We know that a positively charged particle is attracted towards the negatively charged plate and a negatively charged plate and a negatively charged particle is attracted towards the positively charged plate.
Here particle 1 and 2 are attraced positive that mean 1 and 2 are negatively charged. 3 is attracted towards negatively charged plate so it is positively charged.
As the deflection in the path of a charged particle is directly proportional to the charge/mass ratio
 \(y\propto \ \frac { q }{ m } \)
Here, the deflection in particle 3 is maximum. so,  the charge to mass ratio of particle 3 is maximum.

An electrostatic field line cannot be a discontinuous curve,i.e it cannot have breaks.If it has breaks,then it will indicate absence of electric field at the break points.But the electric field vanishes only at infinity.

q₁ = +\(\mu C\), q₂ = +6\(\mu C\), F = 12N
\(q_{1}^{'}=+2-4=-2\mu C;\) \(q_2^{'}=+6-4=2\mu C\) 
F' = ?
\({F'\over F}={(q_1^{'})(q_2^{'})\over q_1q_2}={(-2)(2)\over(2)(6)}=-{1\over 3}\)
\(F'={-F\over 3}={-12\over 3}=-4 N\)(attractive)

(i) Like charges repel and unlike charges attract each other.
(ii) Charge is quantized
(iii) Charge is conserved 
(iv) Charge on a body is not affected by its motion.

Conservation of electric charge means that the total charge on an isolated system remains unchanged with time.

Electric field at a point on the equatorial line of an electric dipole.
Consider an electric dipole consisting of two point charges + q and - q separated by a small distance AB = 21 with centre at O and dipole moment, p = q(2l) as shown in the figure.

Resultant electric field intensity at the"'point Q,
E_Q = E_A + E_B
Here, E_A = \({{1}\over{4\pi{\epsilon}_{0}}}.{{q}\over{({x}^{2}+{l}^{2})}}\)
and E_B = \({{1}\over{4\pi{\epsilon}_{0}}}.{{q}\over{({x}^{2}+{l}^{2})}}\)
On resolving E_A and E_B into two rectangular components, the vectors E_A sin \(\theta\) and E_B sin \(\theta\) are equal in magnitude and opposite to each other and hence, cancel out.
The vectors E_A cos\(\theta\) and E_B cos\(\theta\) are acting along the same direction and hence, add up.
\(\therefore\) E_Q = E_A cos\(\theta\) + E_B cos \(\theta\)= 2E A cos\(\theta\)     \([\because{E}_{A}={E}_{B}]\)
= \({{2}\over{4\pi{\epsilon}_{0}}}.{{q}\over{({x}^{2}+{l}^{2})}}.{{q}\over{({x}^{2}+{l}^{2})^{1/2}}}\)
 \(\left[ \because cos\ \theta={{l}\over{({x}^{2}+{l}^{2})^{1/2}}} \right]\)
= \({{1}\over{4\pi{\epsilon}_{0}}}.{{2ql}\over{({x}^{2}+{l}^{2})^{3/2}}}\)
But, the dipole moment |p| = q X 2l
\(\therefore\)  E_Q = \({{1}\over{4\pi{\epsilon}_{0}}}.{{|p|}\over{({x}^{2}+{l}^{2})^{3/2}}}\)
The direction of E is along QE II BA. i.e . opposite to AB. In vector form, we can rewrite as
E_Q = \({{-p}\over{4\pi{\epsilon}_{0}({x}^{2}+{l}^{2})^{3/2}}}\)
The orientation of the dipole
In stable equilibrium, . p is parallel to E. i.e.
\(\theta=0°\)

In unstable equilibrium, p is anti-parallel to E i.e. 0 = 180°

Electric field intensity due to a long wire : A line charge is in the form of a thin charged rod with uniform linear charge density A (charge per unit length),

To determine the electric field intensity \(\overrightarrow{E}\) at any point P at a perpendicular distance r from the rod, let a right circular closed cylinder of radius r and length I with the infinitely long line of charge as its  axis (as shown in figure). The magnitude of \(\overrightarrow{E}\) at every point on the curved surface of the cylinder is the same as all such points are at the same distance from the line charge. Also \(\overrightarrow{E}\) and unit vector \(\hat{n}\) it normal to curved surface are in the same direction so, \(\theta\)= 0°.
\(\therefore\) Contribution of curved surface of cylinder towards electric flux,
\(\oint_s \overrightarrow{E} \overrightarrow{ds} = \oint_s \overrightarrow{E} \hat{n}{ds}\)
= \(E \oint_s ds = E(2\pi r l)\)
where ( \(2\pi r l\) ) is area of the curved surface of the cylinder.
On the ends of cylinder, the angle between electric field \(\overrightarrow{E}\) and  \(\hat{n}\) is 90°. So it will not contribute to electric flux on cylinder.
\(\Longrightarrow\) \(E \oint_s ds = E(2\pi r l)\)
Charge enclosed in the cylinder linear charge density length
q = \(\lambda \)I
According to Gauss's theorem \(\oint_s \overrightarrow{E} \overrightarrow{ds} = \frac{q}{\epsilon_0}\)
\(\Longrightarrow\)   \((2\pi r l)\times E = \frac{\lambda l}{\epsilon_0}\)
\( E = \frac{\lambda l}{2\pi\epsilon_0rl}\)
\( E = \frac{l}{r}\)
(As \(\frac{\lambda}{2\pi\epsilon_0r} \) = constant)

Here,
Charge at P, q = \(1\mu C={ 10 }^{ -6 }C\)
Charge at \({ A,q }_{ 1 }=2\times { 10 }^{ -8 }C\)
Charge at \({ B,q }_{ 2 }=-{ 10 }^{ -8 }C\)
\(Ap=10\quad cm=0.1m,BP=5cm=0.05m,\angle APB={ 90 }^{ o },F=?\)
Force at P due to q₁ charge at \(A,{ F }_{ 1 }=\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \frac { { q }_{ 1 }q }{ { AP }^{ 2 } } ,\) along AP produced
\(=\frac { 9\times { 10 }^{ 9 }\times 2\times { 10 }^{ -8 }\times { 10 }^{ -6 } }{ { \left( 0.1 \right) }^{ 2 } } =18\times { 10 }^{ -3 }N\)
Force at P due to q2 charge at \(B,{ F }_{ 2 }=\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \frac { { q }_{ 2 }q }{ { BP }^{ 2 } } ,\) , along PB produced \(=\frac { 9\times { 10 }^{ 9 }\times -{ 10 }^{ -8 }\times { 10 }^{ -6 } }{ { \left( 0.05 \right) }^{ 2 } } =-36\times { 10 }^{ -3 }N\)
As, angle between F₁and F₂ is \({ 90 }^{ o }\)
\({ \therefore }\)  Resultant Force,
\(F=\sqrt { { { F }_{ 1 } }^{ 2 }+{ { F }_{ 2 } }^{ 2 } } =\sqrt { { \left( 18\times { 10 }^{ -3 } \right) }^{ 2 }+{ \left( -36\times { 10 }^{ -3 } \right) }^{ 2 } } \)
\(\\ \sqrt { \left( 324+1296 \right) \times { 10 }^{ -6 } } =\sqrt { 1620\times { 10 }^{ -6 } }\)
\( \\ =40.2\times { 10 }^{ -3 }\)
\(\\ =4.0\times { 10 }^{ -2 }N\) 

(a) Electric flux is defined as the number of electric field lines passing through an area normal to the surface.
Alternatively
Surface integral of the electric field is defined as the electric flux through a closed surface
\(\phi= \oint \overrightarrow{E}. \overrightarrow{ds}\)
SI unit :  \(\frac{N.m^2}{C}\) or volt. metre
(a)


Outward flux through the Gaussian surface, is
\(2EA =\sigma A/\epsilon_0\)
\(\therefore\)  \(E =\sigma A/2\epsilon_0\)
Vectorically \(\overrightarrow{E} = \frac{\sigma}{2\epsilon_0}\hat{n}\),
where \(\hat{n}\) is a unit vector normal to the plane, away from it.  
Hence, electric field is independent of the distance from the sheet.
(c) For positively charged sheet  \(\longrightarrow\) away from the sheet  
For negatively charged sheet \(\longrightarrow\) towards the plane sheet