Gauss’s Law MCQ Quiz - Objective Question with Answer for Gauss’s Law - Download Free PDF

The correct answer is: Option 2) σ / (2ε₀)

Concepts:

Gauss’s Law: The total electric flux through a closed surface is equal to the net charge enclosed divided by the permittivity of free space (ε₀).

Mathematical form: Φ = Q_enclosed / ε₀

To find the electric field of an infinite sheet, a cylindrical Gaussian surface (also called a “pillbox”) is used, with its flat surfaces parallel to the sheet and extending equal distances on either side.

Derivation:

Let the surface charge density be σ (charge per unit area).

Charge enclosed by the Gaussian surface = σ × A, where A is the area of the pillbox facing the sheet.

Electric flux Φ = E × A (from one side) + E × A (from the other side) = 2EA

By Gauss’s Law: 2EA = σA / ε₀

⇒ E = σ / (2ε₀)

Explanation:

• Option 1: Gauss law is true for any open surface.

This statement is incorrect. Gauss's law is valid for a **closed surface** and applies to any surface that fully encloses a volume. It is used to calculate the electric flux through a closed surface, which involves the charge enclosed within that surface. An open surface does not satisfy the condition for Gauss's law.

• Option 2: Gauss law includes the sum of all charges enclosed by the surface for calculation of electric flux through the surface.

This statement is correct. Gauss's law states that the total electric flux through a closed surface is directly proportional to the sum of the charges enclosed within that surface. The mathematical form is given by: Φ = (1/ε₀) * Σ Q_enc, where Q_enc is the sum of the enclosed charges.

• Option 3: Gauss law can be used to calculate the magnetic field due to steady current.

This statement is incorrect. Gauss's law applies to electric fields and is related to the distribution of electric charge. To calculate magnetic fields, Ampere's law is used, which relates the magnetic field to the current flowing through a loop. Gauss's law does not apply to the magnetic field in this context.

• Option 4: Gauss law is not based on the inverse square dependence on distance contained in Coulomb's law.

This statement is incorrect. Gauss's law is actually consistent with Coulomb's law. Coulomb's law describes the force between two point charges, which is inversely proportional to the square of the distance between them. Gauss's law, in its integral form, also leads to the same inverse square law for point charges, confirming the relationship between electric flux and charge distribution.

Correct Answer: Option 2) Gauss law includes the sum of all charges enclosed by the surface for calculation of electric flux through the surface.

Calculation:

According to Gauss's Law, the electric flux (Φ) through a closed surface is directly proportional to the charge enclosed (Q):

Φ = Q / ε₀

Where:

Φ is the electric flux,

Q is the charge enclosed,

ε₀ is the permittivity of free space.

The initial flux from a cube with side 1 m is Φ.

The side of the cube is increased to 3 m, and the charge enclosed is reduced to one-third of its original value (Q' = Q / 3).

Using the flux relation, we get:

Φ' = Q' / ε₀ = (Q / 3) / ε₀ = (1/3) × (Q / ε₀)

Since the original flux is Φ = Q / ε₀, we substitute this back to get:

Φ' = Φ / 3

Thus, the electric flux for the larger cube with side 3 m and one-third the original enclosed charge is Φ / 3.

Calculation:
For the ball to be just suspended, the net force acting on it must be zero:

F_electric = F_gravity - F_buoyant

Substituting the expressions for each force:

qE = ρ(4/3 π r³)g - σ(4/3 π r³)g

Isolating q, we get:

q = (4/3 π r³)(ρ - σ)g / E

The correct expression for the charge on the metal ball is:

q = (4 π r³(ρ - σ)g) / 3E

Calculation:

For any θ, let us first find the flux inside a cone of half angle θ. we know that for such a cone, solid angle subtended at centre is

Ω = 2π [1 – cosθ]

⇒ Flux through 1 cone = \(\phi_{0}=\frac{\Omega}{4 \pi} \cdot \frac{Q}{\varepsilon_{0}}=\frac{Q}{2 \varepsilon_{0}}[1-\cos \theta]\)

⇒ Flux through curved surface

= \(\frac{Q}{\varepsilon_{0}}-2 \phi_{0}\)

= \(\frac{Q}{\varepsilon_{0}}-\frac{Q}{\varepsilon_{0}}[1-\cos \theta]=\frac{Q}{\varepsilon_{0}} \cos \theta\) 

⇒ \(\phi=\frac{Q}{\varepsilon_{0}} \cdot \frac{\sqrt{3}}{2}\)

And \(\frac{\phi}{\sqrt{n}}=\frac{Q}{\varepsilon_{0}} \cdot \frac{1}{2}\)

⇒ \(\sqrt{n}=\sqrt{3}\)

⇒ n = 3

CONCEPT:

• Gauss's Law: Gauss's law for the electric field describes the static electric field generated by a distribution of electric charges. 
  • It states that the electric flux through any closed surface is proportional to the total electric charge enclosed by this surface. 

\(\phi_E = \frac{Q}{\epsilon_o}\)

Where ϕ_E = electric flux through a closed surface S enclosing any volume V, Q = total charge enclosed with V, and ϵ_o = electric constant

• Electric flux is the flow of the electric field through a given area.

• Electric flux is proportional to the number of electric field lines going through a virtual surface.

\({\phi}_E= E\cdot S = EScos\theta\)

Where E = electric field, S = are of the surface, E = magnitude, θ = angle between the electric field lines and the normal (perpendicular) to S, and ϕ_E = flux the electric field through a closed cylindrical surface.

EXPLANATION:

• According to Gauss's law

 \(ϕ_E = \frac{Q_(enclosed )}{ϵ_o}\)

• If the radius of the Gaussian surface is doubled, the outward electric flux will remain the same.
• This is because electric flux depends only on the charge enclosed by the surface.

​option 4 is correct.

CONCEPT:

Gauss's law:

• According to Gauss law, the total electric flux linked with a closed surface called Gaussian surface is \(\frac{1}{ϵ_o}\) the charge enclosed by the closed surface.

\(\Rightarrow ϕ=\frac{Q}{ϵ_o}\)

Where ϕ = electric flux linked with a closed surface, Q = total charge enclosed in the surface, and ϵo = permittivity

Important points:

1. Gauss’s law is true for any closed surface, no matter what its shape or size.
2. The charges may be located anywhere inside the surface.

EXPLANATION:

Gauss's law:

• According to Gauss law, the total electric flux linked with a closed surface called Gaussian surface is \(\frac{1}{ϵ_o}\) the charge enclosed by the closed surface.
• So if the total charge enclosed in a closed surface is Q, then the total electric flux associated with it will be given as,

\(\Rightarrow ϕ=\frac{Q}{ϵ_o}\)     -----(1)

• By equation 1 it is clear that the total flux linked with the closed surface in which a certain amount of charge is placed does not depend on the shape and size of the surface. Hence, option 4 is correct.

The correct answer is option 1) i.e. Φ/2

CONCEPT:

Gauss's Law for electric field: It states that the total electric flux emerging out of a closed surface is directly proportional to the charge enclosed by this closed surface. It is expressed as:

\(ϕ = \frac{q}{ϵ_0}\)

Where ϕ is the electric flux, q is the charge enclosed in the closed surface and ϵ₀ is the electric constant.

EXPLANATION:

Given that:

Consider a charge 'q' placed inside a cube of side 'a'.

The electric flux according to Gauss's law, 

\(\Rightarrow ϕ = \frac{q}{ϵ_0}\)

If the charge enclosed is halved, then  

\(\Rightarrow q' =\frac{q}{2}\)

Therefore, the new electric flux associated with this, 

\(\Rightarrow ϕ' =\frac{q'}{\epsilon_0}= \frac{\frac{q}{2}}{ϵ_0} = \frac{1}{2}\times \frac{q}{\epsilon_0} \)

\(\Rightarrow \phi '= \frac{1}{2} \times \phi = \frac{\phi}{2}\)

The electric flux emerging from a closed surface is independent of the shape or dimensions of the closed surface.

• According to gauss law, for any enclosed surface the flux associated is 1/ϵ_o times the total charge enclosed within the surface.
• It is mathematically expressed as : \(\oint \vec{E}.d\vec{S}=\frac{q}{\epsilon _o} \) 
• The flux associated with a surface is independent of the shape and size of the surface.
• It only depends on the number of charges enclosed within that surface.
• The surface should remain closed.
• Thus for any shape, if the number of charges enclosed within the surface is the same then the flux will also be the same.

CONCEPT:

Gauss's law:

• According to Gauss law, the total electric flux linked with a closed surface called the Gaussian surface is \(\frac{1}{ϵ_o}\) the charge enclosed by the closed surface.

\(⇒ ϕ=\frac{Q}{ϵ_o}\)

Where ϕ = electric flux linked with a closed surface, Q = total charge enclosed in the surface, and ϵo = permittivity

Important points:

1. Gauss’s law is true for any closed surface, no matter what its shape or size.
2. The charges may be located anywhere inside the surface.

Linear charge density:

• It is defined as the quantity of charge per unit length.
• Its SI unit is C/m.
• If ΔQ charge is contained in the line element Δl, the linear charge density λ will be,

​\(\Rightarrow \lambda=\frac{\Delta Q}{\Delta l}\)

CALCULATION:

Given charge per meter length on the wire = Q C/m, the radius of the wire = 3 mm, and side of the cube = 2 m

• The diagram of this case will be given as,

• The 2 m length of the wire will be enclosed in the cube.

From figure 1 it is clear that the total charge enclosed in the cube is,

⇒ Q_C = 2Q     -----(1)

Since the cube is a closed surface so the total electric flux through the cube is given as,

\(⇒ ϕ=\frac{Q_C}{ϵ_o}\)

\(⇒ ϕ=\frac{2Q}{ϵ_o}\)

• Hence, option 3 is correct.

CONCEPT

• Electric Flux: It is defined as the number of electric field lines passing through the perpendicular unit area.

• Electric Flux = (Φ) = EA_⊥ [E = electric field, A = perpendicular area]
• Electric flux (Φ) = EA cos θ [where θ is the angle between area plane and electric field]

• The flux is maximum when the angle is 0°
• Gauss Law: According to gauss’s law, total electric flux through a closed surface enclosing a charge is 1/ϵ₀ times the magnitude of charge enclosed.

\(i.e.~{{\mathbf{\Phi }}_{net}}=\frac{\left( {{Q}_{in}} \right)}{{{\epsilon }_{0}}}\)

\(i.e.~\oint \vec{E}\cdot d\vec{S}=\frac{{{Q}_{in}}}{{{\epsilon }_{0}}}\)

Where, Φ = electric flux, Q_in = charge enclosed the sphere, ϵ₀ = permittivity of space (8.85 × 10^-12 C²/Nm²), dS = surface area

EXPLANATION

According to gauss’s law,

The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ϵ₀.

\(\oint \vec{E}\cdot d\vec{S}=\frac{{{Q}_{in}}}{{{\epsilon }_{0}}}\)

Electric flux on a closed surface only depends on the enclosed charge.

Concept:

• Gauss' Law: This law gives the relation between the distribution of electric charge and the resulting electric field.
• According to this law, the total charge Q enclosed in a closed surface is proportional to the total flux ϕ  enclosed by the surface.

ϕ α Q

• The Gauss law formula is expressed by 

ϕ = Q / ϵ0

Here, 

Q = charge enclosed by the surface

 ϵ₀ = permittivity of free space

Explanation:

From the above explanation, we can see that according to Gauss law

\(\phi \text{ }=\text{ }\frac{Q}{{{\varepsilon }_{0}}}\)

Whereas, electric flux through small area dA is the product of the electric field and the area its passing through (dA)

\(\therefore \phi =\text{ }\oint{E.dA}=\text{ }\frac{Q}{{{\varepsilon }_{0}}}\text{ }\)

Thus the given equation represents Gauss’s Law for electricity

The correct answer is option 3) i.e. Zero if the charge is placed outside Earth and \(\frac{q}{\epsilon_0}\) if the charge is placed inside the Earth

CONCEPT:

• Gauss's Law for electric field: It states that the total electric flux emerging out of a closed surface is directly proportional to the charge enclosed by this closed surface. It is expressed as:

\(ϕ = \frac{q}{ϵ_0}\)

Where ϕ is the electric flux, q is the charge enclosed in the closed surface and ϵ0 is the electric constant.

CALCULATION:

• Earth can be considered as a closed surface.

According to Gauss's law, the total electric flux emerging out of a closed surface is \(ϕ = \frac{q}{ϵ_0}\).

• So if the charge q is placed inside Earth, the electric flux through Earth will be \( \frac{q}{ϵ_0}\)
• If the charge is placed outside the Earth, then the flux through Earth will be zero as q = 0.

Therefore, the correct answer is zero if the charge is placed outside Earth and \(\frac{q}{\epsilon_0}\) if the charge is placed inside the Earth.

CONCEPT:

• Gauss Law: According to gauss’s law, total electric flux through a closed surface enclosing a charge is 1/ϵ0 times the magnitude of charge enclosed.

\(i.e.~{{\mathbf{\Phi }}_{net}}=\frac{\left( {{Q}_{in}} \right)}{{{\epsilon }_{0}}}\)

\(i.e.~\oint \vec{E}\cdot d\vec{S}=\frac{{{Q}_{in}}}{{{\epsilon }_{0}}}\)

Where, Φ = electric flux, Qin = charge enclosed the sphere, ϵ0 = permittivity of space (8.85 × 10-12 C2/Nm2), dS = surface area

EXPLANATION:

• If 'φ' is Electric flux through a closed surface 'S', 'q' is total charge enclosed by 'S' and є_o is the permittivity of free space, then these three are related by Gauss' Law formula, φ = q/єo

So option 2 is correct.

CONCEPT:

Gauss's law:

• According to Gauss law, the total electric flux linked with a closed surface called Gaussian surface is \(\frac{1}{ϵ_o}\) the charge enclosed by the closed surface.

\(\Rightarrow ϕ=\frac{Q}{ϵ_o}\)

Where ϕ = electric flux linked with a closed surface, Q = total charge enclosed in the surface, and ϵo = permittivity

Important points:

1. Gauss’s law is true for any closed surface, no matter what its shape or size.
2. The charges may be located anywhere inside the surface.

EXPLANATION:

• Gauss’s law is true for any closed surface, no matter what its shape or size. Hence, option 2 is correct.