Electrostatics MCQ Quiz - Objective Question with Answer for Electrostatics - Download Free PDF

Last updated on Jul 9, 2025

Latest Electrostatics MCQ Objective Questions

Electrostatics Question 1:

If there are more than one charge distribution in Gaussian surface, the net charge is __________.

  1. algebraic sum of all the odd charges
  2. algebraic sum of all the even charges
  3. difference of all the charges
  4. algebraic sum of all the individual charges

Answer (Detailed Solution Below)

Option 4 : algebraic sum of all the individual charges

Electrostatics Question 1 Detailed Solution

Concept:

According to Gauss's Law, the total electric flux through a closed surface (Gaussian surface) is proportional to the net charge enclosed within that surface.

If there are multiple charges inside the surface, the net charge is determined by the algebraic sum of all the individual charges (both positive and negative).

This includes all types of charges, regardless of being odd or even.

Final Answer: algebraic sum of all the individual charges

Electrostatics Question 2:

Which of the following is one of the boundary conditions for time varying fields?

  1. The tangential component of magnetic field intensity is discontinuous across the surface except for a perfect conductor.
  2. The normal component of electric flux density is discontinuous at the boundary if the surface charge density is zero.
  3. The normal component of magnetic flux density is continuous at the boundary.
  4. The tangential component of electric field intensity is discontinuous at the surface.

Answer (Detailed Solution Below)

Option 3 : The normal component of magnetic flux density is continuous at the boundary.

Electrostatics Question 2 Detailed Solution

Explanation:

Boundary Conditions for Time-Varying Fields

Definition: Boundary conditions in electromagnetics describe how electromagnetic field quantities behave at the interface between two different media. For time-varying fields, these boundary conditions are derived from Maxwell's equations and are essential for solving electromagnetic problems involving interfaces.

Correct Option:

The correct option is:

Option 3: The normal component of magnetic flux density is continuous at the boundary.

Explanation of Correct Option:

The boundary condition on the magnetic flux density (B) arises from Gauss's law for magnetism, which states:

∇ • B = 0

This implies that the net magnetic flux through any closed surface is zero, meaning magnetic monopoles do not exist. At the boundary between two media, the integral form of Gauss's law for magnetism can be expressed as:

∮ B • dA = 0

When applied to a small Gaussian pillbox that straddles the boundary, the contributions to the flux integral come from the two faces of the pillbox (one in medium 1 and the other in medium 2). The result is:

B₁ₙ = B₂ₙ

Here, B₁ₙ and B₂ₙ are the normal components of the magnetic flux density in the two media. This shows that the normal component of B is continuous across the boundary, regardless of the properties of the media. This condition is fundamental and applies to all interfaces, making Option 3 correct.

Important Note: While the magnetic flux density's normal component is continuous, the tangential component of B may be discontinuous if there is a surface current at the boundary.

Additional Information:

The continuity of the normal component of magnetic flux density ensures that there are no magnetic monopoles, aligning with Maxwell's equations' fundamental principles. This boundary condition is vital for solving problems involving magnetic fields in different media, such as in transformers, inductors, and waveguides.

Analysis of Other Options

To ensure a comprehensive understanding, let’s evaluate the other options:

Option 1: The tangential component of magnetic field intensity is discontinuous across the surface except for a perfect conductor.

This statement is incorrect. The tangential component of the magnetic field intensity (H) is continuous across the boundary unless there is a surface current density (K) present. If a surface current exists, the tangential components of H in the two media are related by:

(H₂ - H₁)ₜ = K

For a perfect conductor, the tangential component of H at the surface is zero, but this condition applies only to perfect conductors, not general boundaries. Thus, Option 1 is not universally correct.

Option 2: The normal component of electric flux density is discontinuous at the boundary if the surface charge density is zero.

This statement is incorrect. The boundary condition for the normal component of the electric flux density (D) is given by:

D₂ₙ - D₁ₙ = ρₛ

Here, ρₛ is the surface charge density at the boundary. If ρₛ is zero, the normal component of D is continuous across the boundary. The statement in Option 2 contradicts this condition by asserting discontinuity even when the surface charge density is zero.

Option 4: The tangential component of electric field intensity is discontinuous at the surface.

This statement is incorrect. The tangential component of the electric field intensity (E) is continuous across the boundary unless there is a time-varying magnetic field present. The boundary condition for the tangential component of E is derived from Faraday's law:

∇ × E = -∂B/∂t

In the absence of time-varying magnetic fields, the tangential component of E is continuous. However, in the presence of time-varying magnetic fields, the tangential component may vary, but this is a specific case and not a general rule. Thus, Option 4 is not universally correct.

Conclusion:

The correct boundary condition for time-varying fields, as described, is that the normal component of magnetic flux density (B) is continuous at the boundary, making Option 3 correct. This condition is derived from Gauss's law for magnetism and is fundamental to the behavior of magnetic fields in different media. The analysis of other options highlights common misconceptions and emphasizes the importance of understanding Maxwell's equations and their implications for boundary conditions.

Electrostatics Question 3:

A positive point charge of 10–8 C is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm. The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved by a distance of 10 cm further away from the center of the sphere along the radial direction. Taking  = 9 × 10Nm2 /C(where ∈0 is the permittivity of free space), which of the following statements is/are correct: :

  1. Before the grounding, the electrostatic potential of the sphere is 450 V.
  2. Charge flowing from the sphere to the ground because of grounding is 5 × 10–9C.
  3. After the grounding is removed, the charge on the sphere is −5 × 10–9C.
  4. The final electrostatic potential of the sphere is 300 V. 

Answer (Detailed Solution Below)

Option :

Electrostatics Question 3 Detailed Solution

Calculation:

We have net potential at center is kQ / r + k(-q) / R = 0

⇒ q = QR / r

→ Initial potential = (9 × 109 × 10-8) / (1 / 5) = 450 volt (A)

→ q = 10-8 × 10 / 20 = 5 × 10-9 C flows (B) to ground

→ -q = -5 × 10-9 C (C)

→ 9 × 109 [(-5 × 10-9) / (1 / 10) + (10-8 × 10) / 3] = 9 × 10 [(-5) + (10 / 3)] = 90 × (-5 / 3) = -150 V

Electrostatics Question 4:

Consider a spherical region of radius  centered at the origin of the coordinate system. Three point charges each of magnitude 𝑄 are placed at (0,0,𝑅),(0,𝑅,0) and (√2𝑅,0,0). What is the magnitude of the average electric field over the spherical region due to these charges in units of 

  1. 0

Answer (Detailed Solution Below)

Option 4 :

Electrostatics Question 4 Detailed Solution

Solution:

The total electric field at the origin is the vector sum of the fields due to each charge:

Etotal = E1 + E2 + E3 = (Q / 4πε0R2) × ( z + Y + (1/2)X )

The magnitude of the average electric field is:

|Eavg| = (3/2) × ( Q / 4πε0R2 )

Electrostatics Question 5:

A small bob of mass 100 mg and charge +10 C is connected to an insulating string of length 1 m. It is brought near to an infinitely long non conducting sheet of charge density '' as shown in figure. If string subtends an angle of 45° with the sheet at equilibrium the charge density of sheet will be :
(Given,  = 8.85 × 10–12  and acceleration due to gravity, g = 10 m/s2

  1.  0.885 nC/m
  2. 17.7 nC/m2 
  3.  885 nC/m2
  4. 1.77 nC/m2

Answer (Detailed Solution Below)

Option 4 : 1.77 nC/m2

Electrostatics Question 5 Detailed Solution

Calculation:

The FBD is shown below

From equilibrium condition:

⇒ q × (σ / 2 ε0) = mg

⇒ σ = (2 ε0 mg) / q

⇒ σ = (2 × 8.85 × 10-12 × 100 × 10-6 × 10) / (10 × 10-6)

⇒ σ = 17.7 × 10-10 C/m2

⇒ σ = 1.77 nC/m2

∴ Surface charge density σ = 1.77 nC/m2.

Top Electrostatics MCQ Objective Questions

Electric flux is a _______ field, and its density is a _______ field.

  1. Vector, vector
  2. Scalar, vector
  3. Vector, scalar
  4. Scalar, scalar

Answer (Detailed Solution Below)

Option 2 : Scalar, vector

Electrostatics Question 6 Detailed Solution

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Concept:

Electric Flux:

  • It is defined as the number of electric field lines associated with an area element.
  • Electric flux is a scalar quantity, because it's the dot product of two vector quantities, electric field and the perpendicular differential area.
     ϕ = E.A = EA cosθ 
  • The SI unit of the electric flux is N-m2/C.

 

Electric flux density (D) is a vector quantity because it is simply the product of the vector quantity electric field and the scalar quantity permittivity of the medium, i.e.

Its unit is Coulomb per square meter.

The force between two charges is 200 N. If the distance between the charges is doubled, the force will be _______.

  1. 400 N
  2. 100 N
  3. 50 N
  4. 200 N

Answer (Detailed Solution Below)

Option 3 : 50 N

Electrostatics Question 7 Detailed Solution

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Concept:

Coulomb's law: 

It states that the magnitude of the electrostatic force F between two point charges q1 and q2 is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance r between them. 

  • It is represented mathematically by the equation:

Where ϵ0 is the permittivity of free space (8.854 × 10-12 C2 N-1 m-2).

The value of 

Calculation:

So, initial the force between two charges q1 and q2 is 200 N. 

 -- (1)

If new distance r' = 2 r

New Force is

     ---(2)

From (1) and (2)

or

So, the correct option is 50 N.

The potential inside a charged hollow sphere is __________.

  1. Zero
  2. Same as that on the surface
  3. Less than that on the surface
  4. None of these

Answer (Detailed Solution Below)

Option 2 : Same as that on the surface

Electrostatics Question 8 Detailed Solution

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  • The electric field inside a conducting sphere is zero, so the potential remains constant at the value it reaches the surface.
  • When a conductor is at equilibrium, the electric field inside it is constrained to be zero.
  • Since the electric field is equal to the rate of change of potential, this implies that the voltage inside a conductor at equilibrium is constrained to be constant at the value it reaches the surface of the conductor.
  • A good example is the charged conducting sphere, but the principle applies to all conductors at equilibrium.

What is the work done to move a unit charge from one point to another in an electric circuit carrying some current, between those two points? 

  1. Pressure distance
  2. Potential difference
  3. Potential distance
  4. Pressure difference

Answer (Detailed Solution Below)

Option 2 : Potential difference

Electrostatics Question 9 Detailed Solution

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Concept:

Potential difference:
 
  • The electric potential difference between two points in an electric circuit carrying some current is the work done to move a unit charge from one point to the other.
  • The standard metric unit on electric potential difference is the volt, abbreviated V. 
  • The potential difference is the work done per unit charge.

Mathematically, it is defined as:

V = Potential difference

W = Work done

Q = electric charge.

The S.I unit of work is joule and that of the charge is the coulomb.

Two charges of + 4 μC and -16 μC are separated from each other by a distance of 0.6 m. At what distance should a third charge of + 6 μC be placed from + 4 μC so that net force exerts on it will be zero?

  1. 0.4 m
  2. 0.6 m
  3. 1.2 m
  4. 0.3 m

Answer (Detailed Solution Below)

Option 2 : 0.6 m

Electrostatics Question 10 Detailed Solution

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CONCEPT:

Coulomb’s law: When two charged particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.

 

Force (F) ∝ q1 × q2

Where K is a constant = 9 × 109 Nm2/C2 

Calculations:

Consider new charge + 6μC is placed d m apart from old +4 μC charge and (x + 0.6) m apart from -16 μC charge.

Let,
qA = + 4 μC at point A
qB = - 16 μC at point B
qC = + 6 μC at point C

Since, net force on charge qC will zero.

∴ |FCA| = |FCB|

From above concept,

4d2 = (0.6 + d)2

4d2 = 0.36 + d2 + 1.2d

3d2 - 1.2d - 0.36 = 0

d1 = - 0.2 m

d2 = + 0.6 m

Hence, according to option + 0.6 m distance a third charge of + 6 μC be placed from + 4 μC so that the net force exerted on it will be zero.

The unit of dielectric strength is given by

  1. V/m
  2. V2/m
  3. m/V
  4. m/V2

Answer (Detailed Solution Below)

Option 1 : V/m

Electrostatics Question 11 Detailed Solution

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Dielectric Materials: The insulating materials which are poor conductor of electric current are called dielectric materials.

When the dielectric is placed in the electric field no current flows through them but exhibits electric dipole i.e. there is the separation of positive and negative electrically charged entities on a molecular or atomic level.

These are used in capacitors, power line and electric insulation, switch bases and light receptacles.

Dielectric strength: The maximum voltage that can be applied to a given material without causing it to break down is called dielectric strength.

It is measured in volts per unit thickness of the material.

EXPLANATION

The dielectric strength of an insulating material is measured in volts (V) per unit thickness (m) of the material (V/m). So option 1 is correct.

What is the unit of Electric field intensity?

  1. Coulomb/Meter square
  2. Coulomb/Joule
  3. Newton/Coulomb
  4. Faraday/ Coulomb

Answer (Detailed Solution Below)

Option 3 : Newton/Coulomb

Electrostatics Question 12 Detailed Solution

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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.

 Newton/Coulomb

Where F = force and qo = small test charge

The magnitude of the electric field is

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.

A dielectric material is placed in vacuum in a uniform electric field of E = 4 V/m. What is the electric field inside the material if the relative permittivity of dielectric material is 2?

  1. Zero
  2. 4 V/m
  3. 2 V/m
  4. 8 V/m

Answer (Detailed Solution Below)

Option 3 : 2 V/m

Electrostatics Question 13 Detailed Solution

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Dielectric constantr) is defined as the ratio of the electric permittivity of the material to the electric permittivity of free space.

The presence of the dielectric reduces the effective electric field.

The relative permittivity of the vacuum is 1.

As the relative permittivity increases by a factor of 2, the electric field decreases by a factor of 2.

Therefore, the electric field = 2 V/m

Electric intensity at any point in an electric field is equal to the ___________ at that point.

  1. electric flux
  2. magnetic flux density
  3. potential gradient
  4. none of them

Answer (Detailed Solution Below)

Option 3 : potential gradient

Electrostatics Question 14 Detailed Solution

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The total electric field at any point is equal to the vector sum of the separate electric fields that each point charge would create in the absence of the others. That is,

The electric field is nothing but the potential gradient of that particular point.

Two charges q1 and q2 are placed on the x-axis, with q1 at x = a, and q2 at x = 3a. A third charge Q is placed at the origin. The net force on charge Q is zero if q2 is equal to:

  1. 3 q1
  2. -3 q1
  3. 9 q1
  4. -9 q1

Answer (Detailed Solution Below)

Option 4 : -9 q1

Electrostatics Question 15 Detailed Solution

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Coulomb’s law:

When two charged particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.

Force (F) ∝ q1 × q2

Where K is a constant = 9 × 109 Nm2/C2

Calculation:

The given situation can be visualized as shown:

The force on Q because of q1 will be:

Similarly, the force on Q because of q2 will be:

For the net charge at Q to be zero, we write:

q2 = -9q1

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