Electric intensity at any point between two identical conducting plates connected to a DC supply source is

uniform directed from positive to negative plates

directly proportional to distance between the plates

inversely proportional to the square of the distance between the plates

inversely proportional to the distance between the plates

A Gaussian sphere closes an electric dipole within it. Then the total flux through the sphere is

half that due to a single charge

dependent of the position of the dipole

The total electric flux through a closed surface depends

on the value of the net charge only

on the location of the charge only

on the shape of the closed surface only

both on the location of the charge and the shape of the surface

A Gaussian encloses a positive charge Q placed at its center. Then electric intensity

at every point on the surface is same but it acts radially outwards

at very point on the surface is different

is same and acts radially inwards

As per Gauss law ∫ x y E d S = q i n ϵ 0 \int_x^yEd\ S\ =\ \frac{q_{in}}{\epsilon_0} ∫ x y E d S = ϵ 0 q i n Which of the following is true about this

Electric flux on the closed surface due to outside charge is always zero

This is valid for symmetrical surface only

E is the electric field to the charge inside the surface

A point charge is placed at the center of a spherical Gaussian surface. The electric flux is changed if:

a second point charge is placed just outside the sphere

the sphere is replaced by a cube of the same volume

the sphere is replaced by a cube of one-tenth the volume

the point charge is moved off center (but still inside the original sphere)

Choose the INCORRECT statement:

According to Gauss' law, if a closed surface encloses no charge, then the electric field must vanish everywhere on the surface

Gauss' law can be derived from Coulomb's law

Gauss' law states that the net number of lines crossing any closed surface in an outward direction is proportional to the net charge enclosed within the surface

Coulomb's law can be derived from Gauss' law and symmetry

A round wastepaper basket with a 0.15 m radius opening is in a Uniform electric field of 300 N/C, perpendicular to the opening. The total flux through the sides and bottom, in N m 2 /C, is:

can't tell without knowing the areas of the sides and bottom

A hollow conductor is positively charged. A small uncharged metal ball is lowered by a silk thread through a small opening in the top of the conductor and allowed to touch its inner surface. After the ball is removed, it will have:

a charge whose sign depends on what part of the inner surface it touched

If the flux of the electric field through a closed surface is zero,

The electric field may be zero everywhere on the surface and The net charge inside the surface must be zero

The electric field must be zero everywhere on the surface

The electric field may be zero everywhere on the surface

The net charge inside the surface must be zero

A charge ‘Q’ is placed at the centre of a hemispherical surface of radius ‘R’. The flux of electric field due to charge ‘Q’, through the surface of hemisphere is

Q 2 ϵ 0 \frac{Q}{2\epsilon_0} 2 ϵ 0 Q

Q 4 ϵ 0 \frac{Q}{4\epsilon_0} 4 ϵ 0 Q

Q 4 π ϵ 0 \frac{Q}{4\pi\epsilon_0} 4 π ϵ 0 Q

Q 2 π ϵ 0 \frac{Q}{2\pi\epsilon_0} 2 π ϵ 0 Q

A charge Q is enclosed in a dielectric of strength k. The maximum number of electric lines of force are

Q ϵ 0 \frac{Q}{\epsilon_0} ϵ 0 Q

Q k ϵ 0 \frac{Q}{k\epsilon_0} k ϵ 0 Q

The electric flux entering and leaving an enclosed surface is f 1 and f 2 respectively, the electric charge inside the enclosed surface will be:

( ϕ 1 − ϕ 2 ) ϵ 0 (ϕ_1-ϕ_2)ϵ_0 ( ϕ 1 − ϕ 2 ) ϵ 0

( ϕ 1 + ϕ 2 ) ϵ 0 \left(\phi_1+\phi_2\right)\epsilon_0 ( ϕ 1 + ϕ 2 ) ϵ 0

( ϕ 1 + ϕ 2 ) ϵ 0 \frac{\left(\phi_1+\phi_2\right)}{\epsilon_0} ϵ 0 ( ϕ 1 + ϕ 2 )

( ϕ 1 − ϕ 2 ) ϵ 0 \frac{\left(\phi_1-\phi_2\right)}{\epsilon_0} ϵ 0 ( ϕ 1 − ϕ 2 )

Chapter 3: Online Test 3: Gauss’s law:

Authored by RR GROUP

Physics

12th Grade

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Used 17+ times

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OPEN ENDED QUESTION

30 sec • Ungraded

Name

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NGSS.HS-LS4-2

OPEN ENDED QUESTION

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MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Electric field intensity due to an infinite plane sheet of charge is

σ/εo

q/2εo

σ/2εo

q/εo

Tags

NGSS.HS-PS2-4

NGSS.HS-PS3-5

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Electric flux lines due to an infinite sheet of charge is

converging

radial

uniform and perpendicular to the sheet

uniform and parallel to the sheet

Tags

NGSS.HS-PS2-4

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Total flux across a closed surface enclosing charge is independent of

shape of the closed surface

volume of the enclosure

actual spatial argument of charges within the surface

all the above

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The total electric flux over any closed surface is

εo

q2/εo

εo/q

q/εo

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Electric intensity due to an infinitely long plane sheet of a conductor at a point close to its surface is

independent of r

proportional to 1/r2

proportional to 1/r

inversely proportional to 1/r

Tags

NGSS.HS-PS2-4

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