Coulomb's law states that electrostatic force is directly proportional to the product of the magnitude of the two point charges and is inversely proportional to the square of the distance between the two point charges.
\({ { F } }=\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \)

The total electric charge in the universe is constant and charge can neither be created nor be destroyed. In any physical process, the net change in charge is always be zero.

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) This device consists of a long thick copper rod passing from top of the building to the ground. The upper end of the rod has a sharp spike or a sharp needle as shown in Figure 1.64 (a) and (b).
(ii) The lower end of the rod is connected to the copper plate which is buried deep into the ground. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike.
(iii) Since the induced charge density on thin sharp spike is large, it results in a corona discharge.
(iv) This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud.

(v) The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the earth.
(vi) The lightning arrester does not stop the lightning; rather it diverts the lightning to the ground safely.

(i) A dielectric is a non-conducting material and has no free electrons. The electrons in a dielectric are bound within the atoms. Ebonite, glass and mica are some examples of dielectrics.
(ii) When an external electric field is applied, the electrons are not free to move anywhere but they are realigned in a specific way. A dielectric is made up of either polar molecules or non-polar molecules.

(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) A closed surface is present in the region of the non-uniform electric field as shown in Figure (a). The total electric flux over this closed surface is written as
\({ \Phi }_{ E }=\oint { \overset { \rightarrow }{ E } } .d\overset { \rightarrow }{ A } \quad \quad \quad \quad ...(1)\)
(ii) Note the difference between equations \({ \Phi }_{ E }=\int { \overset { \rightarrow }{ E } . } d\overset { \rightarrow }{ A } \) and (1). The integration in equation (1) is a closed surface integration and for each areal element, the outward normal is the direction of d\(\overset { \rightarrow }{ A } \) as shown in the Figure (b). 

(iii) The total electric flux over a closed surface can be negative, positive or zero. In the Figure (b), it is shown that in one area element, the angle between d\(\overset { \rightarrow }{ A } \) and \(\overset { \rightarrow }{ E } \) is less than 90°, then the electric flux is positive and in another areal element, the angle between d\(\overset { \rightarrow }{ A } \) and \(\overset { \rightarrow }{ E } \) is greater than 90°, then the electric flux is negative.
(iv) In general, the electric flux is negative if the electric field lines enter the closed surface and positive if the electric field lines leave the closed surface.

(i) Suppose the electric field is not uniform and the area A is not flat (Figure), then the entire area is divided into n small area segments \(\Delta { \overset { \rightarrow }{ A } }_{ 1 },\Delta { \overset { \rightarrow }{ A } }_{ 2 },\Delta { \overset { \rightarrow }{ A } }_{ 3 }.....\Delta { \overset { \rightarrow }{ A } }_{ n },\) such that each area element is almost flat and the electric field through each area element is considered to be uniform.
(ii) The electric flux for the entire area A is approximately written as
\({ \Phi }_{ E }={ \overset { \rightarrow }{ E } }_{ 1 }.\Delta { \overset { \rightarrow }{ A } }_{ 1 },{ \overset { \rightarrow }{ E } }_{ 2 }.\Delta { \overset { \rightarrow }{ A } }_{ 2 },{ \overset { \rightarrow }{ E } }_{ 3 }.\Delta { \overset { \rightarrow }{ A } }_{ 3 }.....{ \overset { \rightarrow }{ E } }_{ n }.\Delta { \overset { \rightarrow }{ A } }_{ n },\)
\(\sum _{ i=1 }^{ n }{ { \overset { \rightarrow }{ E } }_{ 1 }.\Delta { \overset { \rightarrow }{ A } }_{ 1 } } \quad \quad \quad ....(1)\)

(iii) By taking the limit \({ \overset { \rightarrow }{ A } }_{ 1 }\rightarrow 0\) (for all i) the summation in equation (1) becomes integration. The total electric flux for the entire area is given by
\({ \Phi }_{ E }=\int { { \overset { \rightarrow }{ E } } } .d\overset { \rightarrow }{ A } \quad\quad ....(2)\)
(iv) From Equation (2), it is clear that the electric flux for a given surface depends on both the electric field pattern on the surface area and orientation of the surface with respect to the electric field.

(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).

(i) The potential energy difference per unit charge is given by
\(\frac { \Delta U }{ q' } =\frac { q'\int _{ R }^{ P }{ (-\overset { \rightarrow }{ E } ) } .d\overset { \rightarrow }{ r } }{ q' } =\int _{ R }^{ P }{ \overset { \rightarrow }{ E } } .d\overset { \rightarrow }{ r } \quad ...(1)\)
(ii) The above equation (1) is independent of q'. The quantity \(\frac { \Delta U }{ q' } =\int _{ R }^{ P }{ \overset { \rightarrow }{ E } } .d\overset { \rightarrow }{ r } \) is called electric potential difference between P and R and is denoted as V_P - V_R = ∆V.
(iii) In other words the electric potential difference is also defined as the work done by an external force to bring unit positive charge from point R to point P.
\({ V }_{ p }-{ V }_{ R }=\Delta V=\int _{ R }^{ P }{ \overset { \rightarrow }{ E } } .d\overset { \rightarrow }{ r } \)
(iv) The electric potential energy difference can be written as ∆U = q' ∆V.

(i) Microwave oven works on the principle of torque acting on an electric dipole. The food we consume has water molecules which are permanent electric dipoles.
(ii) Oven produces microwaves that are oscillating electromagnetic fields and produce torque on the water molecules.
(iii) Due to this torque on each water molecule, the molecules rotate very fast and produce thermal energy. Thus, heat generated is used to cook the food.

If the electric field is not uniform, then the force experienced by +q is different from that experienced by -q. In addition to the torque, there will be net force acting on the dipole.

(i) Suppose a number of point charges are distributed in space, to find the electric field at some point P due to this collection of point charges, superposition principle is used.
(ii) 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 of electric fields.
(iii) Consider a collection of point charges q₁, q₂, q₃,,....q_n located at various points in space. The total electric field at some point P due to all these n charges is given by
\({ \overset { \rightarrow }{ E } }_{ tot }={ \overset { \rightarrow }{ E } }_{ 1 }+{ \overset { \rightarrow }{ E } }_{ 2 }+{ \overset { \rightarrow }{ E } }_{ 3 }+......+{ \overset { \rightarrow }{ E } }_{ n }\quad ...(1)\)
\({ \overset { \rightarrow }{ E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left\{ \frac { { q }_{ 1 } }{ { r }_{ 1p }^{ 2 } } { \hat { r } }_{ 1p }+\frac { { q }_{ 2 } }{ { r }_{ 2p }^{ 2 } } { \hat { r } }_{ 2p }+\frac { { q }_{ 3 } }{ { q }_{ 3p }^{ 2 } } { \hat { r } }_{ 3p }+...\frac { { q }_{ n } }{ { r }_{ nP }^{ 2 } } { \hat { r } }_{ nP } \right\} (2)\)
(ill) Here r_1p, r_2p,r_3p,········r_nP are the distances between the point P and the charges q_1P, q_2P, q_3p..... q_nP respectively. Also \({ \hat { r } }_{ 1p },{ \hat { r } }_{ 2p },{ \hat { r } }_{ 3p }\quad ......{ \hat { r } }_{ nP }\) are the unit vectors directed from q_1p, q_2p,q_3p ...... q_nP respectively to P. Equation (2) can be re-written as,
\({ \overset { \rightarrow }{ E } }_{ tot }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \sum _{ i=1 }^{ n }{ \left( \frac { { q }_{ i } }{ { r }_{ ip }^{ 2 } } \hat { r } ip \right) } ....(3)\)
(iv) For example in Figure, the resultant electric field due to three point charges q₁,q₂,q₃, at point P is shown
Note that the relative lengths of the electric field vectors for the charges depend on relative distances of the charges to the point P.