of an optical microscope except that in electron microscope focussing of electron beam is done by the electrostatic or magnetic lenses.
(ii) The electron beam passing across a suitably arranged either electric or magnetic fields undergoes divergence or convergence thereby focussing of the beam is done.
(iii) The electrons emitted from the source are accelerated by high potentials.
(iv) The beam is made parallel by magnetic condenser lens; When the beam passes through the sample whose magnified image is needed, the beam carries the image of the sample.
(v) With the help of magnetic objective lens and magnetic projector lens system, the magnified image is obtained on the screen. These electron microscopes are being used in almost all brands of science.

(i) In metals, the electrons in the outer most shells are loosely bound to the nucleus. Even at room temperature, there are a large number of free electrons which are moving inside the metal in a random manner. Through they move freely inside the metal they cannot leave the surface of the metal. The reason is that when free electrons reach the surface of the metal they are attracted by the positive nuclei of the metal. It is attractive pull which will not allow free electrons to leave the metallic surface at room temperature.
(ii) In order to leave the metallic surface, the free electrons must cross a potential barrier created by the positive nuclei of the metal. The potential barrier. which prevents free electrons from leading the metallic surface is called surface barrier.
(iii) Whenever an additional energy is given to the free electrons, they will have sufficient energy to cross the surface barrier. And they escape from the metallic surface. The liberation of electrons from any surface of a substance is called electron emission.
(iv) The minimum energy needed for an electron to escape from the metal surface is called work function of that metal.
(a) Thermionic emission
(i) When a metal is heated to a high temperature, the free electrons on the surface of the metal get sufficient energy in the form of thermal energy so that they are emitted from the metallic surface. This type of emission is known a thermonic emission.

(ii) The intensity of the thermionic emission (the number of electrons emitted) depends on the metal used and its temperature.
(iii) Examples: cathode ray tubes, electron microscopes, X-ray tubes etc.

(b) Field emission
(i) Electric field emission occurs when a very strong electric field is applied across the metal.
(ii) This strong field pulls the free electrons and helps them to overcome the surface barrier of the metal.
(iii) Ex: Field ermssion scanning electron microscopes, Field-emission display etc.

(c) Photo electric emission
(i) When an electromagnetic radiation of suitable frequency is incident on the surface of the metal, the energy is transferred from the radiation to the free electrons.
(ii) Hence, the free electrons get sufficient energy to cross the surface barrier and the photo electric emission takes place.
(iii) The number of electrons emitted depend on the intensity of the incident radiation.
(iv) Examples: Photo diodes, photo electric cells etc. 

(d) Secondary emission
(i) When a beam of fast-moving electrons strikes the surface of the metal, the kinetic energy of the striking electrons is transferred to the free electrons on the metal surface.
(ii) Thus the free electrons get sufficient kinetic energy so that the secondary emission of electron occurs.
(iii) Examples: Image intensifiers, photo multiplier tubes etc.

(a) Capacitor in series
(i) Consider three capacitors of capacitance C₁, C₂ and C₃ connected in series with a battery of voltage V as shown in the Figure (a).
(ii) As soon as the battery is connected to the capacitors in series, the electrons of charge -Q are transferred from negative terminal to the right plate of C₃ which pushes the electrons of same amount -Q from left plate of C₃ to the right plate of C₂ due to electrostatic induction.

(iii) Similarly, the left plate of C₂ pushes the charges of -Q to the right plate of C₁ which induces the positive charge +Q on the left plate of C₁.
(iv) At the same time, electrons of charge -Q are transferred from left plate of C₁ to positive terminal of the battery.
(v) By these processes, each capacitor stores the same amount of charge Q.
(vi) The capacitances of the capacitors are in general different so that the voltage across each capacitor is also different and are denoted as V₁, V₂ and V₃ respectively.
(vii) The sum voltage across capacitor must be equal to the voltage of the battery.
V = V₁ + V₂ + V₃     ....(1)
Since, Q = CV
We have V = \(\frac { Q }{ { C }_{ 1 } } +\frac { Q }{ { C }_{ 2 } } +\frac { Q }{ { C }_{ 3 } } \)
\(=Q\left[ \frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \right] ...(2)\)
(viii) If three capacitors in series are considered to form an equivalent single capacitor C_s shown in Figure (b), then we have \(V=\frac { Q }{ { C }_{ s } } \). Substituting this expression into equation (2), we get
\(\frac { Q }{ { C }_{ s } } =Q\left( \frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \right) \)
\(\frac { 1 }{ { C }_{ s } } =\frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \)            ...(3)
(ix) Thus, the inverse of the equivalent capacitance C_s of three capacitors connected in series is equal to the sum of the inverses of each capacitance. This equivalent capacitance C_s, is always less than the smallest individual capacitance in the series.
(b) Capacitor in parallel
i) Consider three capacitors of capacitance C₁, C₂ and C₃ connected in parallel with a battery of voltage V as shown in Figure (a).
ii) Since corresponding sides of the capacitors are connected to the same positive and negative terminals of the battery, the voltage across each capacitor is equal to the battery's voltage.

iii) Since capacitance of the capacitors is different, the charge stored in each capacitor is not the same. Let the charge stored in the three capacitors be Q₁, Q₂, and Q₃ respectively.
iv) According to the law of conservation of total charge, the sum of these three charges is equal to the charge Q transferred by the battery,
Q = Q₁ + Q₂ + Q₃ ...(4)  
Now, since Q = CV, we have
Q = C₁V + C₂V + C₃V ...(5)
(v) If these three capacitors are considered to form a single capacitance C, which stores the total charge Q as shown in the Figure (b), then we can write Q = C_pV. Substituting this in equation (2), we get
C_pV = C₁V + C₂V + C₃V 
C_{p =}  C₁+ C₂ + C₃   ...(6)
vi) Thus, the equivalent capacitance of capacitors connected in parallel is equal to the sum of the individual capacitances.
vii) The equivalent capacitance C_p, in a parallel connection is always greater than the largest individual capacitance. In a parallel connection, it is equivalent as area of each capacitance adds to give more effective area such that total capacitance increases.

Effect of dielectrics in capacitors:
Suppose dielectrics like mica, glass or paper are introduced between the plates, then the capacitance of the capacitor is altered. The dielectric can be inserted into the plates in two different ways.
(i) when the capacitor is disconnected from the battery.
(ii) when the capacitor is connected to the battery.
(i) When the capacitor is disconnected from the battery
Consider a capacitor with two parallel plates each of cross-sectional area A and are separated by a distance d. The capacitor is charged by a battery of voltage V_o and the charge stored is Q_o. The capacitance of the capacitor without the dielectric is,
\(C_{0}=\frac{Q_{0}}{V_{0}}\)        ....(i)
The battery is then disconnected from the capacitor and the dielectric is inserted between the plates. This is shown in Figure.

The introduction of dielectric between the plates will decrease the electric field. Experimentally it is found that the modified electric field is given by,
\(E=\frac{E_{0}}{\varepsilon_{r}}\)    ....(2)
Here E_o is the electric field inside the capacitors when there is no dielectric and \(\varepsilon_{\mathrm{r}}\) is the relative permittivity of the dielectric or simply known as the dielectric constant. Since \(\varepsilon_{\mathrm{r}}\) > 1, the electric field E < E_o.
As a result, the electrostatic potential difference between the plates (V = Ed) is also reduced. But at the same time, the charge Q_o will remain constant once the battery is disconnected.
Hence the new potential difference is
\(V=E d=\frac{E_{0}}{\varepsilon_{r}} d=\frac{V_{0}}{\varepsilon_{r}}\)         .......(3)
We know that capacitance is inversely proportional to the potential difference. Therefore as V decreases, C increases.
Thus new capacitance in the presence of a dielectric is
\(C=\frac{Q_{0}}{V}=\varepsilon_{r} \frac{Q_{0}}{V_{0}}=\varepsilon_{r} C_{0}\)           ......(4)
Since \(\varepsilon_{\mathrm{r}}\) > 1, we have C > C_o. Thus insertion of the dielectric increases the capacitance.
We know that, C_o =\(\frac{\varepsilon_{\mathrm{o}} A}{d}\)  ...........(5)
Equation (4) ⇒ \(C=\frac{\varepsilon_{r} \varepsilon_{0} A}{d}=\frac{\varepsilon A}{d} \)           ..........(6)
where \(\varepsilon=\varepsilon_{\mathrm{r}} \varepsilon_{\mathrm{o}}\) is the permittivity of the dielectric medium.
The energy stored in the capacitor before the insertion of a dielectric is given by,
\(U_{0}=\frac{1}{2} \frac{Q_{0}^{2}}{C_{0}}\)           .......(7)
After the dielectric is inserted, the charge Q_o remains constant but the capacitance is increased. As a result, the stored energy is decreased.
\(U=\frac{1}{2} \frac{Q_{0}^{2}}{C}=\frac{1}{2} \frac{Q_{0}^{2}}{\varepsilon_{r} C_{0}}=\frac{U_{0}}{\varepsilon_{r}}\)        ..........(8)
Since \(\varepsilon_{\mathrm{r}}\) > 1we get U < U_o. There is a decrease in energy because, when the dielectric is inserted, the capacitor spends some energy in pulling the dielectric inside.
(ii) When the battery remains connected to the capacitor:
When the battery of voltage v_o remains connected to the capacitor and the dielectric is inserted into the capacitor, then
(a) The potential difference v_o across the plates remains constant.
(b) The charge stored in the capacitor is increased by a factor \(\varepsilon_{\mathrm{r}}\). (Experimentally found).

\({Q}=\varepsilon_{r} Q_{0}\)    .......(1)
Due to this increased charge, the capacitance is also increased. The new capacitance is,
\(C=\frac{Q}{V_{0}}=\varepsilon_{r} \frac{Q_{0}}{V_{0}}=\varepsilon_{r} C_{0}\)        ......(2)
However the reason for the increase in capacitance in this case where the battery remains connected is different from the case when the battery is disconnected before introducing the dielectric.
The energy stored in the capacitor before the insertion of a dielectric is given by,
\(U_{0}=\frac{1}{2} C_{0} V_{0}^{2}\)        ....(4)
After the dielectric is inserted, the capacitance is increased; hence the stored energy is also increased.
\( U=\frac{1}{2} C V_{0}^{2}=\frac{1}{2} \varepsilon_{r} C_{0} V_{0}^{2}=\varepsilon_{r} U_{0} \)         .....(5)
\(Since \ \varepsilon_{r}>1\ we \ have \ U>U_{o}.\)
Note: Here we have not used the expression \(U_o=\frac{1}{2}\frac{Q_0^2}{C_0}\)because here, both charge and capacitance are changed, whereas in equation (4), V_o remains constant.
Since voltage between the capacitor V_o is constant, the electric field between the plates also remains constant .The energy density is given by,
\(u=\frac{1}{2} \varepsilon E_{0}^{2}\)      ..(6)
where ε is the permittivity of the given dielectric material.

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.

(i) In photo emissive cell emission of electrons from a metal cathode is due to irradiation of light or other radiations.
(ii) In photo voltaic cell, sensitive element made of semiconductor is used which generates voltage proportional to the intensity of light or other radiations.
(iii) In photo conductive cell, the resistance of the semiconductor charges in accordance with the radiant energy incident on it.

There is a smallest value of frequency for which the incident light can eject the photo electrons out of the metal. Light of frequencies smaller than v_o cannot eject photoelectrons, no matter how much high is its intensity, Such a frequency is called 'threshold frequency' or 'critical frequency'.

The high voltage in Van de Graaff generator is used to accelerate protons and deuterons for the nuclear disintegrations and other applications.

The electric field due to a collection of point charges at an arbitrary point is simply equal to the vector sum of the electric fields created by the individual point charges. This is called superposition principle of electric fields.

(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.

Laws of photoelectric effect:
(i) For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
(ii) For a given frequency of incident light, the number of photoelectrons emitted is directly proportional to the intensity of the incident light. The saturation current is also directly proportional to the intensity of incident light.
(iii) Maximum kinetic energy of the photoelectrons is independent of the intensity of the incident light.
(iv) Maximum kinetic energy of the photoelectrons from a given metal is directly proportional to the frequency of incident light.
(v) There is no time lag between the incidence of light and the ejection of photoelectrons.

An alpha particle contains 2 protons and 2 neutrons. Therefore, the mass M of the alpha particle is 4 times that of a proton (m_p) (or a neutron) and its charge q is twice that of a proton (+e).
The de Broglie wavelength associated with it is
\(\lambda=\frac { h }{ \sqrt { 2MqV } } =\frac { h }{ \sqrt { 2\times (4m_{ p })\times (2e)\times V } } \)
\(=\frac { 6.626\times { 10 }^{ -34 } }{ \sqrt { 2\times 4\times 1.67\times 10^{ -27 }\times 2\times 1.6\times { 10 }^{ -19 }\times 400 } } \)
\(=\frac { 6.626\times 10^{ -34 } }{ 4\times 20\times { 10 }^{ -23 }\sqrt { 1.67\times 1.6 } } \) = 0.00507 \(\mathring { A }\)

Energy stored in the capacitor
i) Capacitor not only stores the charge but also it stores energy. When a battery is connected to the capacitor, electrons of total charge - Q are transferred from one plate to the other plate. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor.
ii) To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
dW = V dQ
Where \(V=\frac { Q }{ C } \)  .....(1)
iii) The total work done to charge a capacitor is
\(W=\int _{ 0 }^{ Q }{ \frac { Q }{ C } } dQ=\frac { { Q }^{ 2 } }{ 2C } \quad \quad ....(2)\)
This work done is stored as electrostatic potential energy (U_E) in the capacitor.
\({ U }_{ E }=\frac { { Q }^{ 2 } }{ 2C } =\frac { 1 }{ 2 } { CV }^{ 2 },\quad (\therefore Q=CV)\quad ....(3)\)
(iv) This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor.Substituting  \(C=\frac { { \varepsilon }_{ 0 }A }{ d } \) and V = Ed. 
\(U=\frac { 1 }{ 2 } \left( \frac { { \varepsilon }_{ 0 }A }{ d } \right) { (Ed) }^{ 2 }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }(Ad){ E }^{ 2 }\quad \quad \quad \quad \quad ...(4)\)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy density \({ u }_{ E }=\frac { U }{ Volume } \)       
Equation (4) ⇒ \({ u }_{ E }=\frac{1}{2}{ \varepsilon }_{ 0 }{ E }^{ 2 }\).....(5)
(v) From equation (5),
(a) We infer that the energy is stored in the electric field existing between the plates of the capacitor. Once the capacitor is allowed to discharge, the energy is retrieved.
(b) The energy density depends only on the electric field and not on the size of the plates of the capacitor. 
(c) This is true for the electric field due to any type of charge configuration.

Electric potential due to a point charge:
Consider a positive charge q kept fixed at the origin. Let P be a point at distance r from the charge q. This is shown in Figure.

Electrostatic potential at a point P
The electric potential at the point P is
\(V=\int _{ \infty }^{ r }{ \left( -\vec { E } \right) d\vec { r } } =-\int _{ \infty }^{ r }{ \vec { E } } .d\vec { r } \)  ...(1)
Electric field due to positive point charge q is
\(\vec { E } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }^{ 2 } } \hat { r } \)
\(V=\frac { -1 }{ 4\pi { \varepsilon }_{ 0 } } \int _{ \infty }^{ r }{ \frac { q }{ { r }^{ 2 } } \hat { r } .d\vec { r } } \)
The infinitesimal displacement vector, \(d\vec { r } =dr\hat { r } \) and using \(\hat { r } \).\(\hat { r } \) = 1, we have
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int _{ \infty }^{ r }{ \frac { q }{ { r }^{ 2 } } \hat { r } .dr\hat { r } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int _{ \infty }^{ r }{ \frac { q }{ { r }^{ 2 } } dr } } \)
After the integration,
\(V=-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } q{ \left\{ -\frac { 1 }{ r } \right\} }_{ \infty }^{ r }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ r } \)
Hence, the electric potential due to a point charge q at a distance r is
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ r } \)   ....(2)