272170
If a charge $q$ is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :
1 $\frac{q}{{{\varepsilon }_{0}}}$
2 $\frac{q}{2{{\varepsilon }_{0}}}$
3 $\frac{q}{4{{\varepsilon }_{0}}}$
4 $\frac{q}{2\pi {{\varepsilon }_{0}}}$
Explanation:
(None) Total flux through complete spherical surface is $\frac{q}{{{\epsilon }_{0}}}$.
So flux, through curved surface will be $\frac{q}{2{{\epsilon }_{0}}}$.
The flux through flat surface will be zero
NCERT Page-25 / N-22
Electric Charges and Fields
272171
Number of electric lines of force that radiate outwards from one coulomb of charge in vacuum is
1 $1.13\times {{10}^{11}}$
2 $0.61\times {{10}^{11}}$
3 $1.13\times {{10}^{10}}$
4 $0.61\times {{10}^{9}}$
Explanation:
(a) Here, $q=1C,{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}~{{m}^{-2}}$ Number of lines of force $=$ Electric force
$=\frac{q}{{{\varepsilon }_{0}}}=\frac{1}{8.85\times {{10}^{-12}}}=1.13\times {{10}^{11}}$
NCERT Page-25 / N-22
Electric Charges and Fields
272172
In the figure the net electric flux through the area $A$ is $\phi =\vec{E}\cdot \vec{A}$ when the system is in air. On immersing the system in water the net electric flux through the area
1 becomes zero
2 remains same
3 increases
4 decreases
Explanation:
(d) Since electric field $\vec{E}$ decreases inside water, therefore flux $\phi =\vec{E}\cdot \vec{A}$ also decreases.
NCERT Page-26 / N-22
Electric Charges and Fields
272178
The electric field in a region of space is given by, $\vec{E}={{E}_{0}}i+2{{E}_{0}}j$ where ${{E}_{o}}=100~N/C$. The flux of the field through a circular surface of radius $0.02~m$ parallel to the $Y-Z$ plane is nearly:
1 $0.125N{{m}^{2}}/C$
2 $0.02N{{m}^{2}}/C$
3 $0.005N{{m}^{2}}/C$
4 $3.14N{{m}^{2}}/C$
Explanation:
(a) $\vec{E}={{E}_{0}}i+2{{E}_{0}}j$
Given, ${{E}_{0}}=100~N/c$ So, $\vec{E}=100i+200j$
Radius of circular surface $=0.02~m$
Area $=\pi {{r}^{2}}=\frac{22}{7}\times 0.02\times 0.02$
$=1.25\times {{10}^{-3}}im{{}^{2}}$ [Loop is parallel to Y-Z plane]
Now, flux $\left( \phi \right)=EAcos\theta $
$=\left( 100i+200j \right)\cdot 1.25\times {{10}^{-3}}icos{{\theta }^{\circ }}\left[ \theta ={{0}^{\circ }} \right]$
$=125\times {{10}^{-3}}N{{m}^{2}}/C=0.125N{{m}^{2}}/C$
272170
If a charge $q$ is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :
1 $\frac{q}{{{\varepsilon }_{0}}}$
2 $\frac{q}{2{{\varepsilon }_{0}}}$
3 $\frac{q}{4{{\varepsilon }_{0}}}$
4 $\frac{q}{2\pi {{\varepsilon }_{0}}}$
Explanation:
(None) Total flux through complete spherical surface is $\frac{q}{{{\epsilon }_{0}}}$.
So flux, through curved surface will be $\frac{q}{2{{\epsilon }_{0}}}$.
The flux through flat surface will be zero
NCERT Page-25 / N-22
Electric Charges and Fields
272171
Number of electric lines of force that radiate outwards from one coulomb of charge in vacuum is
1 $1.13\times {{10}^{11}}$
2 $0.61\times {{10}^{11}}$
3 $1.13\times {{10}^{10}}$
4 $0.61\times {{10}^{9}}$
Explanation:
(a) Here, $q=1C,{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}~{{m}^{-2}}$ Number of lines of force $=$ Electric force
$=\frac{q}{{{\varepsilon }_{0}}}=\frac{1}{8.85\times {{10}^{-12}}}=1.13\times {{10}^{11}}$
NCERT Page-25 / N-22
Electric Charges and Fields
272172
In the figure the net electric flux through the area $A$ is $\phi =\vec{E}\cdot \vec{A}$ when the system is in air. On immersing the system in water the net electric flux through the area
1 becomes zero
2 remains same
3 increases
4 decreases
Explanation:
(d) Since electric field $\vec{E}$ decreases inside water, therefore flux $\phi =\vec{E}\cdot \vec{A}$ also decreases.
NCERT Page-26 / N-22
Electric Charges and Fields
272178
The electric field in a region of space is given by, $\vec{E}={{E}_{0}}i+2{{E}_{0}}j$ where ${{E}_{o}}=100~N/C$. The flux of the field through a circular surface of radius $0.02~m$ parallel to the $Y-Z$ plane is nearly:
1 $0.125N{{m}^{2}}/C$
2 $0.02N{{m}^{2}}/C$
3 $0.005N{{m}^{2}}/C$
4 $3.14N{{m}^{2}}/C$
Explanation:
(a) $\vec{E}={{E}_{0}}i+2{{E}_{0}}j$
Given, ${{E}_{0}}=100~N/c$ So, $\vec{E}=100i+200j$
Radius of circular surface $=0.02~m$
Area $=\pi {{r}^{2}}=\frac{22}{7}\times 0.02\times 0.02$
$=1.25\times {{10}^{-3}}im{{}^{2}}$ [Loop is parallel to Y-Z plane]
Now, flux $\left( \phi \right)=EAcos\theta $
$=\left( 100i+200j \right)\cdot 1.25\times {{10}^{-3}}icos{{\theta }^{\circ }}\left[ \theta ={{0}^{\circ }} \right]$
$=125\times {{10}^{-3}}N{{m}^{2}}/C=0.125N{{m}^{2}}/C$
272170
If a charge $q$ is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :
1 $\frac{q}{{{\varepsilon }_{0}}}$
2 $\frac{q}{2{{\varepsilon }_{0}}}$
3 $\frac{q}{4{{\varepsilon }_{0}}}$
4 $\frac{q}{2\pi {{\varepsilon }_{0}}}$
Explanation:
(None) Total flux through complete spherical surface is $\frac{q}{{{\epsilon }_{0}}}$.
So flux, through curved surface will be $\frac{q}{2{{\epsilon }_{0}}}$.
The flux through flat surface will be zero
NCERT Page-25 / N-22
Electric Charges and Fields
272171
Number of electric lines of force that radiate outwards from one coulomb of charge in vacuum is
1 $1.13\times {{10}^{11}}$
2 $0.61\times {{10}^{11}}$
3 $1.13\times {{10}^{10}}$
4 $0.61\times {{10}^{9}}$
Explanation:
(a) Here, $q=1C,{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}~{{m}^{-2}}$ Number of lines of force $=$ Electric force
$=\frac{q}{{{\varepsilon }_{0}}}=\frac{1}{8.85\times {{10}^{-12}}}=1.13\times {{10}^{11}}$
NCERT Page-25 / N-22
Electric Charges and Fields
272172
In the figure the net electric flux through the area $A$ is $\phi =\vec{E}\cdot \vec{A}$ when the system is in air. On immersing the system in water the net electric flux through the area
1 becomes zero
2 remains same
3 increases
4 decreases
Explanation:
(d) Since electric field $\vec{E}$ decreases inside water, therefore flux $\phi =\vec{E}\cdot \vec{A}$ also decreases.
NCERT Page-26 / N-22
Electric Charges and Fields
272178
The electric field in a region of space is given by, $\vec{E}={{E}_{0}}i+2{{E}_{0}}j$ where ${{E}_{o}}=100~N/C$. The flux of the field through a circular surface of radius $0.02~m$ parallel to the $Y-Z$ plane is nearly:
1 $0.125N{{m}^{2}}/C$
2 $0.02N{{m}^{2}}/C$
3 $0.005N{{m}^{2}}/C$
4 $3.14N{{m}^{2}}/C$
Explanation:
(a) $\vec{E}={{E}_{0}}i+2{{E}_{0}}j$
Given, ${{E}_{0}}=100~N/c$ So, $\vec{E}=100i+200j$
Radius of circular surface $=0.02~m$
Area $=\pi {{r}^{2}}=\frac{22}{7}\times 0.02\times 0.02$
$=1.25\times {{10}^{-3}}im{{}^{2}}$ [Loop is parallel to Y-Z plane]
Now, flux $\left( \phi \right)=EAcos\theta $
$=\left( 100i+200j \right)\cdot 1.25\times {{10}^{-3}}icos{{\theta }^{\circ }}\left[ \theta ={{0}^{\circ }} \right]$
$=125\times {{10}^{-3}}N{{m}^{2}}/C=0.125N{{m}^{2}}/C$
272170
If a charge $q$ is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :
1 $\frac{q}{{{\varepsilon }_{0}}}$
2 $\frac{q}{2{{\varepsilon }_{0}}}$
3 $\frac{q}{4{{\varepsilon }_{0}}}$
4 $\frac{q}{2\pi {{\varepsilon }_{0}}}$
Explanation:
(None) Total flux through complete spherical surface is $\frac{q}{{{\epsilon }_{0}}}$.
So flux, through curved surface will be $\frac{q}{2{{\epsilon }_{0}}}$.
The flux through flat surface will be zero
NCERT Page-25 / N-22
Electric Charges and Fields
272171
Number of electric lines of force that radiate outwards from one coulomb of charge in vacuum is
1 $1.13\times {{10}^{11}}$
2 $0.61\times {{10}^{11}}$
3 $1.13\times {{10}^{10}}$
4 $0.61\times {{10}^{9}}$
Explanation:
(a) Here, $q=1C,{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}~{{m}^{-2}}$ Number of lines of force $=$ Electric force
$=\frac{q}{{{\varepsilon }_{0}}}=\frac{1}{8.85\times {{10}^{-12}}}=1.13\times {{10}^{11}}$
NCERT Page-25 / N-22
Electric Charges and Fields
272172
In the figure the net electric flux through the area $A$ is $\phi =\vec{E}\cdot \vec{A}$ when the system is in air. On immersing the system in water the net electric flux through the area
1 becomes zero
2 remains same
3 increases
4 decreases
Explanation:
(d) Since electric field $\vec{E}$ decreases inside water, therefore flux $\phi =\vec{E}\cdot \vec{A}$ also decreases.
NCERT Page-26 / N-22
Electric Charges and Fields
272178
The electric field in a region of space is given by, $\vec{E}={{E}_{0}}i+2{{E}_{0}}j$ where ${{E}_{o}}=100~N/C$. The flux of the field through a circular surface of radius $0.02~m$ parallel to the $Y-Z$ plane is nearly:
1 $0.125N{{m}^{2}}/C$
2 $0.02N{{m}^{2}}/C$
3 $0.005N{{m}^{2}}/C$
4 $3.14N{{m}^{2}}/C$
Explanation:
(a) $\vec{E}={{E}_{0}}i+2{{E}_{0}}j$
Given, ${{E}_{0}}=100~N/c$ So, $\vec{E}=100i+200j$
Radius of circular surface $=0.02~m$
Area $=\pi {{r}^{2}}=\frac{22}{7}\times 0.02\times 0.02$
$=1.25\times {{10}^{-3}}im{{}^{2}}$ [Loop is parallel to Y-Z plane]
Now, flux $\left( \phi \right)=EAcos\theta $
$=\left( 100i+200j \right)\cdot 1.25\times {{10}^{-3}}icos{{\theta }^{\circ }}\left[ \theta ={{0}^{\circ }} \right]$
$=125\times {{10}^{-3}}N{{m}^{2}}/C=0.125N{{m}^{2}}/C$
