# Physics Questions and Answers – Electric Flux

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Electric Flux”.

1. What is the dimension of electric flux?
a) [M L3 T-3 I-1]
b) [M L2 T-3 I-1]
c) [M L3 T-3 I1]
d) [M L3 T3 I-1]

Explanation: Electric flux=electric field intensity* area. The dimension of field intensity is [M L T-3 I-1] and the dimension of the area is [L2]. Therefore, the dimension of flux = [M L3 T-3 I-1]. This can also be justified that flux=potential*length. By putting the dimensions of potential and length, we can get the same result.

2. What is the unit of electric flux?
a) V/m
b) N/m
c) V*m
d) N/Coulomb

Explanation: Flux is equal to the product of field intensity and area of the surface. But field intensity multiplied by a length gives the unit of electric potential as E=-$$\frac {dV}{dx}$$. Therefore, flux also means electric potential multiplied by length. This gives us the unit V*m. There is another unit of flux N*m2*C-1.

3. Which one is the correct expression of electric flux?
a) ∫ $$\vec{E}.d\vec{s}$$
b) ∫ $$\vec{E^2}.d\vec{s}$$
c) ∫ $$\vec{E}^{-1}.d\vec{s}$$
d) ∫ $$\vec{E}.d\vec{l}$$

Explanation: Electric flux is defined as the number of field lines crossing perpendicularly through a surface area. The number of electric field lines crossing through the unit cross-section area is known as electric field intensity (E). Therefore field lines crossing through small area ds are E.ds. Taking the integral gives the flux in the entire surface.
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4. Electric flux will be maximum if the angle between the field lines and area vector is ______
a) 45 degree
b) 135 degree
c) 90 degree
d) 0 degree

Explanation: We know that the electric field and area both are vector quantity and the electric flux is expressed as $$\vec{E}.\vec{s}$$ (taking dot product). But if the angle between the two vectors is θ then the formula becomes E.s.cosθ.Cosθ will be maximum if theta is zero, in all other cases, the value of Cosθ is less than 1. Therefore flux will be the maximum if the angle is 0 degrees.

5. Flux linked to a surface is said to be positive if the flux lines are coming out of the surface. The statement is ______
a) True
b) False

Explanation: Depending on the direction of electric flux lines i.e. electric field intensity, we can differentiate between positive and negative flux. If the flux lines are going inside a surface, the flux is said to be negative. But if the flux lines are coming out of the surface, the flux is said to be positive.

6. If a charge is placed outside a closed surface, flux due to that charge inside the surface will be ________
a) Positive
b) May be positive or negative, depending on the nature of the charge
c) Negative
d) Zero

Explanation: According to Gauss’s principle, if there is no charge bound inside a surface, net electric flux coming out of the surface will always be 0. In this case, the charge is kept outside the surface, so it will generate no field lines and hence no flux inside the surface. The situation will be the same if an electric dipole is placed inside a surface as dipole has equal positive and negative charges; the total charge inside the surface becomes 0.

7. Two separate charges q and 10q are placed inside two different spheres. In which case, the electric flux will be greater?
a) Flux will be same in both the cases
b) 1st sphere
c) 2nd sphere
d) No flux in any of the spheres

Explanation: The electric field line is directly proportional to the charge bound inside a sphere. We know that, field lines are the measure of electric flux i.e. number of field lines crossing through a surface are know an electric flux. Therefore flux will be lesser in the case of q charge and will be 10 times in case of a 10q charge.

8. Which of the following law explains the relation between the charge inside a surface and electric flux?
a) Gauss’s Law
b) Coulomb’s Law
d) Pascal’s Law

Explanation: Gauss’s Law gives the relation between electric flux and charges inside a surface. It states that electric flux coming out from a closed surface is equal to $$\frac {1}{\varepsilon}$$ times the charge inside the surface. Coulomb’s Law explains the force between two charges and Pascal’s Law is related to fluids.

9. Flux coming out from a balloon of radius 10 cm is 1.0*103 N.m2.C-1-1. If the radius of the balloon is doubled, the flux coming out from the balloon will be _______
a) 0.5 times
b) 2 times
c) Same
d) 4 times

Explanation: If the radius of the balloon is 2 times the initial value, the surface area of the balloon will be 4 times the initial value because of surface area=4π(radius)2. But in this case, flux coming out from the balloon is dependent only upon the charge inside the balloon, not on the area of the surface. As charge enclosed in the balloon is the same in both the cases, the flux will also be the same.