QBManager

PHXII01:ELECTRIC CHARGES AND FIELDS

358291 A circular disc of radius ' $r$ ' is placed along the plane of paper. A uniform electric field $\vec{E}$ is also present in the plane of paper. What amount of electric flux is associated with it?
supporting img

1

E π r^{2}

2 zero

3

2 E π r

4

\frac{π r^{2}}{E}

PHXII01:ELECTRIC CHARGES AND FIELDS

358292 Assertion :
The electric flux emanating out and entering a closed surface ( in vacuum) are $8 \times 10^{3}$ and $2 \times 10^{3} V m$ respectively. The charge enclosed by the surface is $0.053 μ C$ .
Reason :
Gauss's theorem is not relevant here.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$ϕ = Q_{enclosed} / \in_{0}$
$Q_{e n c l o s e d} = \int_{0} ϕ$
$= 8.85 \times 10^{- 12} [8 \times 10^{3} - 2 \times 10^{3}]$
$= 53.10 \times 10^{- 9} C$
$= 0.053 μ C$
Gauss law is very much applicable as some net electric charge is in a closed surface. So correct option is (3).

PHXII01:ELECTRIC CHARGES AND FIELDS

358293 Five electric dipoles of charge ' $q$ ' each are placed inside the shell. What will be the amount of electric flux associated with the shell?

1

5 q ε_{0}

2

\frac{10 q}{ε_{0}}

3 Zero

4

\frac{5 q}{ε_{0}}

Explanation:

Total charge into the shell is zero because a dipole composed of negative and positive charge.
So, $q_{n e t} = 5 (- q + q) = 0$
$ϕ = \frac{q}{ε_{0}} = zero .$

PHXII01:ELECTRIC CHARGES AND FIELDS

358294 If $\oint_{s} \vec{E} \cdot \vec{d S} = 0$ over a surface, then:

1 The magnitude of electric field on the surface is constant.

2 All the charges must necessarily be inside the surface

3 The electric field inside the surface is necessarily uniform

4 The number of flux lines entering the surface must be equal to the number of flux lines leaving it.

Explanation:

For diagram shown:
$\begin{aligned} ϕ = - E d s + E d s \\ \Rightarrow ϕ = 0 \\ \Rightarrow ϕ_{in} = ϕ_{out} \end{aligned}$
supporting img
Hence, number of field lines entering is equal to number of field lines leaving. Correct option is (4).

NEET - 2023

PHXII01:ELECTRIC CHARGES AND FIELDS

358291 A circular disc of radius ' $r$ ' is placed along the plane of paper. A uniform electric field $\vec{E}$ is also present in the plane of paper. What amount of electric flux is associated with it?
supporting img

1

E π r^{2}

2 zero

3

2 E π r

4

\frac{π r^{2}}{E}

PHXII01:ELECTRIC CHARGES AND FIELDS

358292 Assertion :
The electric flux emanating out and entering a closed surface ( in vacuum) are $8 \times 10^{3}$ and $2 \times 10^{3} V m$ respectively. The charge enclosed by the surface is $0.053 μ C$ .
Reason :
Gauss's theorem is not relevant here.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$ϕ = Q_{enclosed} / \in_{0}$
$Q_{e n c l o s e d} = \int_{0} ϕ$
$= 8.85 \times 10^{- 12} [8 \times 10^{3} - 2 \times 10^{3}]$
$= 53.10 \times 10^{- 9} C$
$= 0.053 μ C$
Gauss law is very much applicable as some net electric charge is in a closed surface. So correct option is (3).

PHXII01:ELECTRIC CHARGES AND FIELDS

358293 Five electric dipoles of charge ' $q$ ' each are placed inside the shell. What will be the amount of electric flux associated with the shell?

1

5 q ε_{0}

2

\frac{10 q}{ε_{0}}

3 Zero

4

\frac{5 q}{ε_{0}}

Explanation:

Total charge into the shell is zero because a dipole composed of negative and positive charge.
So, $q_{n e t} = 5 (- q + q) = 0$
$ϕ = \frac{q}{ε_{0}} = zero .$

PHXII01:ELECTRIC CHARGES AND FIELDS

358294 If $\oint_{s} \vec{E} \cdot \vec{d S} = 0$ over a surface, then:

1 The magnitude of electric field on the surface is constant.

2 All the charges must necessarily be inside the surface

3 The electric field inside the surface is necessarily uniform

4 The number of flux lines entering the surface must be equal to the number of flux lines leaving it.

Explanation:

For diagram shown:
$\begin{aligned} ϕ = - E d s + E d s \\ \Rightarrow ϕ = 0 \\ \Rightarrow ϕ_{in} = ϕ_{out} \end{aligned}$
supporting img
Hence, number of field lines entering is equal to number of field lines leaving. Correct option is (4).

NEET - 2023

PHXII01:ELECTRIC CHARGES AND FIELDS

358291 A circular disc of radius ' $r$ ' is placed along the plane of paper. A uniform electric field $\vec{E}$ is also present in the plane of paper. What amount of electric flux is associated with it?
supporting img

1

E π r^{2}

2 zero

3

2 E π r

4

\frac{π r^{2}}{E}

PHXII01:ELECTRIC CHARGES AND FIELDS

358292 Assertion :
The electric flux emanating out and entering a closed surface ( in vacuum) are $8 \times 10^{3}$ and $2 \times 10^{3} V m$ respectively. The charge enclosed by the surface is $0.053 μ C$ .
Reason :
Gauss's theorem is not relevant here.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$ϕ = Q_{enclosed} / \in_{0}$
$Q_{e n c l o s e d} = \int_{0} ϕ$
$= 8.85 \times 10^{- 12} [8 \times 10^{3} - 2 \times 10^{3}]$
$= 53.10 \times 10^{- 9} C$
$= 0.053 μ C$
Gauss law is very much applicable as some net electric charge is in a closed surface. So correct option is (3).

PHXII01:ELECTRIC CHARGES AND FIELDS

358293 Five electric dipoles of charge ' $q$ ' each are placed inside the shell. What will be the amount of electric flux associated with the shell?

1

5 q ε_{0}

2

\frac{10 q}{ε_{0}}

3 Zero

4

\frac{5 q}{ε_{0}}

Explanation:

Total charge into the shell is zero because a dipole composed of negative and positive charge.
So, $q_{n e t} = 5 (- q + q) = 0$
$ϕ = \frac{q}{ε_{0}} = zero .$

PHXII01:ELECTRIC CHARGES AND FIELDS

358294 If $\oint_{s} \vec{E} \cdot \vec{d S} = 0$ over a surface, then:

1 The magnitude of electric field on the surface is constant.

2 All the charges must necessarily be inside the surface

3 The electric field inside the surface is necessarily uniform

4 The number of flux lines entering the surface must be equal to the number of flux lines leaving it.

Explanation:

For diagram shown:
$\begin{aligned} ϕ = - E d s + E d s \\ \Rightarrow ϕ = 0 \\ \Rightarrow ϕ_{in} = ϕ_{out} \end{aligned}$
supporting img
Hence, number of field lines entering is equal to number of field lines leaving. Correct option is (4).

NEET - 2023

PHXII01:ELECTRIC CHARGES AND FIELDS

358291 A circular disc of radius ' $r$ ' is placed along the plane of paper. A uniform electric field $\vec{E}$ is also present in the plane of paper. What amount of electric flux is associated with it?
supporting img

1

E π r^{2}

2 zero

3

2 E π r

4

\frac{π r^{2}}{E}

PHXII01:ELECTRIC CHARGES AND FIELDS

358292 Assertion :
The electric flux emanating out and entering a closed surface ( in vacuum) are $8 \times 10^{3}$ and $2 \times 10^{3} V m$ respectively. The charge enclosed by the surface is $0.053 μ C$ .
Reason :
Gauss's theorem is not relevant here.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$ϕ = Q_{enclosed} / \in_{0}$
$Q_{e n c l o s e d} = \int_{0} ϕ$
$= 8.85 \times 10^{- 12} [8 \times 10^{3} - 2 \times 10^{3}]$
$= 53.10 \times 10^{- 9} C$
$= 0.053 μ C$
Gauss law is very much applicable as some net electric charge is in a closed surface. So correct option is (3).

PHXII01:ELECTRIC CHARGES AND FIELDS

358293 Five electric dipoles of charge ' $q$ ' each are placed inside the shell. What will be the amount of electric flux associated with the shell?

1

5 q ε_{0}

2

\frac{10 q}{ε_{0}}

3 Zero

4

\frac{5 q}{ε_{0}}

Explanation:

Total charge into the shell is zero because a dipole composed of negative and positive charge.
So, $q_{n e t} = 5 (- q + q) = 0$
$ϕ = \frac{q}{ε_{0}} = zero .$

PHXII01:ELECTRIC CHARGES AND FIELDS

358294 If $\oint_{s} \vec{E} \cdot \vec{d S} = 0$ over a surface, then:

1 The magnitude of electric field on the surface is constant.

2 All the charges must necessarily be inside the surface

3 The electric field inside the surface is necessarily uniform

4 The number of flux lines entering the surface must be equal to the number of flux lines leaving it.

Explanation:

For diagram shown:
$\begin{aligned} ϕ = - E d s + E d s \\ \Rightarrow ϕ = 0 \\ \Rightarrow ϕ_{in} = ϕ_{out} \end{aligned}$
supporting img
Hence, number of field lines entering is equal to number of field lines leaving. Correct option is (4).

NEET - 2023