(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.

If the conductor has no net charge, then its motion is the same as usual projectile motion of a mass m which we studied in Kinematics (unit 2, vol-1 XI physics). Here, in this problem, in addition to downward gravitational force, the charge also will experience a downward uniform electrostatic force.
The acceleration of the charged ball due to gravity = -g\(\hat { j } \)
The acceleration of the charged ball due to uniform electric field =\(-\frac { qE }{ m } \hat { j } \)
The total acceleration of charged ball in downward direction \(\vec { a } =-\left( g+\frac { qE }{ m } \right) \vec { j } \)
It is important here to note that the acceleration depends on the mass of the object. Galileo’s conclusion that all objects fall at the same rate towards the Earth is true only in a uniform gravitational field. When a uniform electric field is included, the acceleration of a charged object depends on both mass and charge.
But still the acceleration a = \(\left( g+\frac { qE }{ m } \right) \) is constant throughout the motion. Hence we use kinematic equations to calculate the range, maximum height and time of flight. In fact we can simply replace g by \(g+\frac { qE }{ m } \) in the usual expressions of range, maximum height and time of flight of a projectile.

Note that the time of flight, maximum height, range are all inversely proportional to the acceleration of the object. Since \(\left( g+\frac { qE }{ m } \right) \) >g for charge +q, the quantities T, h_max, and R will decrease when compared to the motion of an object of mass m and zero net charge. Suppose the charge is –q, then \(\left( g-\frac { qE }{ m } \right) \)< g, and the quantities T, h_max and  R will increase. Interestingly the trajectory is still parabolic as shown in the figure.

(a) Electric potential at point P is given by
V_p=\(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }_{ p } } =\frac { 9\times 10^{ 9 }\times 9\times { 10 }^{ -6 } }{ 10 } \)  = 8.1 x 10³ V
Electric potential at point Q is given by
V_Q = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }_{ Q } } =\frac { 9\times 10^{ 9 }\times 9\times { 10 }^{ -6 } }{ 16 } \) = 5.06 x 10³ V
Note that the electric potential at point Q is less than the electric potential at point P. If we put a positive charge at P, it moves from P to Q. However if we place a negative charge at P it will move towards the charge +9μC.
The potential difference between the points P and Q is given by
ΔV = V_p-V_Q = +3.04 x 10³ V
b) Suppose we replace the charge +9 μC by -9 μC, then the corresponding potentials at the points P and Q are,
V_p = -8.1 x 10³ V, V_Q = -5.06 x 10³ V
Note that in this case electric potential at the point Q is higher than at point P.
The potential difference or voltage between the points P and Q is given by
ΔV = V_p-V_Q= -3.04 x 10³V
(c) The electric potential V at a point P due to some charge is defined as the work done by an external force to bring a unit positive charge from infinity to P. So to bring the q amount of charge from infinity to the point P, work done is given as follows.
W = qV
W_Q = 2 x 10^-6 x 5.06 x 10³J = 10.12 x 10^-3 J.

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) In polar molecules, the centers of the positive and negative charges are separated even in the absence of an external electric field.
(ii) They have a permanent dipole moment.
(iii) Examples : H₂O, N₂O, HCI and NH₃.

i) The electric field is the negative gradient of the electric potential. \(E=\frac{-dV}{dx}\). 
ii) In vector form, \(\vec{E}=-\left[\frac{\partial V}{\partial x} \hat{i}+\frac{\partial V}{\partial y} \hat{j}+\frac{\partial V}{\partial z} \hat{k}\right]\)

(i) The work done to move a charge q between any two points A and B, W = q (V_B - V_A). If the points A and B lie on the same equipotential surface, work done is zero because V_B = V_A.
(ii) The electric field is normal to an equipotential surface. 

For a collection of n point charges, the electric dipole moment is defined as follows, \(\vec{p}=\stackrel{i=n} \sum _{i=1}q_i\vec{r}_i\) where,  \(\vec{r}_i\) is the position vector of charge q_i from the origin.

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.

The maximum torque experienced by the dipole is when it is aligned perpendicular to the applied field.
ፒ_max = pE sin 90⁰ = 3.4 x 10^-30 x 3 x 10⁴ Nm
ፒ_max = 10.2 x 10^-26 Nm.

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.

\(U=\frac{1}{4\piε_0}\frac{q_1q_2}{r_{12}}\)
\(i) U=\frac{1}{4\piε_0}\frac{Q(-Q)}{r}=\frac{1}{4\piε_0}[\frac{-Q^2}{r}]\)
\(ii) U=\frac{1}{4\piε_0}\frac{(-Q)(-Q)}{r}=\frac{1}{4\piε_0}[\frac{Q^2}{r}]\)
\(iii) U=\frac{1}{4\piε_0}\frac{Q(2Q)}{r}=\frac{1}{4\piε_0}[\frac{2Q^2}{r}]\)
\(iv) U=\frac{1}{4\piε_0}\frac{Q(-2Q)}{2r}=\frac{1}{4\piε_0}[\frac{-Q^2}{r}]\)
From the values, 1 = 4 < 2 < 3

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

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

Here, q = -3 x 10^-7 C
Charge of one electron, e = -1.6 x 10^-19 C
Number of electrons transferred from wool to
polythenepiece, n = \(\frac{q}{e}\)\(=\frac { -3\times { 10 }^{ -7 }C }{ -1.6\times { 10 }^{ -19 }C } \)
= 1.875 x 10¹²