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PHXII01:ELECTRIC CHARGES AND FIELDS

358278 The electric field in a region is given by $\vec{E} = \frac{3}{5} E_{0} \hat{i} + \frac{3}{2} E_{0} \hat{j} + E_{0} \hat{k}$ with $E_{0} = 2 \times 10^{3} N C^{- 1} .$ Find the flux of this field through a rectangular surface of area $0.2 m^{2}$ parallel to the $Y$ - $Z$ plane.

1

320 N m^{2} C^{- 1}

2

240 N m^{2} C^{- 1}

3

400 N m^{2} C^{- 1}

4 None of these

Explanation:

The flux through the surface is
$ϕ = E_{x} A = \frac{3}{5} \times 2 \times 10^{3} \times 0.2 = 240 N . m^{2} C^{- 1}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358279 A surface has the area vector $\vec{A} = (2 \hat{i} + 3 \hat{j}) m^{2}$ . The flux of an electric field through it if the field is $\vec{E} = 4 \hat{i} \frac{V}{m} :$

1

8 V - m

2

12 V - m

3

20 V - m

4

Zero

Explanation:

$ϕ = \vec{E} . \vec{A} = 4 \hat{i} . (2 \hat{i} + 3 \hat{j}) = 8 V - m$

PHXII01:ELECTRIC CHARGES AND FIELDS

358280 The electric field in a region of space is given by, $\vec{E} = E_{0} \hat{i} + 2 E_{0} \hat{j}$ where $E_{0} = 100 N / C$ . The flux of the field through a circular surface of radius $0.02 m$ parallel to the $Y - Z$ plane is nearly:

1

0.125 N m^{2} / C

2

0.02 N m^{2} / C

3

0.005 N m^{2} / C

4

3.14 N m^{2} / C

Explanation:

$\vec{E} = E_{0} \hat{i} + 2 E_{0} \hat{j}$
Given, $E_{0} = 100 N / C$
$\Rightarrow \vec{E} = 100 \hat{i} + 200 \hat{j}$
Radius of circular surface $= 0.02 m$
$A r e a = π r^{2} = \frac{22}{7} \times 0.02 \times 0.02$
$= 1.25 \times 10^{- 3} \hat{i} m^{2}$
[Loop is parallel to $Y$ - $Z$ plane]
Now, flux $(ϕ) = E A \cos θ$
But $ϕ = 0^{\circ}$
$ϕ = (100 \hat{i} + 200 \hat{j}) (1.25 \times 10^{- 3} \hat{i})$
$ϕ = 125 \times 10^{- 3} N m^{2} / C = 0.125 N m^{2} / C$

MHTCET - 2021

PHXII01:ELECTRIC CHARGES AND FIELDS

358281 The number of electric field lines crossing an area $Δ S$ is $n_{1}$ when $Δ \vec{S} ∥ \vec{E}$ , while number of field lines crossing same area is $n_{2}$ when $Δ S$ makes an angle of $30^{\circ}$ with $\vec{E}$ , then:

1

n_{1} < n_{2}

2

n_{1} > n_{2}

3

n_{1} = n_{2}

4 Cannot say anything

Explanation:

The no.of field lines crossing an area is directly related to electric flux passing through surface. In first case the flux crossing the surface is more then the flux crossing through surface in $2^{n d}$ case, so $n_{1} > n_{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358278 The electric field in a region is given by $\vec{E} = \frac{3}{5} E_{0} \hat{i} + \frac{3}{2} E_{0} \hat{j} + E_{0} \hat{k}$ with $E_{0} = 2 \times 10^{3} N C^{- 1} .$ Find the flux of this field through a rectangular surface of area $0.2 m^{2}$ parallel to the $Y$ - $Z$ plane.

1

320 N m^{2} C^{- 1}

2

240 N m^{2} C^{- 1}

3

400 N m^{2} C^{- 1}

4 None of these

Explanation:

The flux through the surface is
$ϕ = E_{x} A = \frac{3}{5} \times 2 \times 10^{3} \times 0.2 = 240 N . m^{2} C^{- 1}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358279 A surface has the area vector $\vec{A} = (2 \hat{i} + 3 \hat{j}) m^{2}$ . The flux of an electric field through it if the field is $\vec{E} = 4 \hat{i} \frac{V}{m} :$

1

8 V - m

2

12 V - m

3

20 V - m

4

Zero

Explanation:

$ϕ = \vec{E} . \vec{A} = 4 \hat{i} . (2 \hat{i} + 3 \hat{j}) = 8 V - m$

PHXII01:ELECTRIC CHARGES AND FIELDS

358280 The electric field in a region of space is given by, $\vec{E} = E_{0} \hat{i} + 2 E_{0} \hat{j}$ where $E_{0} = 100 N / C$ . The flux of the field through a circular surface of radius $0.02 m$ parallel to the $Y - Z$ plane is nearly:

1

0.125 N m^{2} / C

2

0.02 N m^{2} / C

3

0.005 N m^{2} / C

4

3.14 N m^{2} / C

Explanation:

$\vec{E} = E_{0} \hat{i} + 2 E_{0} \hat{j}$
Given, $E_{0} = 100 N / C$
$\Rightarrow \vec{E} = 100 \hat{i} + 200 \hat{j}$
Radius of circular surface $= 0.02 m$
$A r e a = π r^{2} = \frac{22}{7} \times 0.02 \times 0.02$
$= 1.25 \times 10^{- 3} \hat{i} m^{2}$
[Loop is parallel to $Y$ - $Z$ plane]
Now, flux $(ϕ) = E A \cos θ$
But $ϕ = 0^{\circ}$
$ϕ = (100 \hat{i} + 200 \hat{j}) (1.25 \times 10^{- 3} \hat{i})$
$ϕ = 125 \times 10^{- 3} N m^{2} / C = 0.125 N m^{2} / C$

MHTCET - 2021

PHXII01:ELECTRIC CHARGES AND FIELDS

358281 The number of electric field lines crossing an area $Δ S$ is $n_{1}$ when $Δ \vec{S} ∥ \vec{E}$ , while number of field lines crossing same area is $n_{2}$ when $Δ S$ makes an angle of $30^{\circ}$ with $\vec{E}$ , then:

1

n_{1} < n_{2}

2

n_{1} > n_{2}

3

n_{1} = n_{2}

4 Cannot say anything

Explanation:

The no.of field lines crossing an area is directly related to electric flux passing through surface. In first case the flux crossing the surface is more then the flux crossing through surface in $2^{n d}$ case, so $n_{1} > n_{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358278 The electric field in a region is given by $\vec{E} = \frac{3}{5} E_{0} \hat{i} + \frac{3}{2} E_{0} \hat{j} + E_{0} \hat{k}$ with $E_{0} = 2 \times 10^{3} N C^{- 1} .$ Find the flux of this field through a rectangular surface of area $0.2 m^{2}$ parallel to the $Y$ - $Z$ plane.

1

320 N m^{2} C^{- 1}

2

240 N m^{2} C^{- 1}

3

400 N m^{2} C^{- 1}

4 None of these

Explanation:

The flux through the surface is
$ϕ = E_{x} A = \frac{3}{5} \times 2 \times 10^{3} \times 0.2 = 240 N . m^{2} C^{- 1}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358279 A surface has the area vector $\vec{A} = (2 \hat{i} + 3 \hat{j}) m^{2}$ . The flux of an electric field through it if the field is $\vec{E} = 4 \hat{i} \frac{V}{m} :$

1

8 V - m

2

12 V - m

3

20 V - m

4

Zero

Explanation:

$ϕ = \vec{E} . \vec{A} = 4 \hat{i} . (2 \hat{i} + 3 \hat{j}) = 8 V - m$

PHXII01:ELECTRIC CHARGES AND FIELDS

358280 The electric field in a region of space is given by, $\vec{E} = E_{0} \hat{i} + 2 E_{0} \hat{j}$ where $E_{0} = 100 N / C$ . The flux of the field through a circular surface of radius $0.02 m$ parallel to the $Y - Z$ plane is nearly:

1

0.125 N m^{2} / C

2

0.02 N m^{2} / C

3

0.005 N m^{2} / C

4

3.14 N m^{2} / C

Explanation:

$\vec{E} = E_{0} \hat{i} + 2 E_{0} \hat{j}$
Given, $E_{0} = 100 N / C$
$\Rightarrow \vec{E} = 100 \hat{i} + 200 \hat{j}$
Radius of circular surface $= 0.02 m$
$A r e a = π r^{2} = \frac{22}{7} \times 0.02 \times 0.02$
$= 1.25 \times 10^{- 3} \hat{i} m^{2}$
[Loop is parallel to $Y$ - $Z$ plane]
Now, flux $(ϕ) = E A \cos θ$
But $ϕ = 0^{\circ}$
$ϕ = (100 \hat{i} + 200 \hat{j}) (1.25 \times 10^{- 3} \hat{i})$
$ϕ = 125 \times 10^{- 3} N m^{2} / C = 0.125 N m^{2} / C$

MHTCET - 2021

PHXII01:ELECTRIC CHARGES AND FIELDS

358281 The number of electric field lines crossing an area $Δ S$ is $n_{1}$ when $Δ \vec{S} ∥ \vec{E}$ , while number of field lines crossing same area is $n_{2}$ when $Δ S$ makes an angle of $30^{\circ}$ with $\vec{E}$ , then:

1

n_{1} < n_{2}

2

n_{1} > n_{2}

3

n_{1} = n_{2}

4 Cannot say anything

Explanation:

The no.of field lines crossing an area is directly related to electric flux passing through surface. In first case the flux crossing the surface is more then the flux crossing through surface in $2^{n d}$ case, so $n_{1} > n_{2}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358278 The electric field in a region is given by $\vec{E} = \frac{3}{5} E_{0} \hat{i} + \frac{3}{2} E_{0} \hat{j} + E_{0} \hat{k}$ with $E_{0} = 2 \times 10^{3} N C^{- 1} .$ Find the flux of this field through a rectangular surface of area $0.2 m^{2}$ parallel to the $Y$ - $Z$ plane.

1

320 N m^{2} C^{- 1}

2

240 N m^{2} C^{- 1}

3

400 N m^{2} C^{- 1}

4 None of these

Explanation:

The flux through the surface is
$ϕ = E_{x} A = \frac{3}{5} \times 2 \times 10^{3} \times 0.2 = 240 N . m^{2} C^{- 1}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358279 A surface has the area vector $\vec{A} = (2 \hat{i} + 3 \hat{j}) m^{2}$ . The flux of an electric field through it if the field is $\vec{E} = 4 \hat{i} \frac{V}{m} :$

1

8 V - m

2

12 V - m

3

20 V - m

4

Zero

Explanation:

$ϕ = \vec{E} . \vec{A} = 4 \hat{i} . (2 \hat{i} + 3 \hat{j}) = 8 V - m$

PHXII01:ELECTRIC CHARGES AND FIELDS

358280 The electric field in a region of space is given by, $\vec{E} = E_{0} \hat{i} + 2 E_{0} \hat{j}$ where $E_{0} = 100 N / C$ . The flux of the field through a circular surface of radius $0.02 m$ parallel to the $Y - Z$ plane is nearly:

1

0.125 N m^{2} / C

2

0.02 N m^{2} / C

3

0.005 N m^{2} / C

4

3.14 N m^{2} / C

Explanation:

$\vec{E} = E_{0} \hat{i} + 2 E_{0} \hat{j}$
Given, $E_{0} = 100 N / C$
$\Rightarrow \vec{E} = 100 \hat{i} + 200 \hat{j}$
Radius of circular surface $= 0.02 m$
$A r e a = π r^{2} = \frac{22}{7} \times 0.02 \times 0.02$
$= 1.25 \times 10^{- 3} \hat{i} m^{2}$
[Loop is parallel to $Y$ - $Z$ plane]
Now, flux $(ϕ) = E A \cos θ$
But $ϕ = 0^{\circ}$
$ϕ = (100 \hat{i} + 200 \hat{j}) (1.25 \times 10^{- 3} \hat{i})$
$ϕ = 125 \times 10^{- 3} N m^{2} / C = 0.125 N m^{2} / C$

MHTCET - 2021

PHXII01:ELECTRIC CHARGES AND FIELDS

358281 The number of electric field lines crossing an area $Δ S$ is $n_{1}$ when $Δ \vec{S} ∥ \vec{E}$ , while number of field lines crossing same area is $n_{2}$ when $Δ S$ makes an angle of $30^{\circ}$ with $\vec{E}$ , then:

1

n_{1} < n_{2}

2

n_{1} > n_{2}

3

n_{1} = n_{2}

4 Cannot say anything

Explanation:

The no.of field lines crossing an area is directly related to electric flux passing through surface. In first case the flux crossing the surface is more then the flux crossing through surface in $2^{n d}$ case, so $n_{1} > n_{2}$