358334
In a cuboid of dimension \(2 L \times 2 L \times L\), a charge \(q\) is placed at the center of the surface '\(S\)' having area of \(4 L^{2}\). The flux through the opposite surface to '\(S\)' is given by
1 \(\dfrac{q}{12 \varepsilon_{0}}\)
2 \(\dfrac{q}{2 \varepsilon_{0}}\)
3 \(\dfrac{q}{3 \varepsilon_{0}}\)
4 \(\dfrac{q}{6 \varepsilon_{0}}\)
Explanation:
\(P\) is the centre of the larger cuboid of size \(2 L \times 2 L \times 2 L\, P\) is also the centre of the surface \(S\). By Gauss's law in electrostatics \(\phi=\dfrac{q}{\varepsilon_{0}}\) (for larger cuboid) For opposite surface of \(S\) \(\phi=\dfrac{q}{6 \varepsilon_{0}}\)
Jee - 2023
PHXII01:ELECTRIC CHARGES AND FIELDS
358335
\(C_{1}\) and \(C_{2}\) are two hollow concentric cubes enclosing charges \(2\,Q\) and \(3\,Q\) respectively as shown in figure. The ratio of electric flux passing through \(C_{1}\) and \(C_{2}\) is
1 \(3: 2\)
2 \(5: 2\)
3 \(2: 5\)
4 \(2: 3\)
Explanation:
Flux passing through \(C_{1}, \phi_{1}=\dfrac{2 Q}{\varepsilon_{0}}\)
Flux passing through \(C_{2}, \phi_{2}=\dfrac{3 Q+2 Q}{\varepsilon_{0}}=\dfrac{5 Q}{\varepsilon_{0}}\) So, ratio, \(\phi_{1}: \phi_{2}=\dfrac{2 Q / \varepsilon_{0}}{5 Q / \varepsilon_{0}}=\dfrac{2}{5}=2: 5\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358336
Assertion : If Gaussian surface does not enclose any charge, then flux through the surface is zero. Reason : If Gaussian surface does not enclose any charge, then \(E\) at any point on the Gaussian surface must be zero.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(E\) at any point on Gaussian surface may be due to outside charges also. If Gaussian surface does not enclose any charge then flux through the surface is zero. Option (3) is correct.
PHXII01:ELECTRIC CHARGES AND FIELDS
358337
What is the nature of Gaussian surface involved in Gauss law of electrostatics ?
358334
In a cuboid of dimension \(2 L \times 2 L \times L\), a charge \(q\) is placed at the center of the surface '\(S\)' having area of \(4 L^{2}\). The flux through the opposite surface to '\(S\)' is given by
1 \(\dfrac{q}{12 \varepsilon_{0}}\)
2 \(\dfrac{q}{2 \varepsilon_{0}}\)
3 \(\dfrac{q}{3 \varepsilon_{0}}\)
4 \(\dfrac{q}{6 \varepsilon_{0}}\)
Explanation:
\(P\) is the centre of the larger cuboid of size \(2 L \times 2 L \times 2 L\, P\) is also the centre of the surface \(S\). By Gauss's law in electrostatics \(\phi=\dfrac{q}{\varepsilon_{0}}\) (for larger cuboid) For opposite surface of \(S\) \(\phi=\dfrac{q}{6 \varepsilon_{0}}\)
Jee - 2023
PHXII01:ELECTRIC CHARGES AND FIELDS
358335
\(C_{1}\) and \(C_{2}\) are two hollow concentric cubes enclosing charges \(2\,Q\) and \(3\,Q\) respectively as shown in figure. The ratio of electric flux passing through \(C_{1}\) and \(C_{2}\) is
1 \(3: 2\)
2 \(5: 2\)
3 \(2: 5\)
4 \(2: 3\)
Explanation:
Flux passing through \(C_{1}, \phi_{1}=\dfrac{2 Q}{\varepsilon_{0}}\)
Flux passing through \(C_{2}, \phi_{2}=\dfrac{3 Q+2 Q}{\varepsilon_{0}}=\dfrac{5 Q}{\varepsilon_{0}}\) So, ratio, \(\phi_{1}: \phi_{2}=\dfrac{2 Q / \varepsilon_{0}}{5 Q / \varepsilon_{0}}=\dfrac{2}{5}=2: 5\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358336
Assertion : If Gaussian surface does not enclose any charge, then flux through the surface is zero. Reason : If Gaussian surface does not enclose any charge, then \(E\) at any point on the Gaussian surface must be zero.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(E\) at any point on Gaussian surface may be due to outside charges also. If Gaussian surface does not enclose any charge then flux through the surface is zero. Option (3) is correct.
PHXII01:ELECTRIC CHARGES AND FIELDS
358337
What is the nature of Gaussian surface involved in Gauss law of electrostatics ?
358334
In a cuboid of dimension \(2 L \times 2 L \times L\), a charge \(q\) is placed at the center of the surface '\(S\)' having area of \(4 L^{2}\). The flux through the opposite surface to '\(S\)' is given by
1 \(\dfrac{q}{12 \varepsilon_{0}}\)
2 \(\dfrac{q}{2 \varepsilon_{0}}\)
3 \(\dfrac{q}{3 \varepsilon_{0}}\)
4 \(\dfrac{q}{6 \varepsilon_{0}}\)
Explanation:
\(P\) is the centre of the larger cuboid of size \(2 L \times 2 L \times 2 L\, P\) is also the centre of the surface \(S\). By Gauss's law in electrostatics \(\phi=\dfrac{q}{\varepsilon_{0}}\) (for larger cuboid) For opposite surface of \(S\) \(\phi=\dfrac{q}{6 \varepsilon_{0}}\)
Jee - 2023
PHXII01:ELECTRIC CHARGES AND FIELDS
358335
\(C_{1}\) and \(C_{2}\) are two hollow concentric cubes enclosing charges \(2\,Q\) and \(3\,Q\) respectively as shown in figure. The ratio of electric flux passing through \(C_{1}\) and \(C_{2}\) is
1 \(3: 2\)
2 \(5: 2\)
3 \(2: 5\)
4 \(2: 3\)
Explanation:
Flux passing through \(C_{1}, \phi_{1}=\dfrac{2 Q}{\varepsilon_{0}}\)
Flux passing through \(C_{2}, \phi_{2}=\dfrac{3 Q+2 Q}{\varepsilon_{0}}=\dfrac{5 Q}{\varepsilon_{0}}\) So, ratio, \(\phi_{1}: \phi_{2}=\dfrac{2 Q / \varepsilon_{0}}{5 Q / \varepsilon_{0}}=\dfrac{2}{5}=2: 5\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358336
Assertion : If Gaussian surface does not enclose any charge, then flux through the surface is zero. Reason : If Gaussian surface does not enclose any charge, then \(E\) at any point on the Gaussian surface must be zero.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(E\) at any point on Gaussian surface may be due to outside charges also. If Gaussian surface does not enclose any charge then flux through the surface is zero. Option (3) is correct.
PHXII01:ELECTRIC CHARGES AND FIELDS
358337
What is the nature of Gaussian surface involved in Gauss law of electrostatics ?
358334
In a cuboid of dimension \(2 L \times 2 L \times L\), a charge \(q\) is placed at the center of the surface '\(S\)' having area of \(4 L^{2}\). The flux through the opposite surface to '\(S\)' is given by
1 \(\dfrac{q}{12 \varepsilon_{0}}\)
2 \(\dfrac{q}{2 \varepsilon_{0}}\)
3 \(\dfrac{q}{3 \varepsilon_{0}}\)
4 \(\dfrac{q}{6 \varepsilon_{0}}\)
Explanation:
\(P\) is the centre of the larger cuboid of size \(2 L \times 2 L \times 2 L\, P\) is also the centre of the surface \(S\). By Gauss's law in electrostatics \(\phi=\dfrac{q}{\varepsilon_{0}}\) (for larger cuboid) For opposite surface of \(S\) \(\phi=\dfrac{q}{6 \varepsilon_{0}}\)
Jee - 2023
PHXII01:ELECTRIC CHARGES AND FIELDS
358335
\(C_{1}\) and \(C_{2}\) are two hollow concentric cubes enclosing charges \(2\,Q\) and \(3\,Q\) respectively as shown in figure. The ratio of electric flux passing through \(C_{1}\) and \(C_{2}\) is
1 \(3: 2\)
2 \(5: 2\)
3 \(2: 5\)
4 \(2: 3\)
Explanation:
Flux passing through \(C_{1}, \phi_{1}=\dfrac{2 Q}{\varepsilon_{0}}\)
Flux passing through \(C_{2}, \phi_{2}=\dfrac{3 Q+2 Q}{\varepsilon_{0}}=\dfrac{5 Q}{\varepsilon_{0}}\) So, ratio, \(\phi_{1}: \phi_{2}=\dfrac{2 Q / \varepsilon_{0}}{5 Q / \varepsilon_{0}}=\dfrac{2}{5}=2: 5\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358336
Assertion : If Gaussian surface does not enclose any charge, then flux through the surface is zero. Reason : If Gaussian surface does not enclose any charge, then \(E\) at any point on the Gaussian surface must be zero.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(E\) at any point on Gaussian surface may be due to outside charges also. If Gaussian surface does not enclose any charge then flux through the surface is zero. Option (3) is correct.
PHXII01:ELECTRIC CHARGES AND FIELDS
358337
What is the nature of Gaussian surface involved in Gauss law of electrostatics ?
