358295
If the electric flux entering and leaving an enclosed surface respectively are \(\phi_{1}\) and \(\phi_{2}\), the electric charge inside the surface will be
Any closed surface is equal to the net charge inside the surface divided by \(\varepsilon_{0}\). Therefore, \(\phi=\dfrac{q}{\varepsilon_{0}}\) Let \(-q_{1}\) be the charge, due to which flux \(\phi_{1}\) is entering the surface \(\phi_{1}=\dfrac{-q_{1}}{\varepsilon_{0}}\) or\(-q_{1}=\varepsilon_{0} \phi_{1}\) Let \(+q_{2}\) be the charge, due to which flux \(\phi_{2}\) is leaving the surface \(\therefore \quad \phi_{2}=\dfrac{q_{2}}{\varepsilon_{0}} \text { or } q_{2}=\varepsilon_{0} \phi_{2}\) So, electric charge inside the surface \( = {q_2} - {q_1}\) \( = {\varepsilon _0}{\phi _2} + {\varepsilon _0}{\phi _1}\) \( = {\varepsilon _0}\left( {{\phi _2} + {\phi _1}} \right)\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358296
A gaussian sphere encloses an electric dipole within it. The total flux across the sphere
1 Zero
2 Half that due to a single charge
3 Double that due to a single charge
4 Dependent on the position of the dipole.
Explanation:
Total electric flux \(\phi = \frac{{( + \,q) + \,( - \,q)}}{{{\varepsilon _0}}} = 0\)
KCET - 2006
PHXII01:ELECTRIC CHARGES AND FIELDS
358297
A point charge \(Q\) is placed at one of the vertices of a cubical block. The electric flux flowing through this cube is
1 \(\dfrac{Q}{6 \varepsilon_{0}}\)
2 \(\dfrac{Q}{4 \varepsilon_{0}}\)
3 \(\dfrac{Q}{8 \varepsilon_{0}}\)
4 \(\dfrac{Q}{\varepsilon_{0}}\)
Explanation:
Let \(\phi_{\text {total }}\) be the total electric flux and \(\phi_{\text {cube }}\) be that of flowing through cube. So, when a point charge \(Q\) is placed at one of the vertices of a cubical block, then only \((1 / 8)\) th part of total flux will pass through this cube, therefore \(\phi_{\text {cube }}=\dfrac{\phi_{\text {total }}}{8}=\dfrac{Q}{8 \varepsilon_{0}}\)\(\left[\because \phi_{\text {total }}=\dfrac{Q}{\varepsilon_{0}}\right]\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358298
The electric field in a certain region is acting radially outward and is given by \({E=A r^{2}}\). A charge contained in a sphere of radius \({a}\) centred at the origin of the field, will be given by:
1 \({A \varepsilon_{0} a^{2}}\)
2 \({4 \pi \varepsilon_{0} A a^{4}}\)
3 \({\varepsilon_{0} A a^{3}}\)
4 \({4 \pi \varepsilon_{0} A a^{2}}\)
Explanation:
Flux linked with sphere \({=\vec{E} \cdot d \vec{s}}\) Since electric field is radial, it is always perpendicular to the surface. \({\phi=A r^{2} \cdot\left(4 \pi r^{2}\right)}\) \({=A\left(a^{2}\right)\left(4 \pi a^{2}\right)(r=a)}\) \({=A\left(4 \pi a^{4}\right)}\) from Gauss' law, \({\phi=\dfrac{q_{\text {in }}}{\varepsilon_{.0}} \Rightarrow q_{\text {in }}=\phi \cdot \varepsilon_{0}}\) \({\Rightarrow q_{\text {in }}=4 \pi \varepsilon_{0} A a^{4}}\)
358295
If the electric flux entering and leaving an enclosed surface respectively are \(\phi_{1}\) and \(\phi_{2}\), the electric charge inside the surface will be
Any closed surface is equal to the net charge inside the surface divided by \(\varepsilon_{0}\). Therefore, \(\phi=\dfrac{q}{\varepsilon_{0}}\) Let \(-q_{1}\) be the charge, due to which flux \(\phi_{1}\) is entering the surface \(\phi_{1}=\dfrac{-q_{1}}{\varepsilon_{0}}\) or\(-q_{1}=\varepsilon_{0} \phi_{1}\) Let \(+q_{2}\) be the charge, due to which flux \(\phi_{2}\) is leaving the surface \(\therefore \quad \phi_{2}=\dfrac{q_{2}}{\varepsilon_{0}} \text { or } q_{2}=\varepsilon_{0} \phi_{2}\) So, electric charge inside the surface \( = {q_2} - {q_1}\) \( = {\varepsilon _0}{\phi _2} + {\varepsilon _0}{\phi _1}\) \( = {\varepsilon _0}\left( {{\phi _2} + {\phi _1}} \right)\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358296
A gaussian sphere encloses an electric dipole within it. The total flux across the sphere
1 Zero
2 Half that due to a single charge
3 Double that due to a single charge
4 Dependent on the position of the dipole.
Explanation:
Total electric flux \(\phi = \frac{{( + \,q) + \,( - \,q)}}{{{\varepsilon _0}}} = 0\)
KCET - 2006
PHXII01:ELECTRIC CHARGES AND FIELDS
358297
A point charge \(Q\) is placed at one of the vertices of a cubical block. The electric flux flowing through this cube is
1 \(\dfrac{Q}{6 \varepsilon_{0}}\)
2 \(\dfrac{Q}{4 \varepsilon_{0}}\)
3 \(\dfrac{Q}{8 \varepsilon_{0}}\)
4 \(\dfrac{Q}{\varepsilon_{0}}\)
Explanation:
Let \(\phi_{\text {total }}\) be the total electric flux and \(\phi_{\text {cube }}\) be that of flowing through cube. So, when a point charge \(Q\) is placed at one of the vertices of a cubical block, then only \((1 / 8)\) th part of total flux will pass through this cube, therefore \(\phi_{\text {cube }}=\dfrac{\phi_{\text {total }}}{8}=\dfrac{Q}{8 \varepsilon_{0}}\)\(\left[\because \phi_{\text {total }}=\dfrac{Q}{\varepsilon_{0}}\right]\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358298
The electric field in a certain region is acting radially outward and is given by \({E=A r^{2}}\). A charge contained in a sphere of radius \({a}\) centred at the origin of the field, will be given by:
1 \({A \varepsilon_{0} a^{2}}\)
2 \({4 \pi \varepsilon_{0} A a^{4}}\)
3 \({\varepsilon_{0} A a^{3}}\)
4 \({4 \pi \varepsilon_{0} A a^{2}}\)
Explanation:
Flux linked with sphere \({=\vec{E} \cdot d \vec{s}}\) Since electric field is radial, it is always perpendicular to the surface. \({\phi=A r^{2} \cdot\left(4 \pi r^{2}\right)}\) \({=A\left(a^{2}\right)\left(4 \pi a^{2}\right)(r=a)}\) \({=A\left(4 \pi a^{4}\right)}\) from Gauss' law, \({\phi=\dfrac{q_{\text {in }}}{\varepsilon_{.0}} \Rightarrow q_{\text {in }}=\phi \cdot \varepsilon_{0}}\) \({\Rightarrow q_{\text {in }}=4 \pi \varepsilon_{0} A a^{4}}\)
358295
If the electric flux entering and leaving an enclosed surface respectively are \(\phi_{1}\) and \(\phi_{2}\), the electric charge inside the surface will be
Any closed surface is equal to the net charge inside the surface divided by \(\varepsilon_{0}\). Therefore, \(\phi=\dfrac{q}{\varepsilon_{0}}\) Let \(-q_{1}\) be the charge, due to which flux \(\phi_{1}\) is entering the surface \(\phi_{1}=\dfrac{-q_{1}}{\varepsilon_{0}}\) or\(-q_{1}=\varepsilon_{0} \phi_{1}\) Let \(+q_{2}\) be the charge, due to which flux \(\phi_{2}\) is leaving the surface \(\therefore \quad \phi_{2}=\dfrac{q_{2}}{\varepsilon_{0}} \text { or } q_{2}=\varepsilon_{0} \phi_{2}\) So, electric charge inside the surface \( = {q_2} - {q_1}\) \( = {\varepsilon _0}{\phi _2} + {\varepsilon _0}{\phi _1}\) \( = {\varepsilon _0}\left( {{\phi _2} + {\phi _1}} \right)\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358296
A gaussian sphere encloses an electric dipole within it. The total flux across the sphere
1 Zero
2 Half that due to a single charge
3 Double that due to a single charge
4 Dependent on the position of the dipole.
Explanation:
Total electric flux \(\phi = \frac{{( + \,q) + \,( - \,q)}}{{{\varepsilon _0}}} = 0\)
KCET - 2006
PHXII01:ELECTRIC CHARGES AND FIELDS
358297
A point charge \(Q\) is placed at one of the vertices of a cubical block. The electric flux flowing through this cube is
1 \(\dfrac{Q}{6 \varepsilon_{0}}\)
2 \(\dfrac{Q}{4 \varepsilon_{0}}\)
3 \(\dfrac{Q}{8 \varepsilon_{0}}\)
4 \(\dfrac{Q}{\varepsilon_{0}}\)
Explanation:
Let \(\phi_{\text {total }}\) be the total electric flux and \(\phi_{\text {cube }}\) be that of flowing through cube. So, when a point charge \(Q\) is placed at one of the vertices of a cubical block, then only \((1 / 8)\) th part of total flux will pass through this cube, therefore \(\phi_{\text {cube }}=\dfrac{\phi_{\text {total }}}{8}=\dfrac{Q}{8 \varepsilon_{0}}\)\(\left[\because \phi_{\text {total }}=\dfrac{Q}{\varepsilon_{0}}\right]\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358298
The electric field in a certain region is acting radially outward and is given by \({E=A r^{2}}\). A charge contained in a sphere of radius \({a}\) centred at the origin of the field, will be given by:
1 \({A \varepsilon_{0} a^{2}}\)
2 \({4 \pi \varepsilon_{0} A a^{4}}\)
3 \({\varepsilon_{0} A a^{3}}\)
4 \({4 \pi \varepsilon_{0} A a^{2}}\)
Explanation:
Flux linked with sphere \({=\vec{E} \cdot d \vec{s}}\) Since electric field is radial, it is always perpendicular to the surface. \({\phi=A r^{2} \cdot\left(4 \pi r^{2}\right)}\) \({=A\left(a^{2}\right)\left(4 \pi a^{2}\right)(r=a)}\) \({=A\left(4 \pi a^{4}\right)}\) from Gauss' law, \({\phi=\dfrac{q_{\text {in }}}{\varepsilon_{.0}} \Rightarrow q_{\text {in }}=\phi \cdot \varepsilon_{0}}\) \({\Rightarrow q_{\text {in }}=4 \pi \varepsilon_{0} A a^{4}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII01:ELECTRIC CHARGES AND FIELDS
358295
If the electric flux entering and leaving an enclosed surface respectively are \(\phi_{1}\) and \(\phi_{2}\), the electric charge inside the surface will be
Any closed surface is equal to the net charge inside the surface divided by \(\varepsilon_{0}\). Therefore, \(\phi=\dfrac{q}{\varepsilon_{0}}\) Let \(-q_{1}\) be the charge, due to which flux \(\phi_{1}\) is entering the surface \(\phi_{1}=\dfrac{-q_{1}}{\varepsilon_{0}}\) or\(-q_{1}=\varepsilon_{0} \phi_{1}\) Let \(+q_{2}\) be the charge, due to which flux \(\phi_{2}\) is leaving the surface \(\therefore \quad \phi_{2}=\dfrac{q_{2}}{\varepsilon_{0}} \text { or } q_{2}=\varepsilon_{0} \phi_{2}\) So, electric charge inside the surface \( = {q_2} - {q_1}\) \( = {\varepsilon _0}{\phi _2} + {\varepsilon _0}{\phi _1}\) \( = {\varepsilon _0}\left( {{\phi _2} + {\phi _1}} \right)\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358296
A gaussian sphere encloses an electric dipole within it. The total flux across the sphere
1 Zero
2 Half that due to a single charge
3 Double that due to a single charge
4 Dependent on the position of the dipole.
Explanation:
Total electric flux \(\phi = \frac{{( + \,q) + \,( - \,q)}}{{{\varepsilon _0}}} = 0\)
KCET - 2006
PHXII01:ELECTRIC CHARGES AND FIELDS
358297
A point charge \(Q\) is placed at one of the vertices of a cubical block. The electric flux flowing through this cube is
1 \(\dfrac{Q}{6 \varepsilon_{0}}\)
2 \(\dfrac{Q}{4 \varepsilon_{0}}\)
3 \(\dfrac{Q}{8 \varepsilon_{0}}\)
4 \(\dfrac{Q}{\varepsilon_{0}}\)
Explanation:
Let \(\phi_{\text {total }}\) be the total electric flux and \(\phi_{\text {cube }}\) be that of flowing through cube. So, when a point charge \(Q\) is placed at one of the vertices of a cubical block, then only \((1 / 8)\) th part of total flux will pass through this cube, therefore \(\phi_{\text {cube }}=\dfrac{\phi_{\text {total }}}{8}=\dfrac{Q}{8 \varepsilon_{0}}\)\(\left[\because \phi_{\text {total }}=\dfrac{Q}{\varepsilon_{0}}\right]\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358298
The electric field in a certain region is acting radially outward and is given by \({E=A r^{2}}\). A charge contained in a sphere of radius \({a}\) centred at the origin of the field, will be given by:
1 \({A \varepsilon_{0} a^{2}}\)
2 \({4 \pi \varepsilon_{0} A a^{4}}\)
3 \({\varepsilon_{0} A a^{3}}\)
4 \({4 \pi \varepsilon_{0} A a^{2}}\)
Explanation:
Flux linked with sphere \({=\vec{E} \cdot d \vec{s}}\) Since electric field is radial, it is always perpendicular to the surface. \({\phi=A r^{2} \cdot\left(4 \pi r^{2}\right)}\) \({=A\left(a^{2}\right)\left(4 \pi a^{2}\right)(r=a)}\) \({=A\left(4 \pi a^{4}\right)}\) from Gauss' law, \({\phi=\dfrac{q_{\text {in }}}{\varepsilon_{.0}} \Rightarrow q_{\text {in }}=\phi \cdot \varepsilon_{0}}\) \({\Rightarrow q_{\text {in }}=4 \pi \varepsilon_{0} A a^{4}}\)
