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Test: Electrostatics - 2


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Test: Electrostatics - 2 - Question 1

Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in

Detailed Solution for Test: Electrostatics - 2 - Question 1

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

Test: Electrostatics - 2 - Question 2

What will be the value of a charge q, such that when it is placed at the centre of two equal and like charges Q, the three charges are in equilibrium.

Detailed Solution for Test: Electrostatics - 2 - Question 2

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

Test: Electrostatics - 2 - Question 3

A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -

Detailed Solution for Test: Electrostatics - 2 - Question 3

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.
 

Test: Electrostatics - 2 - Question 4

In space of horizontal EF(E = (mg)/q) exist as shown in figure and a mass m attached at the end of a light rod. If mass m is released from the position shown in figure find the angular velocity of the rod when it passes through the bottom most position.

                       

Detailed Solution for Test: Electrostatics - 2 - Question 4

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

Test: Electrostatics - 2 - Question 5

Find the force experienced by the semicircular rod charged with a charge q, placed as shown in figure. Radius of the wire is R and the infinitely long line of charge with linear density l is passing through its centre and perpendicular to the plane of wire.

Detailed Solution for Test: Electrostatics - 2 - Question 5

Fnet​=∫dqEcosθ

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

Test: Electrostatics - 2 - Question 6

A wheel having mass m has charges +q and –q on diametrically opposite points. It remains in equilibrium on a rough inclined plane in the presence of uniform vertical electric field E =

                           

Detailed Solution for Test: Electrostatics - 2 - Question 6

*Multiple options can be correct
Test: Electrostatics - 2 - Question 7

Select the correct alternative :

Detailed Solution for Test: Electrostatics - 2 - Question 7

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.
 

*Multiple options can be correct
Test: Electrostatics - 2 - Question 8

Two equal negative charges –q are fixed at the point (0, a) and (0,– a) on the y-axis. A charge +Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will :

Detailed Solution for Test: Electrostatics - 2 - Question 8

The net force is given as,
Fn​=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.

 

*Multiple options can be correct
Test: Electrostatics - 2 - Question 9

Mid way between the two equal and similar charges, we placed the third equal and similar charge. Which of the following statements is correct, concerned to the equilibrium along the line joining the charges ?

Detailed Solution for Test: Electrostatics - 2 - Question 9


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).
 

*Multiple options can be correct
Test: Electrostatics - 2 - Question 10

A negative point charge placed at the point A is

                   

Detailed Solution for Test: Electrostatics - 2 - Question 10

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.
 

*Multiple options can be correct
Test: Electrostatics - 2 - Question 11

Two fixed charges 4Q (positive) and Q (negative) are located at A and B, the distance AB being 3 m.

                    

Detailed Solution for Test: Electrostatics - 2 - Question 11

(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.
 

*Multiple options can be correct
Test: Electrostatics - 2 - Question 12

Select the correct statement : (Only force on a particle is due to electric field)

Detailed Solution for Test: Electrostatics - 2 - Question 12

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 .

Test: Electrostatics - 2 - Question 13

Consider a conductor with a spherical cavity in it. A point charge q0 is placed at the centre of cavity and a point charge Q is placed outside conductor.

Statement-1 : Total charge induced on cavity wall is equal and opposite to the charge inside.
Statement-2 : If cavity is surrounded by a Gaussian surface, where all parts of Gaussian surface are
located inside the conductor,

Test: Electrostatics - 2 - Question 14

The figure shows a nonconducting ring which has positive and negative charge non uniformly distributed on it such that the total charge is zero. Which of the following statements is true ?

Detailed Solution for Test: Electrostatics - 2 - Question 14



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

*Multiple options can be correct
Test: Electrostatics - 2 - Question 15

 If we use permittivity e, resistance R, gravitational constant G and voltage V as fundamental physical quantities, then -

Detailed Solution for Test: Electrostatics - 2 - Question 15

*Multiple options can be correct
Test: Electrostatics - 2 - Question 16

 A particle of mass m and charge q is thrown in a region where uniform gravitational field and electric field are present. The path of particle

Detailed Solution for Test: Electrostatics - 2 - Question 16

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.

*Multiple options can be correct
Test: Electrostatics - 2 - Question 17

Statement - 1 : A positive point charge initially at rest in a uniform electric field starts moving along electric lines of forces. (Neglect all other forces except electric forces)

Statement - 2 : Electric lines of force represents path of charged particle which is released from rest in it.

Test: Electrostatics - 2 - Question 18

Two equal negative charges are fixed at the points [0, a] and [0, –a] on the y-axis. A positive charge Q is released from rest at the points [2a, 0] on the x-axis. The charge Q will -

Test: Electrostatics - 2 - Question 19

Two small balls, each having equal positive charge Q are suspended by two insulating strings of equal length L from a hook fixed to a stand. If the whole set-up is transferred to a satellite in orbit around the earth,the tension in each string is equal to

Detailed Solution for Test: Electrostatics - 2 - Question 19

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

Test: Electrostatics - 2 - Question 20

An electric dipole is placed at an angle of 30o to a non-uniform electric field. The dipole will experience

Detailed Solution for Test: Electrostatics - 2 - Question 20

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

Test: Electrostatics - 2 - Question 21

Two point charges +8q and –2q are located at x = 0 and x = L respectively. The location of a point onthe x axis at which the net electric field due to these two point charges is zero is:

Detailed Solution for Test: Electrostatics - 2 - Question 21

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

According to conditional EBO+ EBA = 0

So a = L

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

Test: Electrostatics - 2 - Question 22

The figure shows the electric field lines in the vicinity of two point charges. Which one of the followingstatements concerning this situation is true?

Test: Electrostatics - 2 - Question 23

Four charges are placed each at a distance a from origin. The dipole moment of configuration is

Detailed Solution for Test: Electrostatics - 2 - Question 23

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

Test: Electrostatics - 2 - Question 24

Figure shows the electric field lines around an electric dipole. Which of the arrows best represents the electric field at point P?

Detailed Solution for Test: Electrostatics - 2 - Question 24

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

Test: Electrostatics - 2 - Question 25

A uniform electric field , intersects a surface of area A. What is the flux through this area if thesurface lies in the yz plane?

Detailed Solution for Test: Electrostatics - 2 - Question 25

Test: Electrostatics - 2 - Question 26

If 109 electrons move out of a body to another body every second, how much time is approximately required to get a total charge of 1 C on the other body?

Detailed Solution for Test: Electrostatics - 2 - Question 26

we know that 1 colomb = 6.242 x 1018 electron
given 109 electron take 1 sec
⇒ 10electron → 1 sec
⇒ 1 electron →  secs

Test: Electrostatics - 2 - Question 27

A uniform line charge with linear density λ lies along the y-axis. What flux crosses a spherical surface centered at the origin with r = R  

Test: Electrostatics - 2 - Question 28

Electric lines of force never intersect each other because

Detailed Solution for Test: Electrostatics - 2 - Question 28

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.

Test: Electrostatics - 2 - Question 29

A neutral point lies at the center of the line joining the charges when

Detailed Solution for Test: Electrostatics - 2 - Question 29

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.

Test: Electrostatics - 2 - Question 30

The dimensions of   where [∈0 : permittivity of free space ; E: electric field]

Detailed Solution for Test: Electrostatics - 2 - Question 30

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

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