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

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

Test Description

## 15 Questions MCQ Test Daily Test for NEET Preparation - Test: Electric Charges and Fields (October 30)

Test: Electric Charges and Fields (October 30) for NEET 2024 is part of Daily Test for NEET Preparation preparation. The Test: Electric Charges and Fields (October 30) questions and answers have been prepared according to the NEET exam syllabus.The Test: Electric Charges and Fields (October 30) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Electric Charges and Fields (October 30) below.
Solutions of Test: Electric Charges and Fields (October 30) questions in English are available as part of our Daily Test for NEET Preparation for NEET & Test: Electric Charges and Fields (October 30) solutions in Hindi for Daily Test for NEET Preparation course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Electric Charges and Fields (October 30) | 15 questions in 30 minutes | Mock test for NEET preparation | Free important questions MCQ to study Daily Test for NEET Preparation for NEET Exam | Download free PDF with solutions
Test: Electric Charges and Fields (October 30) - 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: Electric Charges and Fields (October 30) - Question 1

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

Test: Electric Charges and Fields (October 30) - Question 2

### 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: Electric Charges and Fields (October 30) - Question 2

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.

 1 Crore+ students have signed up on EduRev. Have you?
Test: Electric Charges and Fields (October 30) - Question 3

### 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: Electric Charges and Fields (October 30) - Question 3

Fnet​=∫dqEcosθ

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

*Multiple options can be correct
Test: Electric Charges and Fields (October 30) - Question 4

Select the correct alternative :

Detailed Solution for Test: Electric Charges and Fields (October 30) - Question 4

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: Electric Charges and Fields (October 30) - Question 5

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: Electric Charges and Fields (October 30) - Question 5

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: Electric Charges and Fields (October 30) - Question 6

A negative point charge placed at the point A is

Detailed Solution for Test: Electric Charges and Fields (October 30) - Question 6

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: Electric Charges and Fields (October 30) - Question 7

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

Detailed Solution for Test: Electric Charges and Fields (October 30) - Question 7

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: Electric Charges and Fields (October 30) - Question 8

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: Electric Charges and Fields (October 30) - Question 8

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

*Multiple options can be correct
Test: Electric Charges and Fields (October 30) - Question 9

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

Detailed Solution for Test: Electric Charges and Fields (October 30) - Question 9

*Multiple options can be correct
Test: Electric Charges and Fields (October 30) - Question 10

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: Electric Charges and Fields (October 30) - Question 11

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: Electric Charges and Fields (October 30) - Question 11

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

Test: Electric Charges and Fields (October 30) - Question 12

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

Detailed Solution for Test: Electric Charges and Fields (October 30) - Question 12

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

Test: Electric Charges and Fields (October 30) - Question 13

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

Detailed Solution for Test: Electric Charges and Fields (October 30) - Question 13

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: Electric Charges and Fields (October 30) - Question 14

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: Electric Charges and Fields (October 30) - Question 14

Test: Electric Charges and Fields (October 30) - Question 15

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

## Daily Test for NEET Preparation

12 docs|366 tests
Information about Test: Electric Charges and Fields (October 30) Page
In this test you can find the Exam questions for Test: Electric Charges and Fields (October 30) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Electric Charges and Fields (October 30), EduRev gives you an ample number of Online tests for practice

## Daily Test for NEET Preparation

12 docs|366 tests