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Test: Electric Flux & Gauss Law


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Test: Electric Flux & Gauss Law - Question 1

If the electric field is given by 26i – k. Calculate the electric flux through a surface of area 20 units lying in x-y plane.

Detailed Solution for Test: Electric Flux & Gauss Law - Question 1

► Electric field (E) = 26i - k
► A = 20k
Electric flux = electric field.area
= E.A
= (26i - k)*(20k)
= (26*0 -1)*20
= -20 units

Test: Electric Flux & Gauss Law - Question 2

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?

Detailed Solution for Test: Electric Flux & Gauss Law - Question 2

Net electric flux (ΦNet) through the cubic surface is given by,

Where, ∈0 = Permittivity of free space

= 8.854 × 10−12 N−1C2 m−2

q = Net charge contained inside the cube = 2.0 μC = 2 × 10−6 C

= 2.2 × 105 N m2 C−1

The net electric flux through the surface is 2.2 ×105 N m2C−1.

Test: Electric Flux & Gauss Law - Question 3

Electric flux through an area dA for an electric field E is given by​:

Detailed Solution for Test: Electric Flux & Gauss Law - Question 3

For constant electric field

Test: Electric Flux & Gauss Law - Question 4

A point charge causes an electric flux of −1.0×103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?

Detailed Solution for Test: Electric Flux & Gauss Law - Question 4

Electric flux is given by  since amount of charge not depends on size and shape so by making radius <  double the amount of charge remain same so electric flux remain same.

Test: Electric Flux & Gauss Law - Question 5

Choose the correct statement:

Detailed Solution for Test: Electric Flux & Gauss Law - Question 5

By definition of Solid Angle,

Where, Ω is represented in Steradians.

Test: Electric Flux & Gauss Law - Question 6

Electric flux is a _______ quantity.

Detailed Solution for Test: Electric Flux & Gauss Law - Question 6

Electric flux is a scalar quantityElectric Flux is dot/scalar product of magnetic field vector and area vector. And the result of a dot/scalar quantity is always a scalar quantity. So electric flux is also a scalar quantity 

Test: Electric Flux & Gauss Law - Question 7

If the electric field is given by 6i+3j+4k calculate the electric flux through a surface of area 20 units lying in y-z plane.

Detailed Solution for Test: Electric Flux & Gauss Law - Question 7

► Area = 20i since it is in the Y-Z plane
► E = 6i + 3j + 4k
► E.S = Flux
= 20*6 = 120 units

Test: Electric Flux & Gauss Law - Question 8

A uniform line charge with linear density λ lies along the y-axis. What flux crosses a spherical surface centred at the origin with r = R?

Detailed Solution for Test: Electric Flux & Gauss Law - Question 8

Line charge density = λ
Radius, r = R
Now,  λdl=dθ
For total charge enclosed by radius R, 0R Q = 0R λdl
Qinc = λR
Now, ϕ = Qinco    {from Gauss law}
= λR/ε0

Test: Electric Flux & Gauss Law - Question 9

Dimensional formula of ØE is​:

Detailed Solution for Test: Electric Flux & Gauss Law - Question 9

Flux = E. A
⇒ F = Eq
⇒ E= F/q= N/C
⇒ Flux= N/C × m²
= Nm2C-1

Test: Electric Flux & Gauss Law - Question 10

Electric flux density is a function of _______.

Detailed Solution for Test: Electric Flux & Gauss Law - Question 10

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

Test: Electric Flux & Gauss Law - Question 11

Which of the following correctly states Gauss law?

Detailed Solution for Test: Electric Flux & Gauss Law - Question 11

The electric flux passing through any closed surface is equal to the total charge enclosed by that surface. In other words, electric flux per unit volume leaving a point (vanishing small volume), is equal to the volume charge density

Test: Electric Flux & Gauss Law - Question 12

Electric flux of a surface is maximum, when

Detailed Solution for Test: Electric Flux & Gauss Law - Question 12

By definition,

For maximum, E.dA must be maximum. This is possible when E || dA.

Test: Electric Flux & Gauss Law - Question 13

The Gaussian surface for a point charge will be

Detailed Solution for Test: Electric Flux & Gauss Law - Question 13

A point charge is single dimensional. The three dimensional imaginary enclosed surface of a point charge will be sphere.

Test: Electric Flux & Gauss Law - Question 14

At what point is the electric field intensity due to a uniformly charged spherical shell is maximum?

Detailed Solution for Test: Electric Flux & Gauss Law - Question 14

By Gauss Law,

Electric Field due to spherical shell shows the following behavior:

Test: Electric Flux & Gauss Law - Question 15

A charge q is placed at the centre of the open end of cylindrical vessel whose height is equal to its radius. The electric flux of electric field of charge q through the surface of the vessel is

Detailed Solution for Test: Electric Flux & Gauss Law - Question 15

We can assume a similar vessel, such that the charge lies at the interfaces of both the vessels. Now, by Gauss law, all the electric flux must pass through these two vessels. As they are kept symmetrically with respect to the charge equal amount of flux would pass through them

From both vessels,

For each vessel,

Test: Electric Flux & Gauss Law - Question 16

The charge enclosed by a spherical Gaussian surface is 8.85 X 10-8 C. What will be the electric flux through the Surface?

Detailed Solution for Test: Electric Flux & Gauss Law - Question 16

We know that the flux linked with a Gaussian surface is given as:

Φ = E.ds = q/ε0

Here,

q = 8.85 x 10-8 C

ε0 = 8.85 x 10-12 F/m

So,

Φ = 8.85x10-8/ 8.85x10-12

Thus, flux will be

Φ = 10-4 Nm2/C

Test: Electric Flux & Gauss Law - Question 17

If the electric field is given by  calculate the electric flux through as surface of area 10 units lying in y-z plane.

Detailed Solution for Test: Electric Flux & Gauss Law - Question 17



Hence,

Test: Electric Flux & Gauss Law - Question 18

The flux associated with a spherical surface is 104 N m2C-1.  What will be the flux, if the radius of the spherical shell is doubled?

Detailed Solution for Test: Electric Flux & Gauss Law - Question 18

If the radius of the spherical shell is double then also flux associated with spherical surface remain unchanged because the charge enclosed in system always remains unchanged.

Test: Electric Flux & Gauss Law - Question 19

The Gaussian surface for a line charge will be

Detailed Solution for Test: Electric Flux & Gauss Law - Question 19

A line charge can be visualized as a rod of electric charges. The three dimensional imaginary enclosed surface of a rod can be a cylinder.

Test: Electric Flux & Gauss Law - Question 20

The electric field intensity due to a sphere (solid or hollow) at an external point varies as:

Detailed Solution for Test: Electric Flux & Gauss Law - Question 20

We calculate this using spherical symmtery in Gauss Law

For spherical Shell,

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