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PHXII01:ELECTRIC CHARGES AND FIELDS

358282 A uniform electrical field $a \hat{i} + b \hat{j}$ intersects a surface of area $A$ . What is the flux through this area if the surface lies in $X Z$ plane?

1

A b

2

A a

3

\frac{A (a b)}{a + b}

4

A \sqrt{a^{2} + b^{2}}

Explanation:

The direction of the field is
Given that $\overset{―}{E} = a \hat{i} + b \hat{j} & \overset{―}{A} = A \hat{j}$
$ϕ = \overset{―}{E} . \overset{―}{A} = A b$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358283 A loop of diameter $d$ is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be $ϕ$ . What is the electric field strength?

1

\frac{4 ϕ}{π d^{2}}

2

\frac{2 ϕ}{π d^{2}}

3

\frac{ϕ}{π d^{2}}

4

\frac{π ϕ d^{2}}{4}

Explanation:

$ϕ = E A \cos 0^{\circ} = E \times \frac{π d^{2}}{4}, ∴ E = \frac{4 ϕ}{π d^{2}}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358284 An electric field is given by, $(6 \hat{i} + 5 \hat{j} + 3 \hat{k}) N / C$ . The electric flux through a surface area $30 \hat{i} m^{2}$ lying in $Y Z -$ plane (in $S I$ unit) is

1 150

2 60

3 180

4 90

Explanation:

Given : $\vec{E} = 6 \hat{i} + 5 \hat{j} + 3 \hat{k} N / C, \vec{A} = 30 \hat{i} m^{2}$
Electric flux, $ϕ = \vec{E} . \vec{A}$
$ϕ = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) \cdot (30 \hat{i}) = 180 N m^{2} / C$

JEE - 2024

PHXII01:ELECTRIC CHARGES AND FIELDS

358285 A uniform electric field $E = 2 \times 10^{3} N C^{- 1}$ is acting along the positive $x$ -axis. The flux of this field through a square of $10 c m$ side whose plane is parallel to the $y z$ plane is

1

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

2

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

3

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

4

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

Explanation:

Here, $E = 2 \times 10^{3} N C^{- 1}$ along $+ x$ -axis surface area,
$S = (10 c m)^{2} = 10^{2} \times 10^{- 4} m^{2} = 10^{- 2} m^{2}$
When plane is parallel to $y z$ plane, $θ = 0^{\circ}$
So $ϕ = E S \cos θ = 2 \times 10^{3} \times 10^{- 2} \cos 0^{\circ}$
$= 20 N C^{- 1} m^{2}$
Conducting Parallel Slabs

PHXII01:ELECTRIC CHARGES AND FIELDS

358282 A uniform electrical field $a \hat{i} + b \hat{j}$ intersects a surface of area $A$ . What is the flux through this area if the surface lies in $X Z$ plane?

1

A b

2

A a

3

\frac{A (a b)}{a + b}

4

A \sqrt{a^{2} + b^{2}}

Explanation:

The direction of the field is
Given that $\overset{―}{E} = a \hat{i} + b \hat{j} & \overset{―}{A} = A \hat{j}$
$ϕ = \overset{―}{E} . \overset{―}{A} = A b$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358283 A loop of diameter $d$ is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be $ϕ$ . What is the electric field strength?

1

\frac{4 ϕ}{π d^{2}}

2

\frac{2 ϕ}{π d^{2}}

3

\frac{ϕ}{π d^{2}}

4

\frac{π ϕ d^{2}}{4}

Explanation:

$ϕ = E A \cos 0^{\circ} = E \times \frac{π d^{2}}{4}, ∴ E = \frac{4 ϕ}{π d^{2}}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358284 An electric field is given by, $(6 \hat{i} + 5 \hat{j} + 3 \hat{k}) N / C$ . The electric flux through a surface area $30 \hat{i} m^{2}$ lying in $Y Z -$ plane (in $S I$ unit) is

1 150

2 60

3 180

4 90

Explanation:

Given : $\vec{E} = 6 \hat{i} + 5 \hat{j} + 3 \hat{k} N / C, \vec{A} = 30 \hat{i} m^{2}$
Electric flux, $ϕ = \vec{E} . \vec{A}$
$ϕ = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) \cdot (30 \hat{i}) = 180 N m^{2} / C$

JEE - 2024

PHXII01:ELECTRIC CHARGES AND FIELDS

358285 A uniform electric field $E = 2 \times 10^{3} N C^{- 1}$ is acting along the positive $x$ -axis. The flux of this field through a square of $10 c m$ side whose plane is parallel to the $y z$ plane is

1

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

2

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

3

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

4

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

Explanation:

Here, $E = 2 \times 10^{3} N C^{- 1}$ along $+ x$ -axis surface area,
$S = (10 c m)^{2} = 10^{2} \times 10^{- 4} m^{2} = 10^{- 2} m^{2}$
When plane is parallel to $y z$ plane, $θ = 0^{\circ}$
So $ϕ = E S \cos θ = 2 \times 10^{3} \times 10^{- 2} \cos 0^{\circ}$
$= 20 N C^{- 1} m^{2}$
Conducting Parallel Slabs

PHXII01:ELECTRIC CHARGES AND FIELDS

358282 A uniform electrical field $a \hat{i} + b \hat{j}$ intersects a surface of area $A$ . What is the flux through this area if the surface lies in $X Z$ plane?

1

A b

2

A a

3

\frac{A (a b)}{a + b}

4

A \sqrt{a^{2} + b^{2}}

Explanation:

The direction of the field is
Given that $\overset{―}{E} = a \hat{i} + b \hat{j} & \overset{―}{A} = A \hat{j}$
$ϕ = \overset{―}{E} . \overset{―}{A} = A b$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358283 A loop of diameter $d$ is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be $ϕ$ . What is the electric field strength?

1

\frac{4 ϕ}{π d^{2}}

2

\frac{2 ϕ}{π d^{2}}

3

\frac{ϕ}{π d^{2}}

4

\frac{π ϕ d^{2}}{4}

Explanation:

$ϕ = E A \cos 0^{\circ} = E \times \frac{π d^{2}}{4}, ∴ E = \frac{4 ϕ}{π d^{2}}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358284 An electric field is given by, $(6 \hat{i} + 5 \hat{j} + 3 \hat{k}) N / C$ . The electric flux through a surface area $30 \hat{i} m^{2}$ lying in $Y Z -$ plane (in $S I$ unit) is

1 150

2 60

3 180

4 90

Explanation:

Given : $\vec{E} = 6 \hat{i} + 5 \hat{j} + 3 \hat{k} N / C, \vec{A} = 30 \hat{i} m^{2}$
Electric flux, $ϕ = \vec{E} . \vec{A}$
$ϕ = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) \cdot (30 \hat{i}) = 180 N m^{2} / C$

JEE - 2024

PHXII01:ELECTRIC CHARGES AND FIELDS

358285 A uniform electric field $E = 2 \times 10^{3} N C^{- 1}$ is acting along the positive $x$ -axis. The flux of this field through a square of $10 c m$ side whose plane is parallel to the $y z$ plane is

1

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

2

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

3

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

4

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

Explanation:

Here, $E = 2 \times 10^{3} N C^{- 1}$ along $+ x$ -axis surface area,
$S = (10 c m)^{2} = 10^{2} \times 10^{- 4} m^{2} = 10^{- 2} m^{2}$
When plane is parallel to $y z$ plane, $θ = 0^{\circ}$
So $ϕ = E S \cos θ = 2 \times 10^{3} \times 10^{- 2} \cos 0^{\circ}$
$= 20 N C^{- 1} m^{2}$
Conducting Parallel Slabs

PHXII01:ELECTRIC CHARGES AND FIELDS

358282 A uniform electrical field $a \hat{i} + b \hat{j}$ intersects a surface of area $A$ . What is the flux through this area if the surface lies in $X Z$ plane?

1

A b

2

A a

3

\frac{A (a b)}{a + b}

4

A \sqrt{a^{2} + b^{2}}

Explanation:

The direction of the field is
Given that $\overset{―}{E} = a \hat{i} + b \hat{j} & \overset{―}{A} = A \hat{j}$
$ϕ = \overset{―}{E} . \overset{―}{A} = A b$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358283 A loop of diameter $d$ is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be $ϕ$ . What is the electric field strength?

1

\frac{4 ϕ}{π d^{2}}

2

\frac{2 ϕ}{π d^{2}}

3

\frac{ϕ}{π d^{2}}

4

\frac{π ϕ d^{2}}{4}

Explanation:

$ϕ = E A \cos 0^{\circ} = E \times \frac{π d^{2}}{4}, ∴ E = \frac{4 ϕ}{π d^{2}}$

PHXII01:ELECTRIC CHARGES AND FIELDS

358284 An electric field is given by, $(6 \hat{i} + 5 \hat{j} + 3 \hat{k}) N / C$ . The electric flux through a surface area $30 \hat{i} m^{2}$ lying in $Y Z -$ plane (in $S I$ unit) is

1 150

2 60

3 180

4 90

Explanation:

Given : $\vec{E} = 6 \hat{i} + 5 \hat{j} + 3 \hat{k} N / C, \vec{A} = 30 \hat{i} m^{2}$
Electric flux, $ϕ = \vec{E} . \vec{A}$
$ϕ = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) \cdot (30 \hat{i}) = 180 N m^{2} / C$

JEE - 2024

PHXII01:ELECTRIC CHARGES AND FIELDS

358285 A uniform electric field $E = 2 \times 10^{3} N C^{- 1}$ is acting along the positive $x$ -axis. The flux of this field through a square of $10 c m$ side whose plane is parallel to the $y z$ plane is

1

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

2

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

3

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

4

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

Explanation:

Here, $E = 2 \times 10^{3} N C^{- 1}$ along $+ x$ -axis surface area,
$S = (10 c m)^{2} = 10^{2} \times 10^{- 4} m^{2} = 10^{- 2} m^{2}$
When plane is parallel to $y z$ plane, $θ = 0^{\circ}$
So $ϕ = E S \cos θ = 2 \times 10^{3} \times 10^{- 2} \cos 0^{\circ}$
$= 20 N C^{- 1} m^{2}$
Conducting Parallel Slabs