Energy stored in the capacitor
\(U=\frac { 1 }{ 2 } { Cv }^{ 2 }\quad \quad ...(1)\)
This is rewritten as using \(C=\frac { { \varepsilon }_{ 0 }A }{ d } \& Ed=V\)
\(U=\frac { 1 }{ 2 } \left( \frac { { \varepsilon }_{ 0 }A }{ d } \right) { (Ed })^{ 2 }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }(Ad)\quad { E }^{ 2 }...(2)\)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy density \({ U }_{ E }=\frac { U }{ Volume } \) From equation (4),
We get
\({ u }_{ E }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }{ E }^{ 2 }\quad \quad \quad \quad \quad ...(3)\)
(iv) The energy density depends only on the electric field and not on the size of the plates of the capacitor.

(i) The total electric flux through the closed surface depends only on the charges enclosed by the surface and the charges present outside the surface will not contribute to the flux and the shape of the closed surface which can be chosen arbitrarily.
(ii) The total electric flux is independent of the location of the charges inside the closed surface.
(iii) To arnve at equation \(\Phi =\oint { \overset { \rightarrow }{ E } .d\overset { \rightarrow }{ A } } =\frac { { Q }_{ encl } }{ { \varepsilon }_{ 0 } } \) is chosen a spherical surface. This imaginary surface is called a Gaussian surface. The shape of the Gaussian surface to be chosen depends on the type of charge configuration and the kind of symmetry existing in that charge configuration. The electric field is spherically symmetric for a point charge, therefore spherical Gaussian surface is chosen. cylindrical and planar Gaussian surfaces can be chosen for other kinds of charge configurations.
(iv) In the L.H.S of equation \(\Phi =\oint { \overset { \rightarrow }{ E } .d\overset { \rightarrow }{ A } } =\frac { { Q }_{ encl } }{ { \varepsilon }_{ 0 } } \) the electric field \(\overset { \rightarrow }{ E } \) is due to charges present inside and outside the Gaussian surface but the charge Q_encl denotes the charges which lie only inside the Gaussian surface.
(v) The Gaussian surface cannot pass through any discrete charge but it can pass through continuous charge distributions. It is because, very close to the discrete charges, the electric field is not well defined.
(vi) Gauss law is another form of Coulomb's law and it is also applicable to the charges in motion. Because of this reason, Gauss law is treated as much more general law than Coulomb's law.

(i) Consider a uniform electric field in a region of space. Let us choose an area A normal to the electric field lines as shown in Figure (a). The electric flux for this case is
Φ_E = EA  ...(1)
(ii) Suppose the same area A is kept parallel to the uniform electric field, then no electric field lines pierce through the area A , as shown in Figure (b). The electric flux for this case is zero.
Φ_E = 0  ...(2)
(iii) If the area is inclined at an angle 8 with the field, then the component of the electric field perpendicular to the area alone contributes to the electric flux. The electric field component parallel to the surface area will not contribute to the electric flux. This is shown in Figure (c). For this case, the electric flux
Φ_E = (E cos θ) A   ....(3)
(iv) Further, θ is also the angle between the electric field and the direction normal to the area. Hence in general, for uniform electric field, the electric flux is defined as

\({ \Phi }_{ E }=\overset { \rightarrow }{ E } .\overset { \rightarrow }{ A } \)= EA cos θ
Here, note that \(\overset { \rightarrow }{ A } \) is the area vector \(\overset { \rightarrow }{ A } \)= A\(\hat{n}\)
(v) Its magnitude is simply the area A and the direction is along the unit vector \(\hat{n}\) perpendicular to the area as shown in Figure. Using this definition for flux \({ \Phi }_{ E }=\overset { \rightarrow }{ E } .\overset { \rightarrow }{ A } \), equations (1) and (2) can be obtained as special cases. 
In Figure (a), θ = 0° so\({ \Phi }_{ E }=\overset { \rightarrow }{ E } .\overset { \rightarrow }{ A } \) = EA
In Figure (b),θ = 90^o so \({ \Phi }_{ E }=\overset { \rightarrow }{ E } .\overset { \rightarrow }{ A } \)= 0

(i) The electric potential at a point P due to a collection of charges q₁, q₂, q₃, ···q_n is equal to sum of the electric potentials due to individual charges.
\({ V }_{ tot }=\frac { k{ q }_{ 1 } }{ { r }_{ 1 } } +\frac { { kq }_{ 2 } }{ { r }_{ 2 } } +\frac { { kq }_{ 3 } }{ { r }_{ 3 } } +...\frac { { kq }_{ n } }{ { r }_{ n } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } { \sum { } }_{ i=1 }^{ n }\frac { { q }_{ i } }{ { r }_{ i } } \)
(ii) where r₁, r₂, r₃ .... r_n are the distances of q₁, q₂, q₃ ..... q_n respectively from P(Figure).

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.

To calculate the total electrostatic potential energy, we use the following procedure. We bring all the charges one by one and arrange them according to the configuration as shown in Figure.
a) Bringing a charge q₁ from infinity to the point A requires no work, because there are no other charges already present in the vicinity of charge q₁.
b) To bring the second charge q₂ to the point B, work must be done against the electric field created by the charge q₁. So the work done on the charge q₂ is W = q₂ V_1B. Here V_1B is the electrostatic potential due to the charge q₁ at point B.
\(U_I=\frac{1}{4\piε_o}\frac{q_1q_2}{r_{12}}\)   .....(1)
Note that the expression is same when q₂ is brought first and then q₁ later.
c) Similarly to bring the charge q₃ to the point C, work has to be done against the total electric field due to both the charges q₁ and q₂. So the work done to bring the charge q₃ = q₃ (V_1C + V_2C). Here V_1C is the electrostatic potential due to charge q₁ at point C and V_2C is the electrostatic potential due to charge q₂ at point C.
The electrostatic potential is
\(U_{II}=\frac{1}{4\piε_o}(\frac{q_1q_2}{r_{13}}+\frac{q_2q_3}{r_{23}})\)    .....(2)
d) Adding equations (1) and (3), the total electrostatic potential energy for the system of three charges q₁, q₂ and q₃ is
U = U_I + U_II
\(U=\frac{1}{4\piε_o}(\frac{q_1q_2}{r_{12}}+\frac{q_2q_3}{r_{13}}+\frac{q_2q_3}{r_{23}})\)  ....(3)

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) 

Case (a)
The magnitude of the electric field at point P is
E_p = \(\frac { 1 }{ 4\pi \varepsilon _{ 0 } } \frac { q }{ { r }^{ 2 } } =\frac { 9\times { 10 }^{ 9 }\times 1\times 10^{ -6 } }{ 4 } \)
= 2.25 x 10³ NC^-1
Since the source charge is positive, the electric field points away from the charge. So the electric field at the point P is given by
\(\bar { { E }_{ p } } \) = 2.25 x 10³ NC^-1
For the point Q
\(|\vec { { E }_{ Q } } |=\frac { 9\times 10^{ 9 }\times 1\times { 10 }^{ -6 } }{ 16 } \) = 0.56 x 10³ NC^-1
Hence \(\vec { { E }_{ Q } } \) = 0.56 x 10³\(\hat { j } \) NC^-1
Case (b)
The magnitude of the electric field at point P
\(\bar { { E }_{ p } } =\frac { kq }{ r^{ 2 } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }^{ 2 } } =\frac { 9\times 10^{ 9 }\times 2\times 10^{ -6 } }{ 4 } \)
= 4.5 x 10³ NC^-1
Since the source charge is negative, the electric field points towards the charge. So the electric field at the point P is given by
\(\vec { E_{ p } } \) = -4.5 x 10³\(\hat { i } \)NC^-1
For the point Q, \(|\vec { { E }_{ Q } } |\frac { 9\times 10^{ 9 }\times 2\times { 10 }^{ -6 } }{ 36 } \)
= 0.5 x 10³ NC^-1
\(\vec { E_{ Q } } \) = 0.5 x 10³\(\hat { i } \)NC^-1
At the point Q the electric field is directed along the positive x-axis.

If the two spheres are neutral, the angle between them will be 0^o when hanged vertically. Since they are positively charged spheres, there will be a repulsive force between them and they will be at equilibrium with each other at an angle of 30° with the vertical. At equilibrium, each charge experiences zero net force in each direction. We can draw a free-body diagram for one of the charged spheres and apply Newton’s second law for both vertical and horizontal directions.
The free-body diagram is shown below

In the x-direction, the acceleration of the charged sphere is zero.
Using Newton’s second law \((\vec { { F }_{ tot }= } m\vec { a } )\), we have
T sinθ\(\hat { i } \) - F_e\(\hat { i } \) =0
T sinθ = F_e    ......(1)
Here T is the tension acting on the charge due to the string and F_e is the electrostatic force between the two charges.
In the y-direction also, the net acceleration experienced by the charge is zero
Tcosθ\(\hat { j } \) - mg\(\hat { j } \) = 0
Tcosθ = mg   ..(2)
By dividing equation (1) by equation (2),
tanθ = \(\frac { { F }_{ e } }{ mg } \)          .....(3)
Since they are equally charged, the magnitude of the electrostatic force is
\({ F }_{ e }=k\frac { { q }^{ 2 } }{ { r }^{ 2 } } \) where k=\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \)
Here r = 2a = 2Lsinθ. By substituting these values in equation (3),
tanθ = k\(\frac { { q }^{ 2 } }{ mg(2Lsin\theta )^{ 2 } } \)       ..........(4)
Rearranging the equation (4) to get q
q = 2 Lsinθ\(\\ \sqrt { \frac { mgtan\theta }{ k } } \)
= 2 x 0.1 x sin 30^o x \(\sqrt { \frac { 10^{ -3 }\times 10\times { tan30 }^{ 0 } }{ 9\times 10^{ 9 } } } \)
q = 8.01 x 10^-8C = 80.1 nC

This is due to the fact that
(i) electric field is radial
(ii) electric field E ∝ \(\frac{1}{R^2}\)

(i) When d battery is disconnected, the capacitance increase from C₀ to C
(ii) New potential difference between the plates
\(V=\frac { { Q }_{ 0 } }{ C } =\frac { { C }_{ 0 }{ V }_{ 0 } }{ C } =\frac { { C }_{ 0 }{ V }_{ 0 } }{ { \varepsilon }_{ r }{ C }_{ 0 } } =\frac { { V }_{ 0 } }{ { \varepsilon }_{ r } } \)

(i) Electric flux through one face \(=\frac { 1 }{ 6 } .\frac { Q }{ { \varepsilon }_{ 0 } } \mu { V }_{ m }\)
(ii) By symmetry the flux through each of the six faces of cube will be same when charge is placed at the centre.
\(\\ \therefore { \phi }_{ E }=\frac { 1 }{ 6 } .\frac { Q }{ { \varepsilon }_{ 0 } } \)
Thus electric flux passing through two opposite faces of the cube
\(=2\times \frac { 1 }{ 6 } .\frac { Q }{ { \varepsilon }_{ 0 } } \)
\(\phi =\frac { 1 }{ 3 } .\frac { Q }{ { \varepsilon }_{ 0 } } \)

When the external electric field applied to a dielectric is very large, it tears the atoms apart so that the bound charges become free charges. Then the dielectric starts to conduct electricity. This is called dielectric breakdown.

Charge q = 50 nC = 50 x 10^-9 C , e = 1.6 x 10^-19 C
No. of electrons \(n=\frac{q}{e}\)
\(\frac { q }{ e } =\frac { 50\times { 10 }^{ -9 } }{ 1.6\times { 10 }^{ -19 } } \)
=  31.25 x 10¹⁰ electrons
To produce 50 nC charge, number of electrons transferred = 31.25 x 10¹⁰ electrons

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.

The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux.

Electrostatic potential energy is the work done to assemble the charges at the given locations brought from infinity. If q₁ and q₂ are the charges to be assembled with a separation of distance r, the electric potential energy is, \(U=\frac{1}{4\piε_o}\frac{q_1q_2}{r}\)

Electric field at the point P at a distance r from the point charge q is the force experienced by a unit positive placed at that point P and is given by
\(\vec { E } =\frac { \vec { F } }{ q_{ 0 } } =\frac { kq }{ { r }^{ 2 } } \hat { r } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }^{ 2 } } \hat { r } \)
where \(\hat{r}\) is the unit vector pointing from q to the point of interest P.

(i) In figure (a), area A₁ encloses the charge Q. So electric flux through this closed surface A₁ is \(\frac { Q }{ { \varepsilon }_{ 0 } } \). But the closed surface A₂ contains no charges inside, so electric flux through A₂ is zero.
(ii) In figure (b), the net charge inside the cube is 3q and the total electric flux in the cube is therefore \(\Phi _{ E }=\frac { 3q }{ { \varepsilon }_{ 0 } } \).
Note that the charge -10 q lies outside the cube and it will not contribute the total flux through the surface of the cube.

Note: A similar problem is solved in XIth Physics volume I, unit 3 section 3.3.2. There are three forces that acts on the mass m:
(i) The downward gravitational force exerted by the Earth (mg)
(ii) The normal force exerted by the inclined surface (N)
(iii) The Coulomb force given by uniform electric field (qE) The free body diagram for the mass m is drawn below.

A convenient inertial coordinate system is located in the inclined surface as shown in the figure. The mass m has zero net acceleration both in x and y-direction.
Along x-direction, applying Newton’s second law, we have
mg sinθ\(\hat { i } \) - qE\(\hat { i } \) = 0
mg sinθ - q E = 0
or, E = \(\\ \frac { mgsin\theta }{ q } \)
Note that the magnitude of the electric field is directly proportional to the mass m and inversely proportional to the charge q. It implies that, if the mass is increased by keeping the charge constant, then a strong electric field is required to stop the object from sliding. If the charge is increased by keeping the mass constant, then a weak electric field is sufficient to stop the mass from sliding down the plane.
The electric field also can be expressed in terms of height and the length of the inclined surface of the plane.
E = \(\frac { mgh }{ qL } \).

(i) q₁ = q₂ = 6.5 x 10^-7C
r = 50 cm = 0.5m
Electrostatic force of repulsion,
\(F=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \)
\(\\ =\frac { 9\times { 10 }^{ 9 }\times { (6.5\times { 10 }^{ -7 }) }^{ 2 } }{ { (0.5) }^{ 2 } } \)
F = 1.521 x 10^-2N
(ii) Now if q₁,q₂ are doubled and r' is halved then F becomes 16 times.
i.e., New force of repulsion, \(F'=16\times \frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \)
F' = 16F
= 16 x 1.521510^-2
F' = 0.24 N.

Length, I = 2a = 4cm = 4 x 10^-2m
Angle, θ = 60°
torque ፔ = 4√3 Nm
Charge, Q = 8 x 10^-9C
We know that, ፔ = pE sine [where p = Q x 2a]
ፔ = (Q x 2a)E sinθ
\(\tau =\frac { \partial }{ Q\times (2a)sin\theta } \)
\(=\frac { 4\sqrt { 3 } }{ 8\times { 10 }^{ -9 }\times 4\times { 10 }^{ -2 }\times { sin\quad 60 }^{ 0 } } \)

∴ Potential energy, U = -pE cosθ
= -Q(2a) x E x cosθ
\(=-8\times { 10 }^{ -9 }\times 4\times { 10 }^{ -2 }\times \frac { 4\sqrt { 3 } \times cos{ 60 }^{ 0 } }{ 8\times { 10 }^{ -9 }\times 4\times { 10 }^{ -2 }sin{ 60 }^{ 0 } } \)
\(U=\frac { -4\sqrt { 3 } }{ \sqrt { 3 } } =-4J\)

Charge of an electron, e = 1.6 x 10^-19C
q = 1C
∴ No of electrons, n = \(\frac{q}{e}\)
\(n=\frac { 1 }{ 1.6\times { 10 }^{ -19 } } \)
= 6.25 x 10¹⁸
n = 6.25 x 10¹⁸ electrons

(i) Electric flux \(\phi =\frac { Total\ enclosed\ charge }{ { \varepsilon }_{ 0 } } \)
Net charge enclosed inside the shell q = 0
∴ electric flux through the shell \(\frac { q }{ { \varepsilon }_{ 0 } } =0\)
(ii) The electric field or net charge inside the spherical conducting shell is zero.
Hence the force on charge \(\frac{Q}{2}\)is zero.
Force on charge at \(A,{ F }_{ A }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { 2Q\left( Q+\frac { Q }{ 2 } \right) }{ { x }^{ 2 } } \)

(i) Consider a positive charge q kept fixed at the origin which produces an electric field \(\overset { \rightarrow }{ E } \) around it.
(ii) A positive test charge q' is brought from point R to point P against the repulsive force between q and q' as shown in Figure. Work must be done to overcome this repulsion. This work done is stored as potential energy.
(iii) The test charge q' is brought from R to P with constant velocity which means that external force used to bring the test charge q' from R to P must be equal and opposite to the coulomb force \(\left( { \overset { \rightarrow }{ E } }_{ ext }=-{ \overset { \rightarrow }{ F } }_{ coloumb } \right) \) 
The work done is
\(W=\int _{ R }^{ P }{ { \overset { \rightarrow }{ F } }_{ ext } } .d\overset { \rightarrow }{ r } \quad \quad \quad ...(1)\)
(iii) Since coulomb force is conservative, work done is independent of the path and it depends only on the initial and final positions of the test charge. If potential energy associated with q' at P is Up and that at R is UR' then difference in potential energy is defined as the work done to bring a test charge q' from point P to R and is given as U_p - U _R = W.
\(\Delta U=\int _{ R }^{ P }{ { \overset { \rightarrow }{ F } }_{ ext } } .d\overset { \rightarrow }{ r } \)
\(Since{ \overset { \rightarrow }{ F } }_{ ext }=-{ \overset { \rightarrow }{ F } }_{ coloumb }=-q'\overset { \rightarrow }{ E } \)
\(\Delta U=\int _{ R }^{ P }{ \left( -q'\overset { \rightarrow }{ E } \right) } .d\overset { \rightarrow }{ r } =q'\int _{ R }^{ P }{ \left( -\overset { \rightarrow }{ E } \right) } .d\overset { \rightarrow }{ r } \)

According to the superposition principle, the total electric potential at a point is equal to the sum of the potentials due to each charge at that point.
Consider the point at which the total potential zero is located at a distance x from the charge +q as shown in the figure.

The total electric potential at P is zero.
V_tot = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { q }{ x } -\frac { 2q }{ (9-x) } \right) \)=0
Which gives \(\frac { q }{ x } -\frac { 2q }{ (9-x) } \)
or \(\frac { 1 }{ x } =\frac { 2 }{ (9-x) } \)
Hence, x = 3m

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

\(Φ=\frac { q_{net} }{ { \varepsilon }_{ 0 } } \)
q_net = - q + q + 2q = 2q
Relative permittivity of water = 80
\(\therefore Φ=\frac { q }{{ \varepsilon }_{ r } { \varepsilon }_{ 0 } } \)
\(=\frac{2q}{{ 80 \times \varepsilon }_{ 0 }}=\frac{q}{{ 40 \varepsilon }_{ 0 }}\)