358299
Five charges \(+q,+5 q,-2 q,+3 q\) and \(-4 q\) are situated as shown in the figure. The electric flux due to this configuration through the surface \(S\) is
1 \(\dfrac{3 q}{\varepsilon_{0}}\)
2 \(\dfrac{q}{\varepsilon_{0}}\)
3 \(\dfrac{5 q}{\varepsilon_{0}}\)
4 \(\dfrac{4 q}{\varepsilon_{0}}\)
Explanation:
According to the Gauss's theorem, net flux is only due to the enclosed charge, So, \(\phi=\dfrac{q_{\text {in }}}{\varepsilon_{0}}\) \(\therefore \phi=\dfrac{+5 q+q-2 q}{\varepsilon_{0}}=\dfrac{4 q}{\varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358300
The total electric flux through a closed spherical surface of radius ' \({r}\) ' enclosing an electric dipole of dipole moment \({2 a q}\) is (Give \({\varepsilon_{0}=}\) permittivity of free space)
1 Zero
2 \({\dfrac{q}{\varepsilon_{0}}}\)
3 \({\dfrac{2 q}{\varepsilon_{0}}}\)
4 \({\dfrac{8 \pi r^{2} q}{\varepsilon_{0}}}\)
Explanation:
Using Gauss's law \({\phi=\oint \vec{E} \cdot d \vec{A}=\dfrac{q_{\text {in }}}{\varepsilon_{0}}}\) As net charge is zero, so \({\phi=}\) zero .
KCET - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358301
Charge is distributed uniformly in some space. The net flux passing through the surface of an imaginary cube of side \(a\) in the space is \(\phi \). The net flux passing through the surface of an imaginary sphere of radius \(a\) in the space will be
1 \(\phi \)
2 \(\frac{3}{{4\pi }}\phi \)
3 \(\frac{{2\pi }}{3}\phi \)
4 \(\frac{{4\pi }}{3}\phi \)
Explanation:
Flux through the cube is \({\phi _1} = \frac{{\rho \times {a^3}}}{{{ \in _0}}} = \phi \) Flux through the sphere is \({\phi _2} = \rho \times \frac{{4\pi {a^3}}}{{3{ \in _0}}}\) \({\phi _2} = \phi \times \frac{4}{3}\pi \)
PHXII01:ELECTRIC CHARGES AND FIELDS
358302
A charge \(q\) is placed at the centre of one of the surface of a cube. The flux linked with the cube is
1 \(\dfrac{q}{8 \varepsilon_{0}}\)
2 \(\dfrac{q}{4 \varepsilon_{0}}\)
3 \(\dfrac{q}{2 \varepsilon_{0}}\)
4 zero
Explanation:
As the charge is at centre of one of the surface, so, for enclosed charge one more cube is to be consider. So electric flux for each surface, \(\phi=\dfrac{q}{\varepsilon_{0}}\)
Flux linked with cube, \(\phi^{\prime}=\dfrac{q}{2 \varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358303
Assertion : Gaussian surface is considered carefully. Reason : The point where electric field to be calculated should be within the surface.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Gaussian surface should be chosen such that it encloses charge and easy to calculate over the surface taking symmetry into account. So correct option is (1).
358299
Five charges \(+q,+5 q,-2 q,+3 q\) and \(-4 q\) are situated as shown in the figure. The electric flux due to this configuration through the surface \(S\) is
1 \(\dfrac{3 q}{\varepsilon_{0}}\)
2 \(\dfrac{q}{\varepsilon_{0}}\)
3 \(\dfrac{5 q}{\varepsilon_{0}}\)
4 \(\dfrac{4 q}{\varepsilon_{0}}\)
Explanation:
According to the Gauss's theorem, net flux is only due to the enclosed charge, So, \(\phi=\dfrac{q_{\text {in }}}{\varepsilon_{0}}\) \(\therefore \phi=\dfrac{+5 q+q-2 q}{\varepsilon_{0}}=\dfrac{4 q}{\varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358300
The total electric flux through a closed spherical surface of radius ' \({r}\) ' enclosing an electric dipole of dipole moment \({2 a q}\) is (Give \({\varepsilon_{0}=}\) permittivity of free space)
1 Zero
2 \({\dfrac{q}{\varepsilon_{0}}}\)
3 \({\dfrac{2 q}{\varepsilon_{0}}}\)
4 \({\dfrac{8 \pi r^{2} q}{\varepsilon_{0}}}\)
Explanation:
Using Gauss's law \({\phi=\oint \vec{E} \cdot d \vec{A}=\dfrac{q_{\text {in }}}{\varepsilon_{0}}}\) As net charge is zero, so \({\phi=}\) zero .
KCET - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358301
Charge is distributed uniformly in some space. The net flux passing through the surface of an imaginary cube of side \(a\) in the space is \(\phi \). The net flux passing through the surface of an imaginary sphere of radius \(a\) in the space will be
1 \(\phi \)
2 \(\frac{3}{{4\pi }}\phi \)
3 \(\frac{{2\pi }}{3}\phi \)
4 \(\frac{{4\pi }}{3}\phi \)
Explanation:
Flux through the cube is \({\phi _1} = \frac{{\rho \times {a^3}}}{{{ \in _0}}} = \phi \) Flux through the sphere is \({\phi _2} = \rho \times \frac{{4\pi {a^3}}}{{3{ \in _0}}}\) \({\phi _2} = \phi \times \frac{4}{3}\pi \)
PHXII01:ELECTRIC CHARGES AND FIELDS
358302
A charge \(q\) is placed at the centre of one of the surface of a cube. The flux linked with the cube is
1 \(\dfrac{q}{8 \varepsilon_{0}}\)
2 \(\dfrac{q}{4 \varepsilon_{0}}\)
3 \(\dfrac{q}{2 \varepsilon_{0}}\)
4 zero
Explanation:
As the charge is at centre of one of the surface, so, for enclosed charge one more cube is to be consider. So electric flux for each surface, \(\phi=\dfrac{q}{\varepsilon_{0}}\)
Flux linked with cube, \(\phi^{\prime}=\dfrac{q}{2 \varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358303
Assertion : Gaussian surface is considered carefully. Reason : The point where electric field to be calculated should be within the surface.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Gaussian surface should be chosen such that it encloses charge and easy to calculate over the surface taking symmetry into account. So correct option is (1).
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII01:ELECTRIC CHARGES AND FIELDS
358299
Five charges \(+q,+5 q,-2 q,+3 q\) and \(-4 q\) are situated as shown in the figure. The electric flux due to this configuration through the surface \(S\) is
1 \(\dfrac{3 q}{\varepsilon_{0}}\)
2 \(\dfrac{q}{\varepsilon_{0}}\)
3 \(\dfrac{5 q}{\varepsilon_{0}}\)
4 \(\dfrac{4 q}{\varepsilon_{0}}\)
Explanation:
According to the Gauss's theorem, net flux is only due to the enclosed charge, So, \(\phi=\dfrac{q_{\text {in }}}{\varepsilon_{0}}\) \(\therefore \phi=\dfrac{+5 q+q-2 q}{\varepsilon_{0}}=\dfrac{4 q}{\varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358300
The total electric flux through a closed spherical surface of radius ' \({r}\) ' enclosing an electric dipole of dipole moment \({2 a q}\) is (Give \({\varepsilon_{0}=}\) permittivity of free space)
1 Zero
2 \({\dfrac{q}{\varepsilon_{0}}}\)
3 \({\dfrac{2 q}{\varepsilon_{0}}}\)
4 \({\dfrac{8 \pi r^{2} q}{\varepsilon_{0}}}\)
Explanation:
Using Gauss's law \({\phi=\oint \vec{E} \cdot d \vec{A}=\dfrac{q_{\text {in }}}{\varepsilon_{0}}}\) As net charge is zero, so \({\phi=}\) zero .
KCET - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358301
Charge is distributed uniformly in some space. The net flux passing through the surface of an imaginary cube of side \(a\) in the space is \(\phi \). The net flux passing through the surface of an imaginary sphere of radius \(a\) in the space will be
1 \(\phi \)
2 \(\frac{3}{{4\pi }}\phi \)
3 \(\frac{{2\pi }}{3}\phi \)
4 \(\frac{{4\pi }}{3}\phi \)
Explanation:
Flux through the cube is \({\phi _1} = \frac{{\rho \times {a^3}}}{{{ \in _0}}} = \phi \) Flux through the sphere is \({\phi _2} = \rho \times \frac{{4\pi {a^3}}}{{3{ \in _0}}}\) \({\phi _2} = \phi \times \frac{4}{3}\pi \)
PHXII01:ELECTRIC CHARGES AND FIELDS
358302
A charge \(q\) is placed at the centre of one of the surface of a cube. The flux linked with the cube is
1 \(\dfrac{q}{8 \varepsilon_{0}}\)
2 \(\dfrac{q}{4 \varepsilon_{0}}\)
3 \(\dfrac{q}{2 \varepsilon_{0}}\)
4 zero
Explanation:
As the charge is at centre of one of the surface, so, for enclosed charge one more cube is to be consider. So electric flux for each surface, \(\phi=\dfrac{q}{\varepsilon_{0}}\)
Flux linked with cube, \(\phi^{\prime}=\dfrac{q}{2 \varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358303
Assertion : Gaussian surface is considered carefully. Reason : The point where electric field to be calculated should be within the surface.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Gaussian surface should be chosen such that it encloses charge and easy to calculate over the surface taking symmetry into account. So correct option is (1).
358299
Five charges \(+q,+5 q,-2 q,+3 q\) and \(-4 q\) are situated as shown in the figure. The electric flux due to this configuration through the surface \(S\) is
1 \(\dfrac{3 q}{\varepsilon_{0}}\)
2 \(\dfrac{q}{\varepsilon_{0}}\)
3 \(\dfrac{5 q}{\varepsilon_{0}}\)
4 \(\dfrac{4 q}{\varepsilon_{0}}\)
Explanation:
According to the Gauss's theorem, net flux is only due to the enclosed charge, So, \(\phi=\dfrac{q_{\text {in }}}{\varepsilon_{0}}\) \(\therefore \phi=\dfrac{+5 q+q-2 q}{\varepsilon_{0}}=\dfrac{4 q}{\varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358300
The total electric flux through a closed spherical surface of radius ' \({r}\) ' enclosing an electric dipole of dipole moment \({2 a q}\) is (Give \({\varepsilon_{0}=}\) permittivity of free space)
1 Zero
2 \({\dfrac{q}{\varepsilon_{0}}}\)
3 \({\dfrac{2 q}{\varepsilon_{0}}}\)
4 \({\dfrac{8 \pi r^{2} q}{\varepsilon_{0}}}\)
Explanation:
Using Gauss's law \({\phi=\oint \vec{E} \cdot d \vec{A}=\dfrac{q_{\text {in }}}{\varepsilon_{0}}}\) As net charge is zero, so \({\phi=}\) zero .
KCET - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358301
Charge is distributed uniformly in some space. The net flux passing through the surface of an imaginary cube of side \(a\) in the space is \(\phi \). The net flux passing through the surface of an imaginary sphere of radius \(a\) in the space will be
1 \(\phi \)
2 \(\frac{3}{{4\pi }}\phi \)
3 \(\frac{{2\pi }}{3}\phi \)
4 \(\frac{{4\pi }}{3}\phi \)
Explanation:
Flux through the cube is \({\phi _1} = \frac{{\rho \times {a^3}}}{{{ \in _0}}} = \phi \) Flux through the sphere is \({\phi _2} = \rho \times \frac{{4\pi {a^3}}}{{3{ \in _0}}}\) \({\phi _2} = \phi \times \frac{4}{3}\pi \)
PHXII01:ELECTRIC CHARGES AND FIELDS
358302
A charge \(q\) is placed at the centre of one of the surface of a cube. The flux linked with the cube is
1 \(\dfrac{q}{8 \varepsilon_{0}}\)
2 \(\dfrac{q}{4 \varepsilon_{0}}\)
3 \(\dfrac{q}{2 \varepsilon_{0}}\)
4 zero
Explanation:
As the charge is at centre of one of the surface, so, for enclosed charge one more cube is to be consider. So electric flux for each surface, \(\phi=\dfrac{q}{\varepsilon_{0}}\)
Flux linked with cube, \(\phi^{\prime}=\dfrac{q}{2 \varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358303
Assertion : Gaussian surface is considered carefully. Reason : The point where electric field to be calculated should be within the surface.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Gaussian surface should be chosen such that it encloses charge and easy to calculate over the surface taking symmetry into account. So correct option is (1).
358299
Five charges \(+q,+5 q,-2 q,+3 q\) and \(-4 q\) are situated as shown in the figure. The electric flux due to this configuration through the surface \(S\) is
1 \(\dfrac{3 q}{\varepsilon_{0}}\)
2 \(\dfrac{q}{\varepsilon_{0}}\)
3 \(\dfrac{5 q}{\varepsilon_{0}}\)
4 \(\dfrac{4 q}{\varepsilon_{0}}\)
Explanation:
According to the Gauss's theorem, net flux is only due to the enclosed charge, So, \(\phi=\dfrac{q_{\text {in }}}{\varepsilon_{0}}\) \(\therefore \phi=\dfrac{+5 q+q-2 q}{\varepsilon_{0}}=\dfrac{4 q}{\varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358300
The total electric flux through a closed spherical surface of radius ' \({r}\) ' enclosing an electric dipole of dipole moment \({2 a q}\) is (Give \({\varepsilon_{0}=}\) permittivity of free space)
1 Zero
2 \({\dfrac{q}{\varepsilon_{0}}}\)
3 \({\dfrac{2 q}{\varepsilon_{0}}}\)
4 \({\dfrac{8 \pi r^{2} q}{\varepsilon_{0}}}\)
Explanation:
Using Gauss's law \({\phi=\oint \vec{E} \cdot d \vec{A}=\dfrac{q_{\text {in }}}{\varepsilon_{0}}}\) As net charge is zero, so \({\phi=}\) zero .
KCET - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358301
Charge is distributed uniformly in some space. The net flux passing through the surface of an imaginary cube of side \(a\) in the space is \(\phi \). The net flux passing through the surface of an imaginary sphere of radius \(a\) in the space will be
1 \(\phi \)
2 \(\frac{3}{{4\pi }}\phi \)
3 \(\frac{{2\pi }}{3}\phi \)
4 \(\frac{{4\pi }}{3}\phi \)
Explanation:
Flux through the cube is \({\phi _1} = \frac{{\rho \times {a^3}}}{{{ \in _0}}} = \phi \) Flux through the sphere is \({\phi _2} = \rho \times \frac{{4\pi {a^3}}}{{3{ \in _0}}}\) \({\phi _2} = \phi \times \frac{4}{3}\pi \)
PHXII01:ELECTRIC CHARGES AND FIELDS
358302
A charge \(q\) is placed at the centre of one of the surface of a cube. The flux linked with the cube is
1 \(\dfrac{q}{8 \varepsilon_{0}}\)
2 \(\dfrac{q}{4 \varepsilon_{0}}\)
3 \(\dfrac{q}{2 \varepsilon_{0}}\)
4 zero
Explanation:
As the charge is at centre of one of the surface, so, for enclosed charge one more cube is to be consider. So electric flux for each surface, \(\phi=\dfrac{q}{\varepsilon_{0}}\)
Flux linked with cube, \(\phi^{\prime}=\dfrac{q}{2 \varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358303
Assertion : Gaussian surface is considered carefully. Reason : The point where electric field to be calculated should be within the surface.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Gaussian surface should be chosen such that it encloses charge and easy to calculate over the surface taking symmetry into account. So correct option is (1).
