Test: Electric Charges & Fields - NEET MCQ

# Test: Electric Charges & Fields - NEET MCQ

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## 20 Questions MCQ Test Physics Class 12 - Test: Electric Charges & Fields

Test: Electric Charges & Fields for NEET 2024 is part of Physics Class 12 preparation. The Test: Electric Charges & Fields questions and answers have been prepared according to the NEET exam syllabus.The Test: Electric Charges & Fields MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Electric Charges & Fields below.
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Test: Electric Charges & Fields - Question 1

### When a negatively charged conductor is connected to earth,

Detailed Solution for Test: Electric Charges & Fields - Question 1

Explanation:

When a negatively charged conductor is connected to the earth, electrons will flow from the conductor to the earth. This is because electrons have a negative charge and they will be repelled from the negatively charged conductor and attracted to the positively charged earth. As electrons flow from the conductor to the earth, the negative charge on the conductor will gradually decrease until it becomes neutral.

• Option A is incorrect because charge flow does occur when a negatively charged conductor is connected to the earth.
• Option B is incorrect because protons have a positive charge and they are not free to move in a conductor.
• Option C is incorrect because electrons flow from the earth to the conductor, not the other way around.

Test: Electric Charges & Fields - Question 2

### A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Detailed Solution for Test: Electric Charges & Fields - Question 2

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Test: Electric Charges & Fields - Question 3

### Two identical conductors of copper and aluminium are placed in an identical electric field. The magnitude of induced charge in the aluminium will be:

Detailed Solution for Test: Electric Charges & Fields - Question 3

As aluminium and copper are metals, their mobile electrons move under the influence of the external field until they reach the surface of the metal and collect there. External electric fields induce surface changes on metal objects that exactly cancel the field within. Since the field applied is same in both case, the induced charge will be the same.
Hence, the magnitude of induced charge in the aluminium will be the equal to that of copper.

Test: Electric Charges & Fields - Question 4

A point charge causes an electric flux of −1.0×103Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?

Detailed Solution for Test: Electric Charges & Fields - Question 4

Electric flux, Φ = −1.0 × 103 N m2/C

r = 10.0 cm

Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., −103 N m2/C.

Electric flux is given by the relation,

qϕ=q∈0

Where,

q = Net charge enclosed by the spherical surface

0 = Permittivity of free space = 8.854 × 10−12 N−1 Cm−2

∴ qq=ϕ∈0

= −1.0 × 103 × 8.854 × 10−12

= −8.854 × 10−9 C

= −8.854 nC

Therefore, the value of the point charge is −8.854 nC.

Test: Electric Charges & Fields - Question 5

A hollow spherical conductor of radius 2m carries a charge of 500 μ C. Then electric field strength at its surface is

Detailed Solution for Test: Electric Charges & Fields - Question 5

Test: Electric Charges & Fields - Question 6

Two point charges A and B, having charges +Q and –Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes

Detailed Solution for Test: Electric Charges & Fields - Question 6

In case I :

In Case II :

From equations (i) and (ii),

Test: Electric Charges & Fields - Question 7

Detailed Solution for Test: Electric Charges & Fields - Question 7

Let m be mass of each ball and q be charge on each ball. Force of repulsion,

In equilibrium
Tcosq = mg ...(i)
Tsinq = F ...(ii)
Divide (ii) by (i), we get,

From figure (a),

Divide (iv) by (iii), we get

Test: Electric Charges & Fields - Question 8

Under the influence of the coulomb field of charge +Q, a charge −q is moving around it in an elliptical orbit. Find out the correct statement(s).

Detailed Solution for Test: Electric Charges & Fields - Question 8

Since the charge –q is moving in elliptical orbit so to make its motion stable the total angular momentum of the charge is constant since it experience a centripetal force from the charge +Q so it follow the motion as the motion of earth around sun.

Test: Electric Charges & Fields - Question 9

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field . Due to the force , its velocity increases from 0 to 6 m s–1 in one second duration. At that instant the direction of the field is
reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

Detailed Solution for Test: Electric Charges & Fields - Question 9

Acceleration

Test: Electric Charges & Fields - Question 10

Three point charges +q, –2q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are

Detailed Solution for Test: Electric Charges & Fields - Question 10

This consists of two dipoles, –q and +q with dipole moment along with the +y-direction and –q and +q along the x-direction.

Along the direction 45° that is along OP, where P is (+a, +a, 0).

Test: Electric Charges & Fields - Question 11

An electric dipole is placed at an angle of 30° with an electric field intensity 2 x 105 N C–1.It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is

Detailed Solution for Test: Electric Charges & Fields - Question 11

Here, q = 30°, E = 2 x 105 N C–1, τ = 4 N m,
l = 2 cm = 0.02 m, q = ?
τ = pE sinq = (ql)E sinq

Test: Electric Charges & Fields - Question 12

A  spherical conductor of radius 10 cm has a charge of 3.2 x 10–7 C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere?

Detailed Solution for Test: Electric Charges & Fields - Question 12

Here, r = 10 cm, q = 3.2 x 10–7 C

Test: Electric Charges & Fields - Question 13

Ionization of a neutral atom is the

Detailed Solution for Test: Electric Charges & Fields - Question 13

It is not possible to remove or add protons/neutrons to an atom, but electrons can be added or removed by an atom easily.
By adding electrons it becomes negatively charged.
By removing electrons it becomes positively charged.

Test: Electric Charges & Fields - Question 14

A hollow cylinder has a charge q coulomb within it. If f φ is the electric flux in units of volt meter associated with the curved surface B, the flux linked with the plane surface A in units of V-m will be

Detailed Solution for Test: Electric Charges & Fields - Question 14

Let φA, φB and φC are the electric flux linked with surface is A, B and C.
According to Gauss theorem,

Test: Electric Charges & Fields - Question 15

A charge Q is situated at the corner of a cube, the electric flux passed through all the six faces of the cube is

Detailed Solution for Test: Electric Charges & Fields - Question 15

As at a corner, 8 cubes can be placed symmetrically, flux linked with each cube (due to a charge Q at the corner)
will be

Now for the faces passing through the edge A, electric field E at a face will be parallel to area of face and so flux for these three faces will be zero. Now as the cube has six faces and flux linked with three faces (through A) is zero, so flux linked with remaining three faces will be
Hence, electric flux passed through all the six faces of the cube is

Test: Electric Charges & Fields - Question 16

The electrostatic force is:

Detailed Solution for Test: Electric Charges & Fields - Question 16

The electrostatic force is an attractive and repulsive force between particles are caused due to their electric charges. The electric force between stationary charged bodies is conventionally known as the electrostatic force.

Test: Electric Charges & Fields - Question 17

A point charge + q is placed at the centre of a  cube of side l. The electric flux emerging from the cube is

Detailed Solution for Test: Electric Charges & Fields - Question 17

Electric flux emerging from the cube does not depend on size of cube.

Test: Electric Charges & Fields - Question 18

A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre

Detailed Solution for Test: Electric Charges & Fields - Question 18

In a uniformly charged hollow conducting sphere

Test: Electric Charges & Fields - Question 19

Conductors are materials

Detailed Solution for Test: Electric Charges & Fields - Question 19

Conductors are materials that permit electrons to flow freely from particle to particle. An object made of a conducting material will permit a charge to be transferred across the entire surface of the object.

Test: Electric Charges & Fields - Question 20

The electric field at a distance 3R/2 rom the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R/2 from the centre of the sphere is

Detailed Solution for Test: Electric Charges & Fields - Question 20

Electric field inside the charged spherical shell is zero as there is no charge inside it.

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## Physics Class 12

105 videos|425 docs|114 tests