(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.

According to the superposition principle, the total electrostatic force on charge q₁ is the vector sum of the forces due to the other charges,
\(\vec { { F }_{ 1 }^{ tot } } =\bar { { F }_{ 12 } } +\bar { { F }_{ 13 } } +\bar { F_{ 14 } } \)
The following diagram shows the direction of each force on the charge q₁.

The charges q₂ and q₄ are equi-distant from q₁. As a result the strengths (magnitude) of the forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 14 } } \) are the same even though their directions are different. Therefore the vectors representing these two forces are drawn with equal lengths. But the charge q₃ is located farther compared to q₂ and q₄. Since the strength of the electrostatic force decreases as distance increases, the strength of the force \(\vec { { F }_{ 13 } } \) is lesser than that of forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 14 } } \). Hence the vector representing the force \(\vec { { F }_{ 13 } } \) is drawn with smaller length compared to that for forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 14 } } \).
From the figure, r₂₁ =\(\sqrt { 2 } \) m = r₄₁ and r₃₁ = 2m
The magnitudes of the forces are given by
F₁₃ = \(\frac { kq^{ 2 } }{ r_{ 31 }^{ 2 } } =\frac { 9\times 10^{ 9 }\times 10^{ -12 } }{ 4 } \)
F₁₃  = 2.25 x 10^-3 N
F₁₂ = \(\frac { kq^{ 2 } }{ r_{ 31 }^{ 2 } } ={ F }_{ 14 }=\frac { 9\times 10^{ 9 }\times 10^{ -12 } }{ 2 } \)
= 4.5 x 10^-3N
From the figure, the angle θ = 45⁰. In terms of the components, we have
\(\vec { { F }_{ 12 } } ={ F }_{ 12 }cos\theta \hat { i } -{ F }_{ 12 }sin\theta \hat { j } \)
= 4.5 x 10^-3 x \(\frac { 1 }{ \sqrt { 2 } } \hat { i-4.5\times { 10 }^{ -3 }\times \frac { 1 }{ \sqrt { 2 } } \hat { j } } \)
 \(\vec { { F }_{ 13 } } =F_{ 13 }\hat { i } \) = 2.25 x 10^-3 N\(\hat { i } \)
\(\vec { { F }_{ 14 } } ={ F }_{ 14 }cos\theta \hat { i } +{ F }_{ 14 }sin\theta \hat { j } \)
= 4.5 x 10^-3 x \(\frac { 1 }{ \sqrt { 2 } } \hat { i+4.5\times { 10 }^{ -3 }\times \frac { 1 }{ \sqrt { 2 } } \hat { j } } \)
Then the total force on q₁ is,
\(\vec { { F }_{ 1 }^{ tot } }={ (F }_{ 12 }cos\theta \hat { i } -{ F }_{ 12 }sin\theta \hat { j } )+{ F }_{ 13 }\hat { i } +{ (F }_{ 14 }cos\theta \hat { i } +{ F }_{ 14 }sin\theta \hat { j } )\)
\(\vec { { F }_{ 1 }^{ tot } } =({ F }_{ 12 }cos\theta +F_{13}+{ F }_{ 14 }cos\theta )\hat { i } +(-{ F }_{ 12 }sin\theta +{ F }_{ 14 }sin\theta )\)\(\hat { j } \)
Since F₁₂ = F₁₄, the j ^th component is zero.
Hence we have
\(\vec { { F }_{ 1 }^{ tot } } =({ F }_{ 12 }cos\theta +F_{13}+{ F }_{ 14 }cos\theta )\hat { i } \)
substituting the values in the above equation,
\(\left( \frac { 4.5 }{ \sqrt { 2 } } +2.25+\frac { 4.5 }{ \sqrt { 2 } } \right) \times10^{-3}\hat { i }=(4.5\sqrt { 2 } +2.25)\times 10^{-3}\hat { i } \)
\(\vec { { F }_{ 1 }^{ tot } } \) = 8.61 x 10^-3 N\(\hat { i } \)
The resultant force is along the positive x-axis.

The proton and the electron attract each other. The magnitude of the electrostatic force between these two particles is given by
\(F_e=\frac { ke^{ 2 } }{ { r }^{ 2 } } =\frac { 9\times 10^{ 9 }\times (1.6\times 10^{ -19 })^{ 2 } }{ (5.3\times 10^{ -11 })^{ 2 } } \)
=\(\frac { 9\times 2.56 }{ 28.09 } \) x 10^-7 = 8.2 x 10^-8 N
The gravitational force between the proton and the electron is attractive. The magnitude of the gravitational force between these particles is
F_G = \(\frac { G{ m }_{ e }{ m }_{ p } }{ { r }^{ 2 } } \)
= \(\frac { 6.67\times 10^{ -11 }\times 9.1\times 10^{ -31 }\times 1.6\times 10^{ -27 } }{ (5.3\times 10^{ -11 })^{ 2 } } \)
= \(\frac { 97.11 }{ 28.09 } \) x 10^-47 = 3.4 x 10^-47N
The ratio of the two forces \(\frac { { F }_{ e } }{ F_{ G } } =\frac { 8.2\times 10^{ -8 } }{ 3.4\times 10^{ -47 } } \)
= 2.41 x 10³⁹
Note that F_e ≈ 10³⁹ F_G
The electrostatic force between a proton and an electron is enormously greater than the gravitational force between them. Thus the gravitational force is negligible when compared with the electrostatic force in many situations such as for small size objects and in the atomic domain. This is the reason why a charged comb attracts an uncharged piece of paper with greater force even though the piece of paper is attracted downward by the Earth. This given figure is shown in below.

Electrostatic attraction between a comb and pieces of papers

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

(i) A non-polar molecule is one in which centers of positive and negative charges coincide. As a result, it has no permanent dipole moment.
(ii) Examples of non-polar molecules are hydrogen (H₂), oxygen (O₂), and carbon di oxide (CO₂).

(i) Both forces obey inverse square law, F∝\(\frac{1}{r^2}\)
(ii) Both forces are proportional to product of masses or charges.
(iii) Both forces are conservative forces.
(iv) Both forces can operate in vacuum.

An equipotential surface is a surface on which all the points are at the same electric potential.

Coulomb's law \(\overrightarrow{F_{21}}=\frac{k q_{1} q_{2}}{r^{2}} \hat{r}_{12}\)
where, q₁ - charge; q₂ - charge
r - distance between the charges
\(\hat{r}_{12}\)- the unit vector directed from charge q₁ to charge q₂
k = Proportionality constant

When the water molecules are aligned in the direction of the electric field, it has minimum potential energy. The work done to rotate the dipole from θ = 0° to 90° is equal to the potential energy difference between these two configurations.
W = ΔU = U(90°) - U(0°)
From the equation U =−pE cosθ = −\(\hat p.\hat E\) ,
we write U = − pE cosθ, Next, we calculate the work done to rotate one water molecule from θ = 0° to 90°.
For one water molecule
W = - pE cos90^o + pE cos0^o  = pE
W=  6.3 x 10^-30 x 3 x 10⁵ = 18.9 x 10^-25J
For 10²² water molecules, the total work done is
W_tot = 18.9 x 10^-25 x 10²² = 18.9 x 10^-3J

(a) The work done to arrange the charges in the corners of the square is independent of the way they are arranged. We can follow any order.
(i) First, the charge +q is brought to the corner P. This requires no work since no charge is already present, W_P = 0
(ii) Work required to bring the charge –q to the corner Q = (-q) x potential at a point Q due to +q located at a point P
W_Q = -q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ a } =-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q^{ 2 } }{ a } \)
(iii) Work required to bring the charge +q to the corner R = q x potential at the point R due to charges at the point P and Q.
W_R = q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( -\frac { q }{ a } +\frac { q }{ \sqrt { 2 } a } \right) \)
= \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }^{ 2 } }{ a } \left( -1+\frac { 1 }{ \sqrt { 2 } } \right) \)
(iv) Work required to bring the fourth charge –q at the position S = q × potential at the point S due the all the three charges at the point P, Q and R.
W_s = - q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { q }{ a } +\frac { q }{ a } -\frac { q }{ \sqrt { 2 } a } \right) \)
W_s = \(-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q^2 }{ a } \left( 2-\frac { 1 }{ \sqrt { 2 } } \right) \)
(b) Work required to bring the charge q′ to the center of the square = q′ x potential at the center point O due to all the four charges in the four corners.
The potential created by the two +q charges are canceled by the potential created by the -q charges which are located in the opposite corners. Therefore the net electric potential at the center O due to all the charges in the corners is zero.
Hence no work is required to bring any charge to the point O. Physically this implies that if any charge q′ when brought close to O, then it moves to the point O without any external force.

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

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

The electric flux of D is less than that of C
The electric flux of C is less than that of B
The electric flux of B is less than that of A

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

The line joining B to C is perpendicular to electric field

So potential of B = potential of C
i.e. V_B = V_c
Distance AB = 4 cm
Potential difference
between A & C = E x AB
= 5 x 10³  x (4 x 10^-2)
= 200 volt.
AC² = AB² + B²
AB² = AC² - BC²
= 25 - 9 = 16
AB = 4cm