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PHXII01:ELECTRIC CHARGES AND FIELDS

358321 Two charges $Q, - 3 Q$ are is kept in the cavity inside an uncharged metal, as shown in the figure. Electric flux through the Gaussian surface $S$ inside the metal shown in figure is
supporting img

1

\frac{2 Q}{ε_{0}}

2 zero

3

\frac{- 2 Q}{ε_{0}}

4

\frac{3 Q}{2 ε_{0}}

Explanation:

Electric field inside the conducting medium is always zero under electrostatic condition. $E = 0 \Rightarrow$ Flux $= 0$ . Option (2) is correct.

PHXII01:ELECTRIC CHARGES AND FIELDS

358322 Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 1 $m$ surrounding the total charge is 100 $V$ - $m$ . The flux over the concentric sphere of radius 2 $m$ will be

1

25 V - m

2

50 V - m

3

100 V - m

4

200 V - m

PHXII01:ELECTRIC CHARGES AND FIELDS

358323 Gauss’s law should be invalid if

1 The velocity of light were not a universal constant

2 The inverse square law were not exactly true

3 There were magnetic monopoles

4 None of these

PHXII01:ELECTRIC CHARGES AND FIELDS

358324 A point charge $q$ is placed at the centre of the cubical box. If $q = 12000 ϵ_{0} C$ , find the flux through shaded area of surface is
supporting img

1

150 N - m^{2} / C

2

200 N - m^{2} / C

3

500 N - m^{2} / C

4

650 N - m^{2} / C

Explanation:

Total flux coming out of the cubical box is given as
$ϕ_{0} = \frac{q}{ϵ_{0}}$
This total flux is emerging equally from each of the six face of the box.
Thus flux through each face can be given as
$ϕ = \frac{q}{6 ϵ_{0}}$
The flux through shaded portion would be one fourth of the flux through each face as through all the four triangular portions flux would be same due to symmetry, thus we have
$ϕ^{'} = \frac{ϕ}{4} = \frac{q}{24 \in_{0}} = \frac{12000 \in_{0}}{24 \in_{0}} = 500 N - m^{2} / C$

PHXII01:ELECTRIC CHARGES AND FIELDS

358321 Two charges $Q, - 3 Q$ are is kept in the cavity inside an uncharged metal, as shown in the figure. Electric flux through the Gaussian surface $S$ inside the metal shown in figure is
supporting img

1

\frac{2 Q}{ε_{0}}

2 zero

3

\frac{- 2 Q}{ε_{0}}

4

\frac{3 Q}{2 ε_{0}}

Explanation:

Electric field inside the conducting medium is always zero under electrostatic condition. $E = 0 \Rightarrow$ Flux $= 0$ . Option (2) is correct.

PHXII01:ELECTRIC CHARGES AND FIELDS

358322 Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 1 $m$ surrounding the total charge is 100 $V$ - $m$ . The flux over the concentric sphere of radius 2 $m$ will be

1

25 V - m

2

50 V - m

3

100 V - m

4

200 V - m

PHXII01:ELECTRIC CHARGES AND FIELDS

358323 Gauss’s law should be invalid if

1 The velocity of light were not a universal constant

2 The inverse square law were not exactly true

3 There were magnetic monopoles

4 None of these

PHXII01:ELECTRIC CHARGES AND FIELDS

358324 A point charge $q$ is placed at the centre of the cubical box. If $q = 12000 ϵ_{0} C$ , find the flux through shaded area of surface is
supporting img

1

150 N - m^{2} / C

2

200 N - m^{2} / C

3

500 N - m^{2} / C

4

650 N - m^{2} / C

Explanation:

Total flux coming out of the cubical box is given as
$ϕ_{0} = \frac{q}{ϵ_{0}}$
This total flux is emerging equally from each of the six face of the box.
Thus flux through each face can be given as
$ϕ = \frac{q}{6 ϵ_{0}}$
The flux through shaded portion would be one fourth of the flux through each face as through all the four triangular portions flux would be same due to symmetry, thus we have
$ϕ^{'} = \frac{ϕ}{4} = \frac{q}{24 \in_{0}} = \frac{12000 \in_{0}}{24 \in_{0}} = 500 N - m^{2} / C$

PHXII01:ELECTRIC CHARGES AND FIELDS

358321 Two charges $Q, - 3 Q$ are is kept in the cavity inside an uncharged metal, as shown in the figure. Electric flux through the Gaussian surface $S$ inside the metal shown in figure is
supporting img

1

\frac{2 Q}{ε_{0}}

2 zero

3

\frac{- 2 Q}{ε_{0}}

4

\frac{3 Q}{2 ε_{0}}

Explanation:

Electric field inside the conducting medium is always zero under electrostatic condition. $E = 0 \Rightarrow$ Flux $= 0$ . Option (2) is correct.

PHXII01:ELECTRIC CHARGES AND FIELDS

358322 Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 1 $m$ surrounding the total charge is 100 $V$ - $m$ . The flux over the concentric sphere of radius 2 $m$ will be

1

25 V - m

2

50 V - m

3

100 V - m

4

200 V - m

PHXII01:ELECTRIC CHARGES AND FIELDS

358323 Gauss’s law should be invalid if

1 The velocity of light were not a universal constant

2 The inverse square law were not exactly true

3 There were magnetic monopoles

4 None of these

PHXII01:ELECTRIC CHARGES AND FIELDS

358324 A point charge $q$ is placed at the centre of the cubical box. If $q = 12000 ϵ_{0} C$ , find the flux through shaded area of surface is
supporting img

1

150 N - m^{2} / C

2

200 N - m^{2} / C

3

500 N - m^{2} / C

4

650 N - m^{2} / C

Explanation:

Total flux coming out of the cubical box is given as
$ϕ_{0} = \frac{q}{ϵ_{0}}$
This total flux is emerging equally from each of the six face of the box.
Thus flux through each face can be given as
$ϕ = \frac{q}{6 ϵ_{0}}$
The flux through shaded portion would be one fourth of the flux through each face as through all the four triangular portions flux would be same due to symmetry, thus we have
$ϕ^{'} = \frac{ϕ}{4} = \frac{q}{24 \in_{0}} = \frac{12000 \in_{0}}{24 \in_{0}} = 500 N - m^{2} / C$

PHXII01:ELECTRIC CHARGES AND FIELDS

358321 Two charges $Q, - 3 Q$ are is kept in the cavity inside an uncharged metal, as shown in the figure. Electric flux through the Gaussian surface $S$ inside the metal shown in figure is
supporting img

1

\frac{2 Q}{ε_{0}}

2 zero

3

\frac{- 2 Q}{ε_{0}}

4

\frac{3 Q}{2 ε_{0}}

Explanation:

Electric field inside the conducting medium is always zero under electrostatic condition. $E = 0 \Rightarrow$ Flux $= 0$ . Option (2) is correct.

PHXII01:ELECTRIC CHARGES AND FIELDS

358322 Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 1 $m$ surrounding the total charge is 100 $V$ - $m$ . The flux over the concentric sphere of radius 2 $m$ will be

1

25 V - m

2

50 V - m

3

100 V - m

4

200 V - m

PHXII01:ELECTRIC CHARGES AND FIELDS

358323 Gauss’s law should be invalid if

1 The velocity of light were not a universal constant

2 The inverse square law were not exactly true

3 There were magnetic monopoles

4 None of these

PHXII01:ELECTRIC CHARGES AND FIELDS

358324 A point charge $q$ is placed at the centre of the cubical box. If $q = 12000 ϵ_{0} C$ , find the flux through shaded area of surface is
supporting img

1

150 N - m^{2} / C

2

200 N - m^{2} / C

3

500 N - m^{2} / C

4

650 N - m^{2} / C

Explanation:

Total flux coming out of the cubical box is given as
$ϕ_{0} = \frac{q}{ϵ_{0}}$
This total flux is emerging equally from each of the six face of the box.
Thus flux through each face can be given as
$ϕ = \frac{q}{6 ϵ_{0}}$
The flux through shaded portion would be one fourth of the flux through each face as through all the four triangular portions flux would be same due to symmetry, thus we have
$ϕ^{'} = \frac{ϕ}{4} = \frac{q}{24 \in_{0}} = \frac{12000 \in_{0}}{24 \in_{0}} = 500 N - m^{2} / C$