Carefully observe each property of electric field lines in the above options.

If we see all the forces acting upon the charge Q due to other charge Q at distance r and other charge q at distance r/2, we get
kQQ/r² + kQ(q)/(r/2)²
This results 
Q² / r²  = -4Qq / r²
Thus we get q = -Q/4

electric field due to non-conducting ring along its axis is given by, 
E=kqz/(Z²+r²)^3/2
where z is the separation between point of observation to the centre of the ring, q is the charge on the ring and r is the radius of the ring.
We know, electric field will be maximum only when z = R/√2, meaning at this point the value of force must be maximum. and if we increase the value of z from R/√2 it will decrease and if we decrease the value of z from R/√2 , it will increase.
condition of stable equilibrium,
A small vertical displacement upwards should cause the resultant force on the particle to be downwards, to return it or a small vertical displacement downward should cause the resultant force on the particle to be downward.
hence, at z > R/√2 or, h > R/√2 system will be in stable equilibrium.
 

According to the work energy theorem we have
W_e​+W_g​=(1​/2)mv²
We have work done by electrostatic force as
W_e​=qElsinθ
and work done by the gravitational force as
W_g​=mg(l−lcosθ)
Thus we get
qElsinθ+mg(l−lcosθ)=(1/2)​mv²
Thus we get
mgsinθ+mgl−mglcosθ=(1/2)​mv²
as θ=45^o, we get
mgl=(1/2)​mv2
also as v=ωl
we get
ω=√2g/​​l

F_net​=∫dqEcosθ

cosθdθ=(λq/2π2ε0​R) ​[sinθ]_−π/2^π/2
= (λq/2π2ε_0​R) ​[1−(−1)]= λq​/π2ε₀​​R

There for the potential at all the points on the axis will be zero.

A satellite is in a state of free fall & hence weightlessness. Thus only electric force is responsible for the Tension

The dipole will have some distance along the electric field, so, option (1) is correct.

Suppose that at point B, where net electric field is zero due to charges 8q and 2q

According to conditional E_BO+ E_BA = 0

So a = L

Thus, at distance 2L from origin, net electric field will be zero       

This question can be solved easily , if you have some knowledge about resolution of vectors.
all four charges are placed , each at a distance 'a ' from the origin.
it means distance between two charges = √{a² + a² } = √2a
We also know, dipole moment is the system of two equal magnitude but opposite nature charges . so, we have to divide the charges as shown in the figure for making dipoles .
Hence, there are three dipoles formed.
Let P = q(√2a)
then, P₁ = P , P₂ = P and P₃ = 2P {as shown in figure.}
now resolve the vectors P₁ ,P₂ and P₃ .
as shown in figure,
vertical components of P₁ and P₂ is cancelled .
and horizontal components of P₁ and P₂ is cancelled by horizontal component of P₃ .
and rest part of dipole = vertical component of P₃ = 2Psin45° j ['j' shows direction of net dipole moment]
hence, net dipole moment = 2q(√2a) × 1/√2 j
= 2qa j

Tangent to the electric field line represents the direction of electric field at that point.

we know that 1 colomb = 6.242 x 10¹⁸ electron
given 10⁹ electron take 1 sec
⇒ 10⁹ electron → 1 sec
⇒ 1 electron →  secs

Electric lines of force never intersect each other because at the point of intersection, two tangents can be drawn to the two lines of force. This means two direction of electric field at the point of intersection, which is not possible.

Neutral point due to a system of two like point charges For this case neutral point is obtained at an internal point along the line joining two like charges.Here neutral point lies outside the line joining two unlike charges and also it lies nearer to the charge which is smaller in magnitude.

Dimensions of permittivity of free space ∈₀ = [q²/Fr²] and 
By definition of Electric Field Intensity; [E] =[ F/q]