Test: Electrostatics - 2 - NEET MCQ

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

A.Statement A is wrong.
B.m= m0​​/√(1− v2/c2) ​​
specific charge q/m=q/m0√(1​−(v2/C2))​​
so, as v increases the specific charge decreases
C. None of the fundamental particles that form the Standard Model have charge without mass. The Standard Model describes all energy and matter that makes up the universe, except gravitation. And gravitation does not have charge, it has mass.
D.Yes. Repulsion is observed only when two bodies have like charges that means the bodies must be charged. Therefore, repulsion is sure test for electrification than attraction.
 

The net force is given as,
F_n​=2Fcosθ
=2×(1/4πε₀)​( −qQ /a²+x² ​)×[x/​(a²+x²)^1/2]
Thus, the restoring force is not linear, So, the motion will not be simple harmonic motion but the motion will be oscillatory.

 

The charge in the middle experiences force along the line. The equilibrium is stable along the line connecting charges while The equilibrium is unstable along the line perpendicular to the line of charges  
∴ Only option B is correct (given consider equilibrium only along line joining charges).
 

If the potential energy of the system is minimum, it will be stable equilibrium, i.e , d²U/dx²​>0 and when potential energy is maximum then it will be unstable equilibrium, i.e, d²U/dx²<0. 
As along y direction no electric field, potential energy is minimum and it will be stable equilibrium along y-axis.
Along x-axis potential energy is maximum due to all charges situated along x-axis.so it will be unstable equilibrium.
 

(A) electric field due to positive charge is away from it and electric field due to negative charge is towards it so there cannot be a point on AB where electric field can be zero
(D) negative charge will experience a repulsion force due to −ve charge and attraction force due to +ve charge, so if pushed away from charges then a net attractive force and if pushed towards AB then a net repulsive force will be there so there will be oscillations.
 

If field is straight then the charge moves along the straight line .if the fielss is a curve then the electrostatic force is not sufficient to change the direction along the curved field line .

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

The acceleration due to gravity is insignificant compared with the acceleration caused by the electric field, so the gravitational field can be ignored. Hence, when a particle of mass m and charge q is thrown in a region where uniform gravitational field and electric field are present, before entering the electric field, the charge follows a straight line path (since, no net force due to any of the field). As soon as it enters the field, the charge begins to follow a parabolic path (since, constant force is always in the same direction). As soon as it leaves the field, the charge follows a straight line path  (since, no net force due to any of the field). The path of a particle may be straight line or may be parabola.

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       

as there are more no of lines that are spreading from q₁ means it have a higher magnitude than q₂

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]