Electric Charges And Fields – Multiple Choice Questions

Q1. Ratio of the permittivity of a medium to the permittivity of free space is known as

• Dielectric medium
• Dielectric ratio
• Dielectric permittivity
• Dielectric constant

Solution

Ratio of the permittivity of medium to the permittivity of free space is known as Dielectric constant.

Q2.

An electric dipole consists of a positive and negative charge of 4 μC. Each placed at a distance of 5 mm. The dipole moment for the system is

• 20 x 10^-8 C-m
• 2 x 10^-9 C-m
• 2 x 10^-8 C-m
• 1 x 10^-8 C-m

Solution

Given: q = 4 μC = 4 × 10^-6; d = 5 mm = 5 ×10^-3

The dipole moment = charge x separation distance between them

p = q × d = 4 × 10^-6 × 5 ×10^-3

p = 2 x 10^-8 C-m

Q3.

Two point charges Q_A = 2 μC and Q_B = -2 μC are located 16 cm apart in vacuum. What is the electric field at the mid point O of the line AB joining the two charges?

• E = 5.6 × 10⁶ N/C along OB
• E = 5.6 × 10⁶ N/C along OA  
• E = 6.5 × 10⁶ N/C along OB
• E = 6.5 × 10⁶ N/C along OA

Solution

Here, q_A = 2 μC = 2 x 10^-6 C and q_B = - 2 μC = -2 x 10^-6 C

AB = 16 cm

r = OA = OB = 8 cm

The electric filed is the one unit positive charge (+1 C) at O is

E_{OA =} E_OB

E = 2E_OA

E = 5.6 x 10⁶ NC^-1 along OB.

Q4.

A scientific instrument used to detect the charge on a body is known as:

• Electrometer
• Electroscope
• Charge detector
• Miller counter

Solution

Electroscope is a scientific instrument that is used for detecting an electric charge and its polarity.

Q5.

The value of charge on a body which carries 30 excess electrons is

• -4.8 × 10 ^-19 C
• -4.8 × 10 ^-18 C
• -48 × 10 ^-18 C
• 4.8 × 10^-18 C

Solution

Here, n = 30 q =?

We know that charge on one electron is e = -1.6 × 10^-19 C

So, q = n × e = 30 (-1.6 × 10^-19 C)

q = -4.8 × 10^-18 C

Q6.

Electric lines of force never intersect each other because

• Lines of force always repel each other
• Lines of force always attract each other
• Point of intersection signifies two direction of electric field
• From the point of intersection four tangent can be drawn which signifies four direction of field

Solution

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.

Q7.

A material in which electrons are loosely bound and can move freely at room temperature is called a/an

• Insulator
• Semiconductor
• Superconductor
• Conductor

Solution

Conductors are those materials in which electrons are loosely bound and can move freely at room temperature.

Q8.

Which of the following statement is true?

• Electric field lines start from negative charge
• Electric field lines intersect with each other.
• Electric field lines for single positive charge are radiated inwards.
• Electric field lines do not form closed loops.

Solution

From the properties of electric field lines we know that

(i) field lines always start from positive charges and end at negative charges,

(ii) two field lines never cross each other,

(iii) electric field lines for single positive charge are radiated outwards,

(iv) electric field lines do not form closed loops

Q9.

If a body contains n₁ electrons and n₂ protons, the total amount of charge on the body is

• (n₂ - n₁)e
• (n₂ + n₁) e
• (n₂ /n₁)e
• (n₂ x n₁)e

Solution

The total charge on the body is the algebraic sum of the positive and negative charges on the body.

Hence, we have q = n₁(-e) + n₂e = (n₂ - n₁)e

Q10.

Two identical metallic spheres, having unequal opposite charges are placed at a distance of 0.90 m apart. After bringing them in contact with each other, they are again placed at the same distance apart. Now, the force of repulsion between them is 0.025 N. Calculate the final charge on each of them.

• 1.5 × 10 ^-6 C
• 1.5 × 10 ^-7 C
• 1.3 × 10 ^-6 C
• 2.5 × 10 ^-8 C

Solution

Let the initial charges be q₁ and q₂, respectively.

After they come in contact, the charges are rearranged such that they acquire the same charge.

let us say that charge on each of them is Q.

They are again brought apart at a distance of 0.9 m. Hence, the force between them will be given as

F = kQ²/r²

0.025 = (9 x 10⁹ x Q²) / 0.9²

Q² = (0.025 x 0.9²)/9 x 10⁹

Q = 1.5 x 10^-6 C

Q11.

Two point charges 2 μC and 8μC are placed 12 cm apart. The position of point from 2 μC charge, where the electric field intensity is zero is:

• 8 cm
• 2 cm
• 12 cm
• 4 cm

Solution

Consider two charges 2 μC and 8 μC at 12 cm 

Let P be the point at which the position where the electric field intensity is zero:

Q12.

The unit of permittivity of free space is

• C²/N-m²
• C²/ (N-m)²
• C/N-m
• N-m²/C²

Solution

Q13.

Two glass rods rubbed with silk are placed close to each other. They will

• Repel each other
• Attract each other
• Neither repel nor attract
• Cannot predict without knowing the amount of charge

Solution

The glass rods will acquire a positive charge, i.e. like charges are acquired by both. So, they will repel each other.

Q14.

The ratio of electric field due to an electric dipole at points situated at a distance r along its axial line and along its equatorial plane is

• 1:2
• 2:1
• 1:4
• 1:1

Solution

Electric field due to an electric dipole at points situated at a distance r along its axial line is

Electric field due to an electric dipole at points situated at a distance r along its equatorial plane is

From (i) and (ii), we get

Q15.

The dimensional formula of µ₀ is

• [M^-1 L^-3 T⁴ A²]
• [M^-1 L^-2 T⁴ A]
• [M L^-3 T⁴ A²]
• [M L^-3 T⁴ A]

Solution

Q16.

To make an uncharged object have a positive charge:

• Remove some electrons from it
• Add some electrons to it
• Remove some neutrons from it
• Add some neutrons to it

Solution

The removal of electrons from a body will effectively give it a net positive charge. Hence, to make a body have positive charge, remove electrons from it.

Q17.

The force between two small charged spheres having charges 3 × 10^-6 C and 4 × 10^-6 C placed 40 cm apart in air is

• 6.75 × 10^-1 N
• 67.5 × 10^-1 N
• 6.75 × 10^-2 N
• 67.5 × 10^-3 N

Solution

Given: q₁ = 3 × 10^-6 C; q₂ = 4 × 10^-6 C; r = 0.4 m

From the coulomb’s law, we get

Q18.

A charged comb attracts pieces of paper. Why does this happen?

• paper also gets charged.
• paper is polarized.
• paper has no charge.
• none of these.

Solution

The charged comb polarizes the pieces of paper that induces a net dipole moment in the direction of field.

Q19.

Quantisation of charge means that

• the charge on a body is a fraction of electronic charge, e
• the total charge on a body is always conserved
• the net charge on a body is always zero
• The charge on a body is an integral multiple of electronic charge, e

Solution

 The quantisation of charge is an integral multiple of electronic charge on a body.

Q20.

The field lines for single positive charge are:

• Radiating outwards
• Parallel
• Circular
• Radiating inwards

Solution

The field lines for single positive charge are radiated outwards.

Q21.

A silk cloth rubbed with a glass rod acquires a charge (-1.6 × 10^-19 C). Then the charge on the glass rod is

• - 1.6 × 10 ^-19 C
• +1.6 × 10 ^-19 C
• - 3.2 × 10 ^-19 C
• 3.2 × 10 ^-19 C

Solution

When one body is rubbed with another an equal and opposite charge is transferred. So, the charge on the glass rod is +1.6 × 10^-19 C.

Q22.

What is the angle between the electric dipole moment and the electric field on the equatorial line of the dipole?

• 0°
• 90°
• 180°
• none of the above

Solution

On equatorial line, the direction of electric field is reversed to that of angle of axial line. Therefore, the angle between dipole moment and electric field is 180°.

Q23.

When a charged rod is brought near the disc of a negatively charged gold leaf electroscope, it is observed that the divergence of leaves decreases.  What reference do you draw about the charge on the rod?

• Rod is positively charged
• Rod is negatively charged
• Rod has no charge
• None of the above

Solution

When a charged rod is brought near the disc of a negatively charged gold leaf electroscope, it is observed that the divergence of leaves decreases and the rod is positively charged because  the electrons move from leaves to the disc with the result that negative charge on leaves decreases.

Q24.

Electric flux through a surface does not mean a flow of electric charges through it.

• True
• False
• Depends on the surface
• Depends on the charge

Solution

The electric flux does not mean the flow of any physical quantity through a surface. It only signifies the amount of field lines which cross the surface.

Q25.

If an excess charge is placed on an isolated conductor, then, that amount of charge

• gets neutralized.
• resides on the surface of conductor.
• move inside the conductor.
• either resides on the surface of conductor or gets neutralized.

Solution

As the charge inside the conductor is zero, therefore extra charge stays on the outer surface of the conductor.

Q26.

A dipole is said to be in unstable equilibrium when angle between electric field and dipole moment is

• 45°
• 90°
• 180°
• 0°

Solution

A dipole is said to be in unstable equilibrium when a maximum torque acts on it. This happens when the angle between the electric field and the dipole moment is 180°. At this angle they are anti parallel to each other.

Q27.

For uniform electric field, field lines are:

• Divergent
• Convergent
• Convergent then divergent
• Parallel and equally spaced

Solution

Uniform electric field is represented by parallel and equally spaced filed lines.

Q28.

A test charge of 3nC is placed at origin and experiences a force of 9 x 10^-4 N. The electric field at origin is:

• 9 x 10⁵ N/C
• 3 x 10⁵ N/C
• 0.3 x 10⁵ N/C
• 1 x 10⁵ N/C

Solution

Q29.

Two point charges +4 μC and +6 μC repel each other with a force of 12 N. If each of them is given an additional charge -5 μC, what will be the new force between them?

• 1 N
• 3 N
• 5 N
• 2 N

Solution

Here, F = 12 N =

After adding an additional charge -5 μC, the charges become:

q₁ = 4 μC - 5 μC = -1 x 10^-6 C

q₂ = 6 μC - 5 μC = 1 x 10^-6 C

So, the new force is attractive in nature as the two charges are opposite to each other.

Q30.

A material in which electrons are tightly bound and cannot move freely at room temperature is called a/an

• Insulator
• Semiconductor
• Superconductor
• Conductor

Solution

Insulators are those materials in which electrons are tightly bound and cannot move freely at room temperature.

Q31.

What will be the magnitude of dipole moment for two charges ±10 nC separated by 10 mm?

• 2 Cm
• 10^-10 Cm
• 2 × 10^-10 Cm
• 20 Cm

Solution

Given: q = ±10 nC = 10^-8 C

2a = distance between two charges = 10 mm = 10^-2 m

Dipole moment, p = q(2a)

p = 10^-8 × 10^-2

p = 10^-10 Cm

Q32.

The S.I unit of electric flux is:

• N/m²C  
• Nm² /C
• m²C/N  
• NC/m²  

Solution

Flux is given as the product of electric field and area.

Therefore, the unit of flux is N/C × m² = Nm²/C

Q33.

‘A force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to other charges’ is the statement of

• Principle of Coulomb
• Principle of conservation of charge
• Principle of superposition
• Principle of conservation of force

Solution

According to the principle of superposition, a force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to other charges.

Q34.

Which among the following is false for induction?

(a) The process of induction happens instantly.

(b) After induction, charges move towards the centre of the conductor.

• Only a
• Only b
• Both a and b are false
• Neither a nor b is false

Solution

The process of charge transfer by induction is an instantaneous process. After induction, the charges remain on the surface and redistribute themselves.

Q35.

Which of the following quantities is a vector?

• Electric flux
• Electric potential
• Electric potential energy
• Electric field intensity

Solution

Electric field intensity is vector quantity.

Q36.

Choose the correct statement from the following

• The total positive charge of the universe is constant
• The total negative charge of the universe is constant
• The total charge of the universe is constant
• The total number of charged particles in the universe remains constant

Solution

From the law of conservation of charge, we can confer that the total charge of the universe remains constant.

Q37.

Trajectory of an electron when it moves perpendicular to the electric field is

• Parabolic
• Circular
• Straight
• Hyperbolic

Solution

Trajectory of an electron when it moves perpendicular to the electric field is parabolic

Q38.

To make an uncharged object have a negative charge:

• Remove some electrons from it
• Add some electrons to it
• Add some neutrons to it
• Remove some neutrons from it

Solution

The addition of electrons to a body give a net negative charge to it. So, to give a negative charge to a body add some electrons to it.

Q39.

The dipole moment of a system consisting of a proton and an electron separated by 4 nm is

• 3.2 × 10⁻²⁸ Cm
• 12.8 × 10⁻²⁸ Cm
• 19.2 × 10⁻²⁸ Cm
• 6.4 × 10⁻²⁸ Cm

Solution

The magnitude of the charge on a proton and an electron is the same (q = 1.6 × 10⁻¹⁹ C).

Hence, the dipole moment of the system is

 

Q40.

At what point is the electric field intensity due to a uniformly charged spherical shell maximum?

• inside the spherical shell
• outside the spherical shell
• at the surface of spherical shell
• at the centre of spherical shell