Electric Flux Questions

The flux is negative if the flux lines are penetrating a surface. But the flux will be positive if the flux lines are released out of the surface. The electric flux is defined as the total number of electric field lines crossing a given area in a unit of time. Thus, if the plane is normal to the flow of the electric field, the total flux is given as:

Ф_p = EA.

The projected area is given as Acosθ when the same plane is tilted at an angle θ. The total flux through this surface is given as

Ф = EA cosθ

Where,

• The magnitude of the electric field = E
• The area of the surface = A
• The angle made by the axis parallel to the direction of flow of the electric field and the plane = θ

Important Questions on Electric Flux

1) What among the following is the dimensional formula of electric flux?

a) [M L² T^-2 I⁰]

b) [M L² T^-1 I^-2]

c) [M L⁴ T^-4 I¹]

d) [M L³ T^-3 I^-1]

Correct Option: (d)

Explanation: The formula for electric flux is electric field intensity* area, and [M L T^-3 I^-1] is the dimension of field intensity and [L²] is the dimension of the area; therefore, [M L³ T^-3 I^-1] is the dimension of flux.

2) The unit of electric flux is _____.

a) V*m

b) N²/m

c) V/m

d) N/Coulomb/m

Correct Option: (a)

Explanation: The formula for electric flux is electric field intensity* area, but the unit of electric potential can be obtained by multiplying the field intensity by a length, that is, E=-dV/dx. Hence, flux is also equal to the electric potential and length product, which means that its unit will be V*m.

3) The correct electric flux expression among the following will be ______.

a) ∫ E→. ds→

b) ∫ E2→.d2s→

c) ∫ 2E→ −1.d2s→

d) ∫ 2E→. 2dl→

Correct Option: (a)

Explanation: The number of field lines perpendicularly crossing a surface area is known as electric flux. Therefore, field lines crossing through small area ds are E.ds, because the electric field intensity (E) is defined as the number of electric field lines crossing through the unit cross-section area. Hence, the flux in the entire surface can be determined by taking the integral value.

4) If the angle between the field lines and area vector is_____, the electric flux will be maximum.

a) 45^o

b) 130^o

c) 60^o

d) 0o

Correct Option: (d)

Explanation: The formula for electric flux becomes the formula E.s.cosθ because both the area and electric field are vector quantities, and we can also express electric flux as E→. s→. Here, θ will be the angle between two vectors. And if theta is zero, the value of electric flux will become maximum.

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

a) True

b) False

Correct Option: (a)

Explanation: We can differentiate between positive and negative flux based on the direction of electric flux lines. The flux is negative if the flux lines are going inside a surface. But the flux will be positive if the flux lines are released out of the surface.

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

a) Positive

b) It will depend on the nature of the charge; hence it can be positive or negative.

c) Zero

d) Negative

Correct Option: (c)

Explanation: Gauss’s principle states that the net electric flux coming out of the surface will always be 0 if there is no charge bound inside that surface. The charge will generate no field lines because it is kept outside the surface; hence there will be no flux inside the surface.

7) In which case will the electric flux be greater if two different spheres with two separate charges q and 20q are placed inside them?

a) No flux in any of the spheres

b) 2^nd sphere

c) 1^st sphere

d) Flux will be the same in both cases

Correct Option: (b)

Explanation: The charge bound inside a sphere is directly proportional to the electric field line. And also, electric flux can be measured using the field lines, i.e. electric flux is known as the number of field lines crossing through a particular surface. Hence, the flux will be 20 times more in the case of a 20q charge and less in the case of a q charge.

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

a) Newton’s Law

b) Faraday’s Law

c) Gauss’s Law

d) Pascal’s Law

Correct Option: (c)

Explanation: The relationship between charges inside a particular surface and electric flux is given by Gauss’s law. This law states that electric flux releasing out from a closed surface will be equal to 1/ε times the charge inside the surface.

9) Which among the following laws states that “The charge enclosed by a surface divided by its permittivity gives the total electric flux through that closed surface”?

a) Gauss’s law

b) Newton’s law

c) Coulomb’s law

d) Snell’s law

Correct Option: (a)

Explanation: Gauss’s law deals with electric flux among the four given laws. And Gauss’s law states that the charge enclosed by a surface divided by its permittivity gives the total electric flux through that closed surface.

10) Electric flux density is directly related to _____.

a) Charge

b) Volume

c) Potential difference

d) Current

Correct Option: (a)

Explanation: Electric flux density is the charge per unit area. Hence it is a charge function and not any of the other values.

Practice Questions

1) Define electric flux.

2) What are the factors that affect electric flux?

3) Define 1 coulomb.

4) What is Faraday’s law of electromagnetic induction?

5) What is Gauss’s law?

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