358331
Two charges of \(5 Q\) and \(-2 Q\) are situated at the points \((3 a, 0)\) and \((-5 a, 0)\) respectively. The electric flux through a sphere of radius ' \(4 a\) ' having center at origin is
1 \(\dfrac{2 Q}{\varepsilon_{0}}\)
2 \(\dfrac{7 Q}{\varepsilon_{0}}\)
3 \(\dfrac{5 Q}{\varepsilon_{0}}\)
4 \(\dfrac{3 Q}{\varepsilon_{0}}\)
Explanation:
From Gauss law, electric flux, \(\phi=\dfrac{q_{\text {enclosed }}}{\varepsilon_{0}}\) As the sphere is of radius \(4 a\) so it wil enclose only \(5 Q\) charge inside it (as \(-2 Q\) charge is at distance which is outside the sphere). Hence, \(\phi=\dfrac{5 Q}{\varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358332
Eight dipoles of charges of magnitude \(e\) are placed inside a cube. The total electric flux coming out of the cube will be
1 \(\frac{{16e}}{{{\varepsilon _0}}}\)
2 \(\frac{e}{{{\varepsilon _0}}}\)
3 \(\frac{{8e}}{{{\varepsilon _0}}}\)
4 Zero
Explanation:
The total flux is \(\phi = \frac{{{q_{total}}}}{{{\varepsilon _0}}} = 0\) (Net charge on dipole is zero)
PHXII01:ELECTRIC CHARGES AND FIELDS
358333
In the figure, a cone of radius \({R}\) is shown. Electric field of intensity \({E_{0}}\) is present perpendicular to the circular cross-section of the cone. The electric flux through the carved surface of the hemisphere is
1 0
2 \({E_{0} \times \pi R^{2}}\)
3 \({E_{0} \times 2 \pi R^{2}}\)
4 \({E_{0} \times 3 \pi R^{2}}\)
Explanation:
The flux through both flat and carved surface are same. \({\phi_{\text {Flat }}=\phi_{\text {carved }}=E_{0} \pi R^{2}}\)
358331
Two charges of \(5 Q\) and \(-2 Q\) are situated at the points \((3 a, 0)\) and \((-5 a, 0)\) respectively. The electric flux through a sphere of radius ' \(4 a\) ' having center at origin is
1 \(\dfrac{2 Q}{\varepsilon_{0}}\)
2 \(\dfrac{7 Q}{\varepsilon_{0}}\)
3 \(\dfrac{5 Q}{\varepsilon_{0}}\)
4 \(\dfrac{3 Q}{\varepsilon_{0}}\)
Explanation:
From Gauss law, electric flux, \(\phi=\dfrac{q_{\text {enclosed }}}{\varepsilon_{0}}\) As the sphere is of radius \(4 a\) so it wil enclose only \(5 Q\) charge inside it (as \(-2 Q\) charge is at distance which is outside the sphere). Hence, \(\phi=\dfrac{5 Q}{\varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358332
Eight dipoles of charges of magnitude \(e\) are placed inside a cube. The total electric flux coming out of the cube will be
1 \(\frac{{16e}}{{{\varepsilon _0}}}\)
2 \(\frac{e}{{{\varepsilon _0}}}\)
3 \(\frac{{8e}}{{{\varepsilon _0}}}\)
4 Zero
Explanation:
The total flux is \(\phi = \frac{{{q_{total}}}}{{{\varepsilon _0}}} = 0\) (Net charge on dipole is zero)
PHXII01:ELECTRIC CHARGES AND FIELDS
358333
In the figure, a cone of radius \({R}\) is shown. Electric field of intensity \({E_{0}}\) is present perpendicular to the circular cross-section of the cone. The electric flux through the carved surface of the hemisphere is
1 0
2 \({E_{0} \times \pi R^{2}}\)
3 \({E_{0} \times 2 \pi R^{2}}\)
4 \({E_{0} \times 3 \pi R^{2}}\)
Explanation:
The flux through both flat and carved surface are same. \({\phi_{\text {Flat }}=\phi_{\text {carved }}=E_{0} \pi R^{2}}\)
358331
Two charges of \(5 Q\) and \(-2 Q\) are situated at the points \((3 a, 0)\) and \((-5 a, 0)\) respectively. The electric flux through a sphere of radius ' \(4 a\) ' having center at origin is
1 \(\dfrac{2 Q}{\varepsilon_{0}}\)
2 \(\dfrac{7 Q}{\varepsilon_{0}}\)
3 \(\dfrac{5 Q}{\varepsilon_{0}}\)
4 \(\dfrac{3 Q}{\varepsilon_{0}}\)
Explanation:
From Gauss law, electric flux, \(\phi=\dfrac{q_{\text {enclosed }}}{\varepsilon_{0}}\) As the sphere is of radius \(4 a\) so it wil enclose only \(5 Q\) charge inside it (as \(-2 Q\) charge is at distance which is outside the sphere). Hence, \(\phi=\dfrac{5 Q}{\varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358332
Eight dipoles of charges of magnitude \(e\) are placed inside a cube. The total electric flux coming out of the cube will be
1 \(\frac{{16e}}{{{\varepsilon _0}}}\)
2 \(\frac{e}{{{\varepsilon _0}}}\)
3 \(\frac{{8e}}{{{\varepsilon _0}}}\)
4 Zero
Explanation:
The total flux is \(\phi = \frac{{{q_{total}}}}{{{\varepsilon _0}}} = 0\) (Net charge on dipole is zero)
PHXII01:ELECTRIC CHARGES AND FIELDS
358333
In the figure, a cone of radius \({R}\) is shown. Electric field of intensity \({E_{0}}\) is present perpendicular to the circular cross-section of the cone. The electric flux through the carved surface of the hemisphere is
1 0
2 \({E_{0} \times \pi R^{2}}\)
3 \({E_{0} \times 2 \pi R^{2}}\)
4 \({E_{0} \times 3 \pi R^{2}}\)
Explanation:
The flux through both flat and carved surface are same. \({\phi_{\text {Flat }}=\phi_{\text {carved }}=E_{0} \pi R^{2}}\)
358331
Two charges of \(5 Q\) and \(-2 Q\) are situated at the points \((3 a, 0)\) and \((-5 a, 0)\) respectively. The electric flux through a sphere of radius ' \(4 a\) ' having center at origin is
1 \(\dfrac{2 Q}{\varepsilon_{0}}\)
2 \(\dfrac{7 Q}{\varepsilon_{0}}\)
3 \(\dfrac{5 Q}{\varepsilon_{0}}\)
4 \(\dfrac{3 Q}{\varepsilon_{0}}\)
Explanation:
From Gauss law, electric flux, \(\phi=\dfrac{q_{\text {enclosed }}}{\varepsilon_{0}}\) As the sphere is of radius \(4 a\) so it wil enclose only \(5 Q\) charge inside it (as \(-2 Q\) charge is at distance which is outside the sphere). Hence, \(\phi=\dfrac{5 Q}{\varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358332
Eight dipoles of charges of magnitude \(e\) are placed inside a cube. The total electric flux coming out of the cube will be
1 \(\frac{{16e}}{{{\varepsilon _0}}}\)
2 \(\frac{e}{{{\varepsilon _0}}}\)
3 \(\frac{{8e}}{{{\varepsilon _0}}}\)
4 Zero
Explanation:
The total flux is \(\phi = \frac{{{q_{total}}}}{{{\varepsilon _0}}} = 0\) (Net charge on dipole is zero)
PHXII01:ELECTRIC CHARGES AND FIELDS
358333
In the figure, a cone of radius \({R}\) is shown. Electric field of intensity \({E_{0}}\) is present perpendicular to the circular cross-section of the cone. The electric flux through the carved surface of the hemisphere is
1 0
2 \({E_{0} \times \pi R^{2}}\)
3 \({E_{0} \times 2 \pi R^{2}}\)
4 \({E_{0} \times 3 \pi R^{2}}\)
Explanation:
The flux through both flat and carved surface are same. \({\phi_{\text {Flat }}=\phi_{\text {carved }}=E_{0} \pi R^{2}}\)
358331
Two charges of \(5 Q\) and \(-2 Q\) are situated at the points \((3 a, 0)\) and \((-5 a, 0)\) respectively. The electric flux through a sphere of radius ' \(4 a\) ' having center at origin is
1 \(\dfrac{2 Q}{\varepsilon_{0}}\)
2 \(\dfrac{7 Q}{\varepsilon_{0}}\)
3 \(\dfrac{5 Q}{\varepsilon_{0}}\)
4 \(\dfrac{3 Q}{\varepsilon_{0}}\)
Explanation:
From Gauss law, electric flux, \(\phi=\dfrac{q_{\text {enclosed }}}{\varepsilon_{0}}\) As the sphere is of radius \(4 a\) so it wil enclose only \(5 Q\) charge inside it (as \(-2 Q\) charge is at distance which is outside the sphere). Hence, \(\phi=\dfrac{5 Q}{\varepsilon_{0}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358332
Eight dipoles of charges of magnitude \(e\) are placed inside a cube. The total electric flux coming out of the cube will be
1 \(\frac{{16e}}{{{\varepsilon _0}}}\)
2 \(\frac{e}{{{\varepsilon _0}}}\)
3 \(\frac{{8e}}{{{\varepsilon _0}}}\)
4 Zero
Explanation:
The total flux is \(\phi = \frac{{{q_{total}}}}{{{\varepsilon _0}}} = 0\) (Net charge on dipole is zero)
PHXII01:ELECTRIC CHARGES AND FIELDS
358333
In the figure, a cone of radius \({R}\) is shown. Electric field of intensity \({E_{0}}\) is present perpendicular to the circular cross-section of the cone. The electric flux through the carved surface of the hemisphere is
1 0
2 \({E_{0} \times \pi R^{2}}\)
3 \({E_{0} \times 2 \pi R^{2}}\)
4 \({E_{0} \times 3 \pi R^{2}}\)
Explanation:
The flux through both flat and carved surface are same. \({\phi_{\text {Flat }}=\phi_{\text {carved }}=E_{0} \pi R^{2}}\)
