Test: Electric Charges and Fields (October 30) - NEET MCQ

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

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.
 

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 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.
 

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.

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.

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