358312
Statement A : Total flux through a closed surface is zero if net charge enclosed by the surface is zero. Statement B : Gauss law is true for any closed surface, no matter what its shape or size is.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Gauss law implies that the total electric flux through a closed surface is zero if no net charge is enclosed by the surface and it is true for any closed surface, independent of its shape and size. So option (3) is correct.
PHXII01:ELECTRIC CHARGES AND FIELDS
358313
Consider a circle of radius \(R\). A point charge is present at a distance \(a\) from its centre and on its axis such that the \(R = a\sqrt 3 \). If electric flux passing through the circle is \(\phi \) then the magnitude of the point charge is
358314
Electric field in a region is given by \(\vec E = - 4x\hat i + 6y\hat j\). The charge enclosed in the cube of side \(1\,m\) oriented as shown in the diagram is given by \({\alpha \epsilon_{0}}\). Find the value of \({\alpha}\)
358315
The flux coming out from a unit positive charge placed in air is
1 \({\varepsilon_{0}}\)
2 \({\varepsilon_{0}^{-1}}\)
3 \({\left(4 \pi \varepsilon_{0}\right)^{-1}}\)
4 \({4 \pi \varepsilon_{0}}\)
Explanation:
Since by Gauss's law \(\phi_{T}=\dfrac{q}{\varepsilon_{0}} \quad \therefore \quad \phi_{T}=\dfrac{1 C}{\varepsilon_{0}}=\varepsilon_{0}^{-1}\). So correct option is (2)
358312
Statement A : Total flux through a closed surface is zero if net charge enclosed by the surface is zero. Statement B : Gauss law is true for any closed surface, no matter what its shape or size is.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Gauss law implies that the total electric flux through a closed surface is zero if no net charge is enclosed by the surface and it is true for any closed surface, independent of its shape and size. So option (3) is correct.
PHXII01:ELECTRIC CHARGES AND FIELDS
358313
Consider a circle of radius \(R\). A point charge is present at a distance \(a\) from its centre and on its axis such that the \(R = a\sqrt 3 \). If electric flux passing through the circle is \(\phi \) then the magnitude of the point charge is
358314
Electric field in a region is given by \(\vec E = - 4x\hat i + 6y\hat j\). The charge enclosed in the cube of side \(1\,m\) oriented as shown in the diagram is given by \({\alpha \epsilon_{0}}\). Find the value of \({\alpha}\)
358315
The flux coming out from a unit positive charge placed in air is
1 \({\varepsilon_{0}}\)
2 \({\varepsilon_{0}^{-1}}\)
3 \({\left(4 \pi \varepsilon_{0}\right)^{-1}}\)
4 \({4 \pi \varepsilon_{0}}\)
Explanation:
Since by Gauss's law \(\phi_{T}=\dfrac{q}{\varepsilon_{0}} \quad \therefore \quad \phi_{T}=\dfrac{1 C}{\varepsilon_{0}}=\varepsilon_{0}^{-1}\). So correct option is (2)
358312
Statement A : Total flux through a closed surface is zero if net charge enclosed by the surface is zero. Statement B : Gauss law is true for any closed surface, no matter what its shape or size is.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Gauss law implies that the total electric flux through a closed surface is zero if no net charge is enclosed by the surface and it is true for any closed surface, independent of its shape and size. So option (3) is correct.
PHXII01:ELECTRIC CHARGES AND FIELDS
358313
Consider a circle of radius \(R\). A point charge is present at a distance \(a\) from its centre and on its axis such that the \(R = a\sqrt 3 \). If electric flux passing through the circle is \(\phi \) then the magnitude of the point charge is
358314
Electric field in a region is given by \(\vec E = - 4x\hat i + 6y\hat j\). The charge enclosed in the cube of side \(1\,m\) oriented as shown in the diagram is given by \({\alpha \epsilon_{0}}\). Find the value of \({\alpha}\)
358315
The flux coming out from a unit positive charge placed in air is
1 \({\varepsilon_{0}}\)
2 \({\varepsilon_{0}^{-1}}\)
3 \({\left(4 \pi \varepsilon_{0}\right)^{-1}}\)
4 \({4 \pi \varepsilon_{0}}\)
Explanation:
Since by Gauss's law \(\phi_{T}=\dfrac{q}{\varepsilon_{0}} \quad \therefore \quad \phi_{T}=\dfrac{1 C}{\varepsilon_{0}}=\varepsilon_{0}^{-1}\). So correct option is (2)
358312
Statement A : Total flux through a closed surface is zero if net charge enclosed by the surface is zero. Statement B : Gauss law is true for any closed surface, no matter what its shape or size is.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Gauss law implies that the total electric flux through a closed surface is zero if no net charge is enclosed by the surface and it is true for any closed surface, independent of its shape and size. So option (3) is correct.
PHXII01:ELECTRIC CHARGES AND FIELDS
358313
Consider a circle of radius \(R\). A point charge is present at a distance \(a\) from its centre and on its axis such that the \(R = a\sqrt 3 \). If electric flux passing through the circle is \(\phi \) then the magnitude of the point charge is
358314
Electric field in a region is given by \(\vec E = - 4x\hat i + 6y\hat j\). The charge enclosed in the cube of side \(1\,m\) oriented as shown in the diagram is given by \({\alpha \epsilon_{0}}\). Find the value of \({\alpha}\)
358315
The flux coming out from a unit positive charge placed in air is
1 \({\varepsilon_{0}}\)
2 \({\varepsilon_{0}^{-1}}\)
3 \({\left(4 \pi \varepsilon_{0}\right)^{-1}}\)
4 \({4 \pi \varepsilon_{0}}\)
Explanation:
Since by Gauss's law \(\phi_{T}=\dfrac{q}{\varepsilon_{0}} \quad \therefore \quad \phi_{T}=\dfrac{1 C}{\varepsilon_{0}}=\varepsilon_{0}^{-1}\). So correct option is (2)
