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

Explanation:

Net flux through the cube:
The net flux is calculated as:
Net flux through cube = ( -q + 3q - q ) / ε₀ = q / ε₀

Flux passing through the same area:
Since the flux through the same area is the same, the values for the positions x = (-a)/2, y = a/2, and z = a/2 are the same.

Concept:

Electric Flux Through a Cube:

The total electric flux through a closed surface is given by Gauss's law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the medium.

\( \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \)

Where:

\( \Phi_E \) is the total electric flux through the surface.

\( Q_{\text{enc}} \) is the total charge enclosed within the surface.

\( \epsilon_0 \) is the permittivity of free space.

Dipoles consist of pairs of equal and opposite charges. When multiple dipoles are placed inside a closed surface like a cube, the net charge enclosed is the sum of all individual charges.

Calculation:

Given,

There are eight dipoles placed inside a cube, each dipole consisting of charges +e" id="MathJax-Element-273-Frame" role="presentation" style="position: relative;" tabindex="0">+e+e $+e$ and −e" id="MathJax-Element-274-Frame" role="presentation" style="position: relative;" tabindex="0">−e−e $-e$ .

The total charge enclosed by the cube is the sum of all individual charges:

\( Q_{\text{enc}} = 8 \times (+e + (-e)) \)

⇒ \( Q_{\text{enc}} = 8 \times 0 \)

⇒ \( Q_{\text{enc}} = 0 \)

Using Gauss's law:

\( \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \)

⇒ \( \Phi_E = \frac{0}{\epsilon_0} \)

⇒ \( \Phi_E = 0 \)

∴ The total electric flux coming out of the cube is zero.

The correct answer is - 2 × 10⁻⁶ Wb

Key Points

• Electric Field (E)
  • The electric field (E) is given as 4 × 10³ N/C.
• Cube's Side Length (x)
  • The side length of the cube (x) is 1 cm, which is 0.01 m when converted to meters.
• Flux Calculation
  • The flux (Φ) through a surface is given by the formula Φ = E × A, where A is the area of the surface.
  • For a cube, the total flux is calculated through the six faces. However, since the electric field is uniform and only one side of the cube is perpendicular to the field, we consider just that side.
  • The area of one face of the cube is A = x² = (0.01 m)² = 1 × 10⁻⁴ m².
  • Thus, the flux through one face is Φ = E × A = 4 × 10³ N/C × 1 × 10⁻⁴ m² = 4 × 10⁻¹ Wb.
  • Since the cube has six faces, the net flux for a uniform field is Φ = 2 × 10⁻⁶ Wb (considering the field lines only pass through two opposite faces).

Additional Information

• Gauss's Law
  • Gauss's Law states that the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the medium (&epsilon₀).
  • Mathematically, it is expressed as Φ = Q/ε₀, where Q is the total charge enclosed.
• Electric Flux
  • Electric flux represents the number of electric field lines passing through a given area.
  • It is a scalar quantity and can be positive or negative depending on the direction of the electric field relative to the surface.

Concept:

Electric Flux:

Electric flux is a measure of the electric field passing through a given area.

The formula for electric flux is Φ = q / ε₀, where q is the charge and ε₀ is the permittivity of free space.

In this case, the square loop is in front of a point charge, and we are asked to calculate the flux through the shaded region.

The total flux through the square is divided into 8 equal parts, and the flux is the same for each part.

Calculation:

Given,

Charge, q = 1 C

Side of the square loop, a = 1 m

The total flux through the square is: Φ(total) = q / ε₀

We divide the square into 8 equal parts, and the flux is the same for each part. Hence, the flux through the shaded portion is:

Total flux through square = \(\frac{q}{\epsilon_{0}}\left(\frac{1}{6}\right)\)

Lets divide square is 8 equal parts.

Flux is same for each part. 

∴ Flux through shaded portion is \(\frac{5}{8}\) (Total flux) 

= \(\frac{5}{8} \times \frac{\mathrm{q}}{\epsilon_{0}} \frac{1}{6}=\frac{5}{48} \frac{1}{\epsilon_{0}}\)

∴ required Ans. is 48 

Note : Distance of charge from square loop is not mentioned we have assume it as \(\frac{a}{2}\).

Given:

Square plate of side a , a point charge Q is placed at a distance\( \frac{a}{2}\) from the square's center.

Electric flux through the shaded area is given by:

\(\Phi = \frac{NQ}{48\epsilon_0} \).

To find the value of N .

Concept:

The total electric flux due to a point charge Q is distributed over a spherical surface of radius r given by:

• \(\Phi_{total} = \frac{Q}{\epsilon_0} \).

For a square plate, only a fraction of the total flux passes through the surface depending on the symmetry and geometry.

The shaded area corresponds to \(\frac{1}{8}\) of the total flux passing through one face of a cube formed around the charge.

Calculation:

The total flux through the cube surface:

\(⇒ \Phi_{cube} = \frac{Q}{\epsilon_0} .\)

Flux through one face of the cube:

\(⇒ \Phi_{face} = \frac{Q}{6\epsilon_0} .\)

Flux through the shaded area (5/8th of one face):

\(⇒ \Phi_{shaded} = \frac{5}{8} \times \frac{Q}{6\epsilon_0} . \\ ⇒ \Phi_{shaded} = \frac{5Q}{48\epsilon_0} .\)

Comparing with \(\Phi = \frac{NQ}{48\epsilon_0}\) :

⇒ N = 5 .

∴ The value of N is 5.

Concept:

Electric flux (ϕ): The number of electric field lines passing through a surface area normally is called electric flux. It is denoted by Φ.

The electric flux through a chosen surface is given by:

\({\rm{\Delta }}ϕ = \vec E.{\rm{\Delta }}\vec S = E{\rm{\Delta }}Scos\theta \)

Where θ is the angle between the electrical field and the positive normal to the surface.

Gauss’s Law: It states that the net electric field through a closed surface equals the net charge enclosed by the surface divided by ϵ0.

\(\oint \vec E.\overrightarrow {ds} = \frac{{{q_{inside}}}}{{{ϵ_0}}}\)

Where E = electric field, ds = small area, qinside = the total charge inside the surface, and ϵ0 = the permittivity of free space.

Calculation:

According to the question, the charge is placed on one of the corners of a cube, and to cover up a charged particle fully we need 8 cubes.

\(\therefore ϕ =\frac{1}{8} \frac{q}{{{\varepsilon _0}}}\)

Therefore the electric flux passing through the given cube is  

 \(\Rightarrow ϕ = \frac{q}{{8{\varepsilon _0}}}\) 

CONCEPT:

Electric flux:

• It is defined as the number of electric field lines associated with an area element.
• Electric flux is a scalar quantity.
• The SI unit of the electric flux is N-m2/C.
• If the electric field is E and the area is A then the electric flux associated with the area is given as,
  \(⇒ ϕ = \vec{E}.\vec{A}\)
  ⇒ ϕ = EA cos θ
  Where θ = angle between the surface and the electric field

EXPLANATION:

• If the electric field is E and the area is A then the electric flux associated with the area is given as,
  ⇒ ϕ = EAcosθ     ---(1)
  Where θ = angle between the surface and the electric field
• By equation 1 it is clear that the electric flux depends on the electric field intensity, area, and the angle between the surface and the electric field. Hence, option 4 is correct.

CONCEPT:

• Electric flux: The number of electric field lines passing through a surface area normally is called electric flux. It is denoted by Φ.

• The electric flux through a chosen surface is given by:

\({\rm{\Delta }}\phi = \vec E.{\rm{\Delta }}\vec S = E{\rm{\Delta }}Scos\theta \)

Where θ is the angle between the electrical field and the positive normal to the surface.

EXPLANATION:

• The number of electric lines of force passing through unit area is called electric flux. So option 1 is correct.
• The electric charge per unit area is called electric surface charge density of that surface.
• The space or region around the electric charge in which electrostatic force can be experienced by other charged particles is called an electric field by that electric charge.

EXTRA POINTS:

• Gauss’s Law: It states that the net electric field through a closed surface equals the net charge enclosed by the surface divided by ϵ0.

\(\oint \vec E.\overrightarrow {ds} = \frac{{{q_{inside}}}}{{{ϵ_0}}}\)

Where E = electric field, ds = small area, qinside = the total charge inside the surface, and ϵ0 = the permittivity of free space.

=Concept:

• Electric flux: It is defined as the total number of electric field lines passing a given area in a unit of time.

• Formula, electric flux,  ϕE = EA cosθ, where, E = electric field, A = surface area, θ = angle between the electric field and the area vector

Calculation: 

Given,

The electric field, E = (1.6 × 104 \(\frac{N}{C}\)) i

Side of the square sheet, a = 5.0 cm = 0.05 m

Surface area, s = a² = (0.05)² = 2.5 × 10^-3 m²

The normal unit vector, r = [(\(\frac{\sqrt3}{2}\))i + \((\frac{1}{2})\)j].

Angle, \(θ=\tan^{-1} (\frac{1/2}{\sqrt{3}/2}) \)

θ = 30º

The formula for the electric flux, ϕ_E = EA cosθ 

ϕE = 1.6 × 10⁴ × 2.5 × 10-3 × cos 30º

ϕE = 34.6 Nm²/C

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:

Electric field:

• It is defined as the electric force per unit of positive charge.
• Formula, Electric field, \(E=\frac{F}{q} \)  where, F = electric force, q = positive charge

• The SI unit of electric field N/C.

Electric flux:

• ​The electric flux is defined as the total number of electric field lines passing a given area in a unit of time.

• Formula, electric flux, ϕ = EA cosθ where, E = electric field lines, A = given surface area, θ = angle between the surface vector and electric field lines. 
• The SI unit of electric flux is N - m2/C.

Calculation:

Given,

The uniform electric field along the y-axis,  E = 4 × 104 ĵ N/C

The side of the square, a = 20 cm

The surface of the square is in the X-Z plane, so the area vector is along the y-axis.

Area, A = a² = (20 ×10^-2)² = 0.04 m²

Using the formula of electric flux, ϕ = EA cosθ 

here, the angle between E and A is zero, so, cos 0º = 1

ϕ =  4 × 104 × 0.04 = 1600 N - m²/C

Hence, the electric flux is 1600 N - m2/C.

Concept:

Electric Flux

• The electric flux denotes the number of electric field lines passing through the given area. 
• It is denoted as
   

\(\phi = \oint E.dA\)

• It is basically the product of the electric field and the surface area perpendicular to the field. 
   

Gauss Law

• The electric flux across a closed surface is given as the charge enclosed divided by permeability (ϵ). 
• The law is given as 

\(\phi = \frac{q_{in}}{\epsilon_0}\)

Calculation:

Given, the charge is at one of the corners.

Now, in order to find the flux through the cube, we need to assume a gaussian surface considering different cubes around it. 

We can see, to cover the whole enclosed charge, eight cubes are required. 

The flux through 8 cubes which enclose this charge q is 

\(\phi = \frac{q_{in}}{\epsilon_0}\)

So, flux through 1 cube will be

\(\phi =\frac{1}{8} \frac{q_{in}}{\epsilon_0}\)

\(\implies \phi = \frac{q_{in}}{8\epsilon_0}\)

So, the correct option is 

\(\dfrac{q}{8\epsilon_0}\)

CONCEPT:

• Flux is a total electric field that passes through the given area.

If we choose a simple surface with area A, the magnetic flux is:

\(ϕ=\oint \overrightarrow{E}.\overrightarrow{dA}\)

where ϕ is the flux, E is the electric field, and dA is the small area.

• 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

CALCULATION:

The magnitude of flux ϕ = Q / ϵ0

the magnitude of the flux through an area element ΔS of a sphere of radius r 

\(ϕ = \frac{Q}{\epsilon_0}\frac{\Delta S}{Sphere\space Area}\)

\(ϕ = \frac{q}{\epsilon_0} \frac{\Delta S}{4 \pi r^2}\)

ϕ = qΔS/(4πεor2)

So the correct answer is option 2.

CONCEPT:

Electric flux:

• It is defined as the number of electric field lines associated with an area element.
• Electric flux is a scalar quantity.
• The SI unit of the electric flux is N-m2/C.
• If the electric field is E and the area is A then the electric flux associated with the area is given as,

\(⇒ ϕ = \vec{E}.\vec{A}\)

⇒ ϕ = EAcosθ

Where θ = angle between the surface and the electric field

CALCULATION:

Given \(\vec{A}=5\hat{i}-6\hat{j}\) and \(\vec{E}=-2\hat{i}+4\hat{j}\)

• If the electric field is E and the area is A then the electric flux associated with the area is given as,

\(⇒ ϕ = \vec{E}.\vec{A}\)     -----(1)

Where θ = angle between the surface and the electric field

By equation 1, the electric flux is given as,

\(⇒ ϕ = \vec{E}.\vec{A}\)

\(⇒ ϕ = (5\hat{i}-6\hat{j}).(-2\hat{i}+4\hat{j})\)

= -10 - 24 

⇒ ϕ = -34 N-m2/C

• Hence, option 3 is correct.

CONCEPT:

Electric flux:

• It is defined as the number of electric field lines associated with an area element.
• Electric flux is a scalar quantity.
• The SI unit of the electric flux is N-m2/C.
• If the electric field is E and the area is A then the electric flux associated with the area is given as,

\(⇒ ϕ = \vec{E}.\vec{A}\)

⇒ ϕ = EAcosθ

Where θ = angle between the surface and the electric field

EXPLANATION:

If the electric field is E and the area is A then the electric flux associated with the area is given as,

⇒ ϕ = EAcosθ     -----(1)

Where θ = angle between the surface and the electric field

• Here the plane of the ring is parallel to the electric field so the angle between the surface and the electric field will be 90°.

So by equation 1, the electric flux associated with the ring will be,

⇒ ϕ = EA×cos90

⇒ ϕ = 0

• Hence, option 1 is correct.