Electric Flux Important Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. Electric flux density is directly related to

1. Potential difference
2. Charge
3. Current
4. Volume

Ans. The correct answer is b. Charge

Explanation: Electric flux density, or the number of electric lines of force traveling across a specific region, is a measurement of the intensity of an electric field produced by a free electric charge. Electric flux density is the quantity of flux perpendicular to the direction of the flux that is moving through a particular area.

Ques. The relationship between the charge inside a surface and electric flux can be explained by which among the following laws?

1. Faraday’s Law
2. Pascal’s Law
3. Newton’s Law
4. Gauss’s Law

Ans. The correct answer is d. Gauss’s Law

Explanation: Gauss's law specifies the relationship between the charges present on a particular surface and the electric flux. According to this law, the amount of electric flux that emerges from a closed surface will be 1/∈₀ times the charge present.

Ques. What will happen to the flux due to a charge inside the surface if that charge is placed outside a closed surface?

1. Zero
2. Positive
3. Negative
4. It will depend on the nature of the charge; hence it can be positive or negative.

Ans. The correct answer is a. Zero

Explanation: According to Gauss's principle, if there is no charge confined inside the surface, the net electric flux leaving the surface will always be 0. Because the charge is held outside the surface, it won't produce any field lines and won't induce any flux inside the surface.

Ques. If the flux lines are released out of the surface, then the flux linked to a surface is said to be positive.

1. True
2. False

Ans. The correct answer is a. True

Explanation: Based on the orientation of the electric flux lines, we may distinguish between positive and negative flux. If the flux lines go into a surface, the flux is negative. However, if the flux lines are allowed to emerge through the surface, the flux will be positive.

Ques. What among the following is the dimensional formula of electric flux?

1. [M L² T^-1 I^-2]
2. [M L³ T^-3 I^-1]
3. [M L⁴ T^-4 I¹]
4. [M L² T^-2 I⁰]

Ans. The correct answer is b. [M L³ T^-3 I^-1]

Explanation: Using the formula of the electric flux Φ = EA cosθ, we will determine the dimensional formula of electric flux as [M L³ T^-3 I^-1]

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Short Answers Questions [2 Marks Questions]

Ques. What is meant by electric flux? Write its SI unit.

Ans. Electric flux is a measure of the distribution of the electric field or the rate at which the electric field flows across a certain area.

It is denoted by the symbol Φ.

The SI unit of electric flux is Volt-meter (V m) or N m² C^-1

Ques. What are the factors that affect electric flux?

Ans. The electric flux through any surface is the total number of electric field lines passing through that surface. It is given by

Φ = EA cosθ

Hence electric flux will depend on

• The magnitude of electric field E
• The area of the surface A
• The angle between the electric field and the area vector (θ)

Ques. What is the Gauss Law?

Ans. According to Gauss Law, the net electric flux through any closed surface is equal to the 1/????0 times the total electric charge enclosed by the surface i.e. 

Φ = Q/∈₀ 

Where

• Φ is the net flux passing through the closed surface
• Q is the net charge enclosed by the closed surface
• ∈₀ is the absolute permittivity of free space.

Ques. Define 1 coulomb.

Ans. One coulomb is the amount of charge that flows through a conductor's cross-section in one second, given a current of one ampere.

Check out: 

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Long Answers Questions [3 Marks Questions]

Ques. Determine the electric flux of a flat square having an area of 10 m² if a uniform electric field of 8000 N/C passing perpendicular to it.

Ans. Given

• Area of the square, A = 10 m²
• Electric field, E = 8000 N/C

Since the electric field lines are passing perpendicular to the plane of the flat square, then the angle between the area vector and electric field is zero i.e. θ = 0°

The formula of electric flux is given by

Φ = EA cosθ

On substituting the values, we get

Φ = 8000 x 10 x cos0°

⇒ Φ = 8 x 10⁴ N m²/C

Ques. A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then what will be the outward electric flux?

Ans. According to Gauss theorem, the net electric flux through any closed surface is equal to the 1/∈₀ times the total electric charge enclosed by the surface i.e. 

Φ = Q/∈₀ 

From the above equation, it is clear that the electric flux will not depend on the radius of the closed surface.

Hence, if the radius of the closed area is doubled, then the outward electric flux will remain the same.

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

Ans. According to Gauss theorem, the net electric flux through the Gaussian surface is equal to the 1/????0 times the total electric charge Q enclosed by the surface i.e. 

Φ = Q/∈₀ 

Given

• The net charge enclosed by the surface, Q = 2.0 μC = 2 x 10^-6 C
• The value of absolute permittivity, ∈₀ = 8.854 x 10^-12 N^-2 m^-2 C²

On substituting the values, we get

Φ = (2 x 10^-6)/(8.854 x 10^-12)

⇒ Φ = 2.26 x 10⁵ N m²/C

Hence, the net electric flux through the surface is 2.26 x 10⁵ N m²/C.

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Very Long Answers Questions [5 Marks Questions]

Ques. Consider a uniform electric field E = 3 x 10³ \(\hat{i}\) N/C. 

1. What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz-plane?
2. What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

Ans. Given

• The magnitude of the electric field, E = 3 x 10³ N/C
• Length of the side of the square, a = 10 cm = 0.1 m

1. The plane of the square is parallel to the yz-plane and the direction of the electric field is along the x-axis.

Therefore, the angle between the area vector of the square and the electric field is zero i.e. θ = 0°

The formula of electric flux is given by

Φ = EA cosθ

Where A is the area of the square.

A = (Side)² = a² = 0.12 = 0.01 m²

On substituting the values, we get

Φ = 3 x 10³ x 0.01 x cos0°

⇒ Φ = 30 N m²/C

2. If the normal to the plane of the square makes a 60° angle with the x-axis, then the angle between the area vector of the square and the electric field is 60° i.e. θ = 60°

The formula of electric flux is given by

Φ = EA cosθ

On substituting the values, we get

Φ = 3 x 10³ x 0.01 x cos60°

⇒ Φ = 30 x 1/2 

⇒ Φ = 15 N m²/C

Ques. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10³ N m²/C.

1. What is the net charge inside the box?
2. If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Ans. Given that the net outward flux through the surface of the box, Φ = 8.0 × 10³ N m²/C

1. According to Gauss theorem, the net electric flux through any closed surface is equal to the 1/∈₀ times the total electric charge enclosed by the surface i.e. 

Φ = Q/∈₀ 

Therefore the net charge enclosed by the box, Q = Φ x ∈₀ 

⇒ Q = 8.0 × 10³ x 8.854 x 10^-12

⇒ Q = 70.832 x 10^-9 C

⇒ Q = 0.07 x 10^-6 C = 0.07 µC

2. The net flux passing out through a body is determined by the body's net charge. If the net flux is zero, it concludes that the net charge inside the body is also zero. The body may have an equal amount of positive and negative charges.

Ques. A point charge causes an electric flux of – 1.0 × 10³ N m²/C to pass through a spherical Gaussian surface of a 10.0 cm radius centered on the charge.

1. If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
2. What is the value of the point charge?

Ans. Given

• The value of the electric flux passing through the surface, Φ = – 1.0 × 10³ N m²/C
• Radius of the Gaussian surface, r = 10 cm = 10 x 10^-2 m

1. According to Gauss theorem, the net electric flux through any closed surface is equal to the 1/∈₀ times the total electric charge enclosed by the surface i.e. 

Φ = Q/∈₀ 

From the above equation, it is clear that the electric flux will not depend on the radius of the closed surface.

Hence, if the radius of the closed area is doubled, then the outward electric flux will remain the same i.e. Φ = – 1.0 × 10³ N m²/C

2. From Gauss theorem, we have

Φ = Q/∈₀ 

Therefore the net charge enclosed by the box, Q = Φ x ∈₀ 

⇒ Q = – 1.0 × 10³ x 8.854 x 10^-12

⇒ Q = - 8.854 x 10^-9 C

⇒ Q = - 8.854 nC

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