Electric Flux MCQ Quiz - Objective Question with Answer for Electric Flux - Download Free PDF
Last updated on Apr 18, 2025
Latest Electric Flux MCQ Objective Questions
Electric Flux Question 1:
A cubical region of side a has its centre at the origin. It encloses three fixed point charges, -q at (0, -a/4, 0), +3q at (0, 0, 0) and -q at (0, +a/4 ,0).
Match the following :
List I | List II |
P) The net electric flux crossing the plane x = +a/2 is equal to | i) the net electric flux crossing the plane y = -a/2 |
Q) The net electric crossing the plane y = +a/2 is more than the net electric flux crossing the plane y = -a/2 | ii) the net electric flux crossing the plane x = -a/2 |
R) The net electric flux crossing the entire region is | iii) the net electric flux crossing the plane x = +a/2 |
S) The net electric flux crossing the plane z = +a/ 2 is equal to | iv) \( \frac { q }{ { \varepsilon }_{ o } } \) |
Answer (Detailed Solution Below)
Electric Flux Question 1 Detailed Solution
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.
Electric Flux Question 2:
For the given diagram E = 4 × 10³ N/C, the net flux through the cube of side x = 1 cm, placed x = 1 cm from the origin as shown is:
(E is acting towards positive x-axis and Φ = 1)
Answer (Detailed Solution Below)
Electric Flux Question 2 Detailed Solution
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.
Electric Flux Question 3:
A square loop of sides a = 1 m is held normally in front of a point charge q = 1C. The flux of the electric field through the shaded region is \(\frac{5}{\mathrm{p}} \times \frac{1}{\varepsilon_{0}} \frac{\mathrm{Nm}^{2}}{\mathrm{C}}\), where the value of p is ____.
Answer (Detailed Solution Below) 48
Electric Flux Question 3 Detailed Solution
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 \(\frac{a}{2}\).
Electric Flux Question 4:
The electric flux through the shaded area of square plate of side a due to point charge placed at distance of 𝑎/2 from it as shown in fig is \(\frac{N Q}{48 G_{0}}\), then 𝑁 is
Answer (Detailed Solution Below) 5
Electric Flux Question 4 Detailed Solution
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.
Electric Flux Question 5:
If a charge q is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :
Answer (Detailed Solution Below)
Electric Flux Question 5 Detailed Solution
Concept:
The electric flux (Φ) through a surface is related to the charge enclosed by the surface via Gauss's Law:
Φ = q / ε0,
where:
- Φ is the electric flux,
- q is the charge enclosed by the surface,
- ε0 is the permittivity of free space.
For a complete spherical surface enclosing a charge q, the total flux through the surface is q / ε0.
When the charge is at the center of a hemispherical surface, only half of the total flux will pass through the curved surface. The remaining flux passes through the flat surface, but since it is a non-conducting hemisphere, the flux through the flat surface will be zero.
Calculation:
The total flux through a complete spherical surface is:
Φ = q / ε0
The flux through the curved surface will be half of this, which is:
Φcurved = q / 2ε0
The flux through the flat surface is zero, so:
Φflat = 0
Remark: The flux through the flat surface is zero, but no option is given for the electric flux passing through the curved surface.
Top Electric Flux MCQ Objective Questions
What is the flux through a cube of side a, if a point charge of q is at one of its corners?
Answer (Detailed Solution Below)
Electric Flux Question 6 Detailed Solution
Download Solution PDFConcept:
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
\(\Rightarrow ϕ = \frac{q}{{8{\varepsilon _0}}}\)
The electric flux depends on the:
Answer (Detailed Solution Below)
Electric Flux Question 7 Detailed Solution
Download Solution PDFCONCEPT:
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.
Number of electric lines of force passing through unit area is called
Answer (Detailed Solution Below)
Electric Flux Question 8 Detailed Solution
Download Solution PDFCONCEPT:
- 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.
A square sheet of 5.0 cm is placed in an electric field E = (1.6 × 104 \(\frac{N}{C}\)) i such that the normal unit vector for the sheet is [(\(\frac{\sqrt3}{2}\))i + \((\frac{1}{2})\)j]. The electric flux through the sheet is:
Answer (Detailed Solution Below)
Electric Flux Question 9 Detailed Solution
Download Solution PDF=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, ϕE = EA cosθ
ϕE = 1.6 × 104 × 2.5 × 10-3 × cos 30º
ϕE = 34.6 Nm2/C
If there existed only one type of charge q on the Earth, then what is the electric flux related to Earth?
Answer (Detailed Solution Below)
Electric Flux Question 10 Detailed Solution
Download Solution PDFThe 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.
Suppose a uniform electric field is given as E = 4 × 104 ĵ N/C (ĵ is unit vector along the y axis), then the flux of this field through a square of 20 cm on a side whose plane is parallel to the xz plane is:
Answer (Detailed Solution Below)
Electric Flux Question 11 Detailed Solution
Download Solution PDFConcept:
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.
If a charge q is placed at one corner of cube, the flux through the cube is
Answer (Detailed Solution Below)
Electric Flux Question 12 Detailed Solution
Download Solution PDFConcept:
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
\(\dfrac{q}{8\epsilon_0}\)
The magnitude of the flux through an area element ΔS of a sphere of radius r, which encloses a point charge q at its centre is _______________.
Answer (Detailed Solution Below)
Electric Flux Question 13 Detailed Solution
Download Solution PDFCONCEPT:
- 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.
If a closed square loop of area \(5\hat{i}-6\hat{j}\) is placed in the electric field of \(-2\hat{i}+4\hat{j}\), then the electric flux will be:
Answer (Detailed Solution Below)
Electric Flux Question 14 Detailed Solution
Download Solution PDFCONCEPT:
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.
A ring is placed in the electric field such that its plane is parallel to the electric field, then the electric flux associated with the ring will be:
Answer (Detailed Solution Below)
Electric Flux Question 15 Detailed Solution
Download Solution PDFCONCEPT:
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.