Electric Field MCQ Quiz - Objective Question with Answer for Electric Field - Download Free PDF

Concept:

• When there are various types of charges in a region, but the total charge is zero, the region is supposed to contain a number of dipoles. 
• Therefore, at points outside the region, the dominant electric field is inversely proportional to the cube of r for a large value of r 
• The electric field due to the dipole is 

\(E = \frac{q}{4 \pi \epsilon_0}\frac{2P}{r^3}\)

Where p = q × l (dipole)

r is the distance from the dipole r >> l

Hence the correct option is option 3

Concept Used:

The magnetic energy stored in a coaxial cable arises due to the magnetic field (B) generated by the steady current flowing through the inner conductor.

The magnetic energy density in a medium is given by:

u = B² / (2μ₀)

The total magnetic energy (U) stored in a volume V is:

U = ∫ u dV

Calculation:

For a coaxial cable:

B(r) = (μ₀ I) / (2πr)

dU = (B² / (2μ₀)) × (2πr dr L)

The magnetic field (B) at a distance r from the axis (inside the cylindrical region) is given by Ampere’s Law :

The energy stored per unit length (dU/dL) in a cylindrical shell of radius r and thickness dr is:

Integrating from r = a to r = b :

⇒ U = ∫[(μ₀ I² / (8π² r²)) × (2πr dr L)] from a to b

⇒ U = (μ₀ I² L / (4π)) ∫(dr / r) from a to b

⇒ U = (μ₀ I² L / (4π)) ln(b / a)

The total magnetic energy stored between the inner and outer conductor is:

⇒ (μ₀ I² L / 4π) ln(b / a)

For b=4 a

2(μ₀ I² L / 4π) ln(2)

Concept Used:

The Poynting vector (S) represents the rate of energy transport per unit area due to electric and magnetic fields .

The Poynting vector is given by:

S = (E × B) / μ₀

For a coaxial cable:

The electric field (E) at the surface of the inner conductor (r = a) is given by: E = λ / (2πε₀a)

The magnetic field (B) at the surface of the inner conductor (r = a) is given by: B = μ₀I / (2πa)

Calculation:

The Poynting vector magnitude at the surface of the inner conductor is:

⇒ S = (E × B) / μ₀

⇒ S = [(λ / (2πε₀a)) × (μ₀I / (2πa))] / μ₀

⇒ S = (λI) / (4π²a²ε₀)

The energy transported per unit time, per unit area at the surface of the inner conductor is:

⇒ (λI) / (4π²a²ε₀)

Concept:

The work done (W) to rotate a dipole of dipole moment p by an angle θ in a uniform electric field E is given by:

W = -pE cos(θ) + pE cos(0)

For a rotation of 180° (θ = 180°), cos(180°) = -1, and the work done becomes:

W = -2pE

The dipole moment p is given by:

p = q × d

Where q is the charge on each center (approximated to the elementary charge e = 1.6 × 10⁻¹⁹ C for the water molecule), and d is the separation of the centers of charges.

Calculation:

Given,

Work done, W = 5 × 10⁻²⁵ J

Electric field, E = 2.5 × 10⁴ N/C

The equation for work done is:

W = -2pE

Substitute the given values:

5 × 10⁻²⁵ = -2p(2.5 × 10⁴)

⇒ 5 × 10⁻²⁵ = -5 × 10⁴ p

⇒ p = -1 × 10⁻²⁹ C·m

The dipole moment p is also given by:

p = q × d

For a water molecule, q = 1.6 × 10⁻¹⁹ C

Substitute the values to solve for d:

d = p / q

d = (1 × 10⁻²⁹) / (1.6 × 10⁻¹⁹)

d ≈ 0.625 × 10⁻¹⁰ m

∴ The approximate separation of the centers of charges is 0.625 × 10⁻¹⁰ m.
Hence, the correct option is 2) 0.625 × 10⁻¹⁰ m.

CONCEPT:

Electric Field Intensity:

• The electric field intensity at any point is the strength of the electric field at the point.
• It is defined as the force experienced by the unit positive charge placed at that point.

\(\vec E = \frac{{\vec F}}{{{q_o}}}\)

Where F = force and qo = small test charge

• The magnitude of the electric field is

\(E = \frac{{kq}}{{{r^2}}}= \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}\)

Where K = constant called electrostatic force constant, q = source charge and r = distance

• The region around a charged particle in which electrostatic force can be experienced by other charges is called the electric field.
  • The electric field is denoted by E.

EXPLANATION:

• Electric field intensity is a vector quantity because it can only be properly defined when its magnitude and direction both are known. So option 2 is correct.

CONCEPT:

Electric field intensity: 

• It is defined as the force experienced by a unit positive test charge in the electric field at any point.
• The positive charge experiences a force in the direction of the electric field and the negative charge experiences a force in the direction opposite of the electric field.

\(⇒ E=\frac{F}{q_{o}}\)    

Where E = electric field intensity, qo = charge on the particle

CALCULATION:

Given q_o = -6C, and F = 60 N towards north

• We know that the electric field intensity is given as,

\(⇒ E=\frac{F}{q_{o}}\)

\(⇒ E=\frac{60}{6}\)

⇒ E = 10 N/C

• The negative charge experiences a force in the direction opposite of the electric field.
• So the direction of the electric field is towards the south. Hence, option 2 is correct.

The correct answer is option 2) i.e. Electric field

CONCEPT: 

• Electric field: The region around an electric charge in which it can influence other charges is known as the electric field. 
  • The electric field intensity (E) at any point is defined as the force experienced by a unit charge placed at that location.

It is given by:

​\(E = \frac{F}{q}\)

EXPLANATION:

The electric field intensity at a point is given by:

 \(E = \frac{F}{q}\) .

• Therefore, the force acting per unit positive test charge at a location is a measure of the intensity of the electric field.

Additional Information

• Electric potential is the difference in potential energy per unit charge between two points in an electric field.
• Coulomb force is the force of interaction between electric charges separated by a distance.
• Gravity is the universal force of attraction acting between all matter. 

CONCEPT:

The electric field in the form of potential can be written as;

\(E = - \frac{{dV}}{{dr}} \) ---- (1)

Where V is the potential.

CALCULATION:

Given :

V =  0.2 m³, V = 5 V

using equation (1) we have,

\(E = - \frac{{dV}}{{dr}} \)

⇒ \(E = - \frac{{d(5)}}{{dr}} \)

⇒ E = 0

Hence, option 3) is the correct answer.

CONCEPT:

• Electric field force: When a charge moves in an electric field, a force will act on it, that is known as electric field force.
  • Due to the electric field, Force is always in the direction of the electric field if the charge is positive.

F = q E

where F is the force due to the electric field, q is the charge, and E is the electric field. 

CALCULATION:

Given that:

E = 104 Newton per Coulomb and q = 1.6 × 10^-19 C

F = q E

F = (1.6 × 10-19) × 104

F = 1.6 × 10-15 Newton

So the correct answer is option 4.

CONCEPT:

Electric field intensity: 

• It is defined as the force experienced by a unit positive test charge in the electric field at any point.

\(⇒ E=\frac{F}{q_{o}}\)    

Where E = electric field intensity, q_o = charge on the particle

EXPLANATION:

• The magnitude of electric force experienced by a charged particle in an electric field is given as,

\(\Rightarrow F=Eq_{o}\)   

• From the above equation, it is clear that the magnitude of electric force experienced by a charged particle in an electric field depends on the magnitude of the charge on the particle.
• Hence, option 1 is correct.

Concept:

Electric Field

• The electric field or electric field intensity at any point is the strength of the electric field at the point.
• It is defined as the force experienced by the unit positive charge placed at that point.

\(⇒ \vec E = \frac{{\vec F}}{{{q_o}}}\)

Where F = force and qo = small test charge

• The magnitude of the electric field is

\(⇒ E = \frac{{kq}}{{{r^2}}}\)

Where K = constant called electrostatic force constant, q = sorce charge and r = distance

Electric Potential

• Electric potential is equal to the amount of work done per unit charge by an external force to move the charge q from infinity to a specific point in an electric field.

\(⇒ V=\frac{W}{q}\)

• The difference in electric potential between two points is called potential difference or voltage.
• The voltage is the electric field divided by the distance.

Explanation:

• The electric field is the force by charge or potential multiplied by the distance.
• SI Unit of electric potential is volt and SI unit of distance is meter.
• Unit of potential by distance is Volt / meter or V m^-1.

So, the correct option is Vm-1. 

CONCEPT:

• The space or region around the electric charge in which electrostatic force can be experienced by other charged particle is called as an electric field by that electric charge.
• The imaginary lines which is used to represent the electric field are called an electric field line.
• The tangent line at a point on the electric field line gives the direction of the electric field at that point.

• The field lines emerge from positive charge and terminate at a negative charge.
• They originate and end at right angles to the surface of the charge.
• Electric field lines do not make loop. 
• The magnitude of the electric field will be maximum where the number of field lines is maximum. 

Explanation:

From the above explanation, we can see that electric lines of force are imaginary lines used to describe the strength of an electric field.

Whereas unlike magnetic field lines, electric field lines don't form a closed loop 

Hence option 4 is incorrect among  all

CONCEPT:

• Electric field: The space or region around the electric charge in which electrostatic force can be experienced by other charged particle is called an electric field by that electric charge.
  • It is denoted by E and the SI unit of the electric field is N/C.
• Electric field force: When a charge moves in an electric field, a force will act on it, which is known as electric field force.
  • Due to the electric field Force, the charged particle always moves in the direction of the electric field.

F = qE

where F is the force due to the electric field, q is the charge, and E is the electric field.

EXPLANATION:

1. The force per unit charge (E = F/q) is called the electric field. So option 1 is correct.
2. The electric charge flowing per unit time is called current.
3. The electric field per unit area is called electric flux.
4. The electrostatic potential energy per unit charge is called electric potential.

Explanation:

Given

We know that,

Electric field due to the uniformly charged large surface is given as:

E = σ/ 2ϵ_o  ----- (1)

From equation (1) it is clear that the electric field due to a uniformly charged surface is independent of distance from the surface.

It only depends on the surface density.

Hence in both cases, the Electric field will be the same, i.e.

E₁ = E2 = σ/ 2ϵo

Hence option 3) is the correct choice.