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
1 \({\dfrac{2 Q}{\varepsilon_{0}}}\)
2 zero
3 \({\dfrac{-2 Q}{\varepsilon_{0}}}\)
4 \({\dfrac{3 Q}{2 \varepsilon_{0}}}\)
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 \(200\,V - m\)
Explanation:
Flux does not depend on the size and shape of the close surface, and so, it remains same.
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
Explanation:
Conceptual Question
PHXII01:ELECTRIC CHARGES AND FIELDS
358324
A point charge \({q}\) is placed at the centre of the cubical box. If \({q=12000 \epsilon_{0} C}\), find the flux through shaded area of surface is
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 \(\phi_{0}=\dfrac{q}{\epsilon_{0}}\) This total flux is emerging equally from each of the six face of the box. Thus flux through each face can be given as \(\phi=\dfrac{q}{6 \epsilon_{0}}\) 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 \({\phi ^\prime } = \frac{\phi }{4} = \frac{q}{{24{ \in _0}}} = \frac{{12000{ \in _0}}}{{24{ \in _0}}} = 500\;N - {m^2}/C\)
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
1 \({\dfrac{2 Q}{\varepsilon_{0}}}\)
2 zero
3 \({\dfrac{-2 Q}{\varepsilon_{0}}}\)
4 \({\dfrac{3 Q}{2 \varepsilon_{0}}}\)
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 \(200\,V - m\)
Explanation:
Flux does not depend on the size and shape of the close surface, and so, it remains same.
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
Explanation:
Conceptual Question
PHXII01:ELECTRIC CHARGES AND FIELDS
358324
A point charge \({q}\) is placed at the centre of the cubical box. If \({q=12000 \epsilon_{0} C}\), find the flux through shaded area of surface is
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 \(\phi_{0}=\dfrac{q}{\epsilon_{0}}\) This total flux is emerging equally from each of the six face of the box. Thus flux through each face can be given as \(\phi=\dfrac{q}{6 \epsilon_{0}}\) 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 \({\phi ^\prime } = \frac{\phi }{4} = \frac{q}{{24{ \in _0}}} = \frac{{12000{ \in _0}}}{{24{ \in _0}}} = 500\;N - {m^2}/C\)
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
1 \({\dfrac{2 Q}{\varepsilon_{0}}}\)
2 zero
3 \({\dfrac{-2 Q}{\varepsilon_{0}}}\)
4 \({\dfrac{3 Q}{2 \varepsilon_{0}}}\)
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 \(200\,V - m\)
Explanation:
Flux does not depend on the size and shape of the close surface, and so, it remains same.
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
Explanation:
Conceptual Question
PHXII01:ELECTRIC CHARGES AND FIELDS
358324
A point charge \({q}\) is placed at the centre of the cubical box. If \({q=12000 \epsilon_{0} C}\), find the flux through shaded area of surface is
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 \(\phi_{0}=\dfrac{q}{\epsilon_{0}}\) This total flux is emerging equally from each of the six face of the box. Thus flux through each face can be given as \(\phi=\dfrac{q}{6 \epsilon_{0}}\) 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 \({\phi ^\prime } = \frac{\phi }{4} = \frac{q}{{24{ \in _0}}} = \frac{{12000{ \in _0}}}{{24{ \in _0}}} = 500\;N - {m^2}/C\)
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
1 \({\dfrac{2 Q}{\varepsilon_{0}}}\)
2 zero
3 \({\dfrac{-2 Q}{\varepsilon_{0}}}\)
4 \({\dfrac{3 Q}{2 \varepsilon_{0}}}\)
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 \(200\,V - m\)
Explanation:
Flux does not depend on the size and shape of the close surface, and so, it remains same.
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
Explanation:
Conceptual Question
PHXII01:ELECTRIC CHARGES AND FIELDS
358324
A point charge \({q}\) is placed at the centre of the cubical box. If \({q=12000 \epsilon_{0} C}\), find the flux through shaded area of surface is
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 \(\phi_{0}=\dfrac{q}{\epsilon_{0}}\) This total flux is emerging equally from each of the six face of the box. Thus flux through each face can be given as \(\phi=\dfrac{q}{6 \epsilon_{0}}\) 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 \({\phi ^\prime } = \frac{\phi }{4} = \frac{q}{{24{ \in _0}}} = \frac{{12000{ \in _0}}}{{24{ \in _0}}} = 500\;N - {m^2}/C\)
