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Electric Charges and Fields

272170 If a charge $q$ is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :

1

\frac{q}{ε_{0}}

2

\frac{q}{2 ε_{0}}

3

\frac{q}{4 ε_{0}}

4

\frac{q}{2 π ε_{0}}

Explanation:

(None)
Total flux through complete spherical surface is $\frac{q}{ϵ_{0}}$ .
So flux, through curved surface will be $\frac{q}{2 ϵ_{0}}$ .
The flux through flat surface will be zero

NCERT Page-25 / N-22

Electric Charges and Fields

272171 Number of electric lines of force that radiate outwards from one coulomb of charge in vacuum is

1

1.13 \times 10^{11}

2

0.61 \times 10^{11}

3

1.13 \times 10^{10}

4

0.61 \times 10^{9}

Explanation:

(a) Here, $q = 1 C, ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ Number of lines of force $=$ Electric force
$= \frac{q}{ε_{0}} = \frac{1}{8.85 \times 10^{- 12}} = 1.13 \times 10^{11}$

NCERT Page-25 / N-22

Electric Charges and Fields

272172 In the figure the net electric flux through the area $A$ is $ϕ = \vec{E} \cdot \vec{A}$ when the system is in air. On immersing the system in water the net electric flux through the area

1 becomes zero

2 remains same

3 increases

4 decreases

Explanation:

(d) Since electric field $\vec{E}$ decreases inside water, therefore flux $ϕ = \vec{E} \cdot \vec{A}$ also decreases.

NCERT Page-26 / N-22

Electric Charges and Fields

272178 The electric field in a region of space is given by, $\vec{E} = E_{0} i + 2 E_{0} j$ where $E_{o} = 100 N / C$ . 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:

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

NCERT Page-26 / N-22

Electric Charges and Fields

272170 If a charge $q$ is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :

1

\frac{q}{ε_{0}}

2

\frac{q}{2 ε_{0}}

3

\frac{q}{4 ε_{0}}

4

\frac{q}{2 π ε_{0}}

Explanation:

(None)
Total flux through complete spherical surface is $\frac{q}{ϵ_{0}}$ .
So flux, through curved surface will be $\frac{q}{2 ϵ_{0}}$ .
The flux through flat surface will be zero

NCERT Page-25 / N-22

Electric Charges and Fields

272171 Number of electric lines of force that radiate outwards from one coulomb of charge in vacuum is

1

1.13 \times 10^{11}

2

0.61 \times 10^{11}

3

1.13 \times 10^{10}

4

0.61 \times 10^{9}

Explanation:

(a) Here, $q = 1 C, ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ Number of lines of force $=$ Electric force
$= \frac{q}{ε_{0}} = \frac{1}{8.85 \times 10^{- 12}} = 1.13 \times 10^{11}$

NCERT Page-25 / N-22

Electric Charges and Fields

272172 In the figure the net electric flux through the area $A$ is $ϕ = \vec{E} \cdot \vec{A}$ when the system is in air. On immersing the system in water the net electric flux through the area

1 becomes zero

2 remains same

3 increases

4 decreases

Explanation:

(d) Since electric field $\vec{E}$ decreases inside water, therefore flux $ϕ = \vec{E} \cdot \vec{A}$ also decreases.

NCERT Page-26 / N-22

Electric Charges and Fields

272178 The electric field in a region of space is given by, $\vec{E} = E_{0} i + 2 E_{0} j$ where $E_{o} = 100 N / C$ . 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:

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

NCERT Page-26 / N-22

Electric Charges and Fields

272170 If a charge $q$ is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :

1

\frac{q}{ε_{0}}

2

\frac{q}{2 ε_{0}}

3

\frac{q}{4 ε_{0}}

4

\frac{q}{2 π ε_{0}}

Explanation:

(None)
Total flux through complete spherical surface is $\frac{q}{ϵ_{0}}$ .
So flux, through curved surface will be $\frac{q}{2 ϵ_{0}}$ .
The flux through flat surface will be zero

NCERT Page-25 / N-22

Electric Charges and Fields

272171 Number of electric lines of force that radiate outwards from one coulomb of charge in vacuum is

1

1.13 \times 10^{11}

2

0.61 \times 10^{11}

3

1.13 \times 10^{10}

4

0.61 \times 10^{9}

Explanation:

(a) Here, $q = 1 C, ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ Number of lines of force $=$ Electric force
$= \frac{q}{ε_{0}} = \frac{1}{8.85 \times 10^{- 12}} = 1.13 \times 10^{11}$

NCERT Page-25 / N-22

Electric Charges and Fields

272172 In the figure the net electric flux through the area $A$ is $ϕ = \vec{E} \cdot \vec{A}$ when the system is in air. On immersing the system in water the net electric flux through the area

1 becomes zero

2 remains same

3 increases

4 decreases

Explanation:

(d) Since electric field $\vec{E}$ decreases inside water, therefore flux $ϕ = \vec{E} \cdot \vec{A}$ also decreases.

NCERT Page-26 / N-22

Electric Charges and Fields

272178 The electric field in a region of space is given by, $\vec{E} = E_{0} i + 2 E_{0} j$ where $E_{o} = 100 N / C$ . 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:

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

NCERT Page-26 / N-22

Electric Charges and Fields

272170 If a charge $q$ is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :

1

\frac{q}{ε_{0}}

2

\frac{q}{2 ε_{0}}

3

\frac{q}{4 ε_{0}}

4

\frac{q}{2 π ε_{0}}

Explanation:

(None)
Total flux through complete spherical surface is $\frac{q}{ϵ_{0}}$ .
So flux, through curved surface will be $\frac{q}{2 ϵ_{0}}$ .
The flux through flat surface will be zero

NCERT Page-25 / N-22

Electric Charges and Fields

272171 Number of electric lines of force that radiate outwards from one coulomb of charge in vacuum is

1

1.13 \times 10^{11}

2

0.61 \times 10^{11}

3

1.13 \times 10^{10}

4

0.61 \times 10^{9}

Explanation:

(a) Here, $q = 1 C, ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ Number of lines of force $=$ Electric force
$= \frac{q}{ε_{0}} = \frac{1}{8.85 \times 10^{- 12}} = 1.13 \times 10^{11}$

NCERT Page-25 / N-22

Electric Charges and Fields

272172 In the figure the net electric flux through the area $A$ is $ϕ = \vec{E} \cdot \vec{A}$ when the system is in air. On immersing the system in water the net electric flux through the area

1 becomes zero

2 remains same

3 increases

4 decreases

Explanation:

(d) Since electric field $\vec{E}$ decreases inside water, therefore flux $ϕ = \vec{E} \cdot \vec{A}$ also decreases.

NCERT Page-26 / N-22

Electric Charges and Fields

272178 The electric field in a region of space is given by, $\vec{E} = E_{0} i + 2 E_{0} j$ where $E_{o} = 100 N / C$ . 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:

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

NCERT Page-26 / N-22

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