358304
A point charge \(Q\) is located just above the centre of the flat face of hemisphere as shown in figure. The electric flux through the flat face and curved face of hemisphere are respectively
Place another identical hemisphere on given one so that flat faces of two coincide. Then total flux linked with cross-section of bottom half (lower hemisphere) is \(\frac{Q}{{2{\varepsilon _o}}}\),as charge enclosed by given hemisphere is zero, so flux linked with flat face of hemisphere is \(\frac{{ - Q}}{{2{\varepsilon _o}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358305
In a region, the intensity of an electric field is given by \(E=8 \hat{i}+8 \hat{j}+\hat{k}\) in \(N C^{-1}\). The electric flux through a surface \(S = 10\hat i\;{m^2}\) in the region is
1 \(10\,N{m^2}\,{C^{ - 1}}\)
2 \(80\,N{m^2}\,{C^{ - 1}}\)
3 \(8\,N{m^2}\,{C^{ - 1}}\)
4 None
Explanation:
\(\phi = \vec E \cdot \vec A = (8\hat i + 8\hat j + \hat k) \cdot (10\hat i) = 80\,N{m^2}{C^{ - 1}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358306
A charge \( + q\) is at a distance \(L/2\) above a square of side \(L\). Then what is the flux linked with the surface?
1 \(\frac{q}{{4{\varepsilon _0}}}\)
2 \(\frac{{2q}}{{3{\varepsilon _0}}}\)
3 \(\frac{q}{{6{\varepsilon _0}}}\)
4 \(\frac{{6q}}{{{\varepsilon _0}}}\)
Explanation:
The given square of side \(L\) may be considered as one of the faces of a cube with edge \(L\). Then given charge \(q\) will be considered to be placed at the centre of the cube. Then according to Gauss's theorem, the magnitude of the electric flux through the faces (six) of the cube is given by \(\phi=\mathrm{q} / \varepsilon_{0}\) Hence, electric flux through one face of the cube for the given square will be \({\phi ^\prime } = \frac{1}{6}\phi = \frac{q}{{6{\varepsilon _0}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358307
Electric flux through the Gaussian surface \({S}\), as shown in the figure, is \({\dfrac{q_{1}+q_{2}}{\varepsilon_{0}}}\). Electric field E on the Gaussian surface is due to
1 \({q_{1}}\) and \({q_{2}}\) only
2 \({q_{3}, q_{4}, q_{5}}\) and \({q_{6}}\) only
3 \({q_{1}, q_{3}}\) and \({q_{5}}\) only
4 all charges
Explanation:
The Field at any point on the surface of the conductor is the resultant of fields produced by all the charges. Correct option is (4).
358304
A point charge \(Q\) is located just above the centre of the flat face of hemisphere as shown in figure. The electric flux through the flat face and curved face of hemisphere are respectively
Place another identical hemisphere on given one so that flat faces of two coincide. Then total flux linked with cross-section of bottom half (lower hemisphere) is \(\frac{Q}{{2{\varepsilon _o}}}\),as charge enclosed by given hemisphere is zero, so flux linked with flat face of hemisphere is \(\frac{{ - Q}}{{2{\varepsilon _o}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358305
In a region, the intensity of an electric field is given by \(E=8 \hat{i}+8 \hat{j}+\hat{k}\) in \(N C^{-1}\). The electric flux through a surface \(S = 10\hat i\;{m^2}\) in the region is
1 \(10\,N{m^2}\,{C^{ - 1}}\)
2 \(80\,N{m^2}\,{C^{ - 1}}\)
3 \(8\,N{m^2}\,{C^{ - 1}}\)
4 None
Explanation:
\(\phi = \vec E \cdot \vec A = (8\hat i + 8\hat j + \hat k) \cdot (10\hat i) = 80\,N{m^2}{C^{ - 1}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358306
A charge \( + q\) is at a distance \(L/2\) above a square of side \(L\). Then what is the flux linked with the surface?
1 \(\frac{q}{{4{\varepsilon _0}}}\)
2 \(\frac{{2q}}{{3{\varepsilon _0}}}\)
3 \(\frac{q}{{6{\varepsilon _0}}}\)
4 \(\frac{{6q}}{{{\varepsilon _0}}}\)
Explanation:
The given square of side \(L\) may be considered as one of the faces of a cube with edge \(L\). Then given charge \(q\) will be considered to be placed at the centre of the cube. Then according to Gauss's theorem, the magnitude of the electric flux through the faces (six) of the cube is given by \(\phi=\mathrm{q} / \varepsilon_{0}\) Hence, electric flux through one face of the cube for the given square will be \({\phi ^\prime } = \frac{1}{6}\phi = \frac{q}{{6{\varepsilon _0}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358307
Electric flux through the Gaussian surface \({S}\), as shown in the figure, is \({\dfrac{q_{1}+q_{2}}{\varepsilon_{0}}}\). Electric field E on the Gaussian surface is due to
1 \({q_{1}}\) and \({q_{2}}\) only
2 \({q_{3}, q_{4}, q_{5}}\) and \({q_{6}}\) only
3 \({q_{1}, q_{3}}\) and \({q_{5}}\) only
4 all charges
Explanation:
The Field at any point on the surface of the conductor is the resultant of fields produced by all the charges. Correct option is (4).
358304
A point charge \(Q\) is located just above the centre of the flat face of hemisphere as shown in figure. The electric flux through the flat face and curved face of hemisphere are respectively
Place another identical hemisphere on given one so that flat faces of two coincide. Then total flux linked with cross-section of bottom half (lower hemisphere) is \(\frac{Q}{{2{\varepsilon _o}}}\),as charge enclosed by given hemisphere is zero, so flux linked with flat face of hemisphere is \(\frac{{ - Q}}{{2{\varepsilon _o}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358305
In a region, the intensity of an electric field is given by \(E=8 \hat{i}+8 \hat{j}+\hat{k}\) in \(N C^{-1}\). The electric flux through a surface \(S = 10\hat i\;{m^2}\) in the region is
1 \(10\,N{m^2}\,{C^{ - 1}}\)
2 \(80\,N{m^2}\,{C^{ - 1}}\)
3 \(8\,N{m^2}\,{C^{ - 1}}\)
4 None
Explanation:
\(\phi = \vec E \cdot \vec A = (8\hat i + 8\hat j + \hat k) \cdot (10\hat i) = 80\,N{m^2}{C^{ - 1}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358306
A charge \( + q\) is at a distance \(L/2\) above a square of side \(L\). Then what is the flux linked with the surface?
1 \(\frac{q}{{4{\varepsilon _0}}}\)
2 \(\frac{{2q}}{{3{\varepsilon _0}}}\)
3 \(\frac{q}{{6{\varepsilon _0}}}\)
4 \(\frac{{6q}}{{{\varepsilon _0}}}\)
Explanation:
The given square of side \(L\) may be considered as one of the faces of a cube with edge \(L\). Then given charge \(q\) will be considered to be placed at the centre of the cube. Then according to Gauss's theorem, the magnitude of the electric flux through the faces (six) of the cube is given by \(\phi=\mathrm{q} / \varepsilon_{0}\) Hence, electric flux through one face of the cube for the given square will be \({\phi ^\prime } = \frac{1}{6}\phi = \frac{q}{{6{\varepsilon _0}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358307
Electric flux through the Gaussian surface \({S}\), as shown in the figure, is \({\dfrac{q_{1}+q_{2}}{\varepsilon_{0}}}\). Electric field E on the Gaussian surface is due to
1 \({q_{1}}\) and \({q_{2}}\) only
2 \({q_{3}, q_{4}, q_{5}}\) and \({q_{6}}\) only
3 \({q_{1}, q_{3}}\) and \({q_{5}}\) only
4 all charges
Explanation:
The Field at any point on the surface of the conductor is the resultant of fields produced by all the charges. Correct option is (4).
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII01:ELECTRIC CHARGES AND FIELDS
358304
A point charge \(Q\) is located just above the centre of the flat face of hemisphere as shown in figure. The electric flux through the flat face and curved face of hemisphere are respectively
Place another identical hemisphere on given one so that flat faces of two coincide. Then total flux linked with cross-section of bottom half (lower hemisphere) is \(\frac{Q}{{2{\varepsilon _o}}}\),as charge enclosed by given hemisphere is zero, so flux linked with flat face of hemisphere is \(\frac{{ - Q}}{{2{\varepsilon _o}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358305
In a region, the intensity of an electric field is given by \(E=8 \hat{i}+8 \hat{j}+\hat{k}\) in \(N C^{-1}\). The electric flux through a surface \(S = 10\hat i\;{m^2}\) in the region is
1 \(10\,N{m^2}\,{C^{ - 1}}\)
2 \(80\,N{m^2}\,{C^{ - 1}}\)
3 \(8\,N{m^2}\,{C^{ - 1}}\)
4 None
Explanation:
\(\phi = \vec E \cdot \vec A = (8\hat i + 8\hat j + \hat k) \cdot (10\hat i) = 80\,N{m^2}{C^{ - 1}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358306
A charge \( + q\) is at a distance \(L/2\) above a square of side \(L\). Then what is the flux linked with the surface?
1 \(\frac{q}{{4{\varepsilon _0}}}\)
2 \(\frac{{2q}}{{3{\varepsilon _0}}}\)
3 \(\frac{q}{{6{\varepsilon _0}}}\)
4 \(\frac{{6q}}{{{\varepsilon _0}}}\)
Explanation:
The given square of side \(L\) may be considered as one of the faces of a cube with edge \(L\). Then given charge \(q\) will be considered to be placed at the centre of the cube. Then according to Gauss's theorem, the magnitude of the electric flux through the faces (six) of the cube is given by \(\phi=\mathrm{q} / \varepsilon_{0}\) Hence, electric flux through one face of the cube for the given square will be \({\phi ^\prime } = \frac{1}{6}\phi = \frac{q}{{6{\varepsilon _0}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358307
Electric flux through the Gaussian surface \({S}\), as shown in the figure, is \({\dfrac{q_{1}+q_{2}}{\varepsilon_{0}}}\). Electric field E on the Gaussian surface is due to
1 \({q_{1}}\) and \({q_{2}}\) only
2 \({q_{3}, q_{4}, q_{5}}\) and \({q_{6}}\) only
3 \({q_{1}, q_{3}}\) and \({q_{5}}\) only
4 all charges
Explanation:
The Field at any point on the surface of the conductor is the resultant of fields produced by all the charges. Correct option is (4).
