272177
A surface has the area vector $\vec{A}=\left( 2i+3j \right){{m}^{2}}$. The flux of an electric field through it if the field is $\vec{E}=4i\frac{V}{m}$ :
272173
A cylinder of radius $R$ and length $l$ is placed in a uniform electric field $E$ parallel to the axis of the cylinder. The total flux over the curved surface of the cylinder is
1 zero
2 $\pi {{R}^{2}}E$
3 $2\pi {{R}^{2}}E$
4 $E/\pi {{R}^{2}}$
Explanation:
(a) For the curved surface, $\theta ={{90}^{\circ }}$
$\therefore \phi =E$ ds $cos{{90}^{\circ }}=0$.
NCERT Page-26 / N-22
Electric Charges and Fields
272174
Positive electric flux indicates that electric lines of force are directed
1 outwards
2 inwards
3 either (a) or (b)
4 None of these
Explanation:
(a)
NCERT Page-26 / N-22
Electric Charges and Fields
272175
Electric flux over a surface in an electric field may be
1 positive
2 zero
3 negative
4 All of these
Explanation:
(d)
NCERT Page-26 / N-22
Electric Charges and Fields
272176
Consider an electric field $\vec{E}={{E}_{0}}x$ where ${{E}_{0}}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is
272177
A surface has the area vector $\vec{A}=\left( 2i+3j \right){{m}^{2}}$. The flux of an electric field through it if the field is $\vec{E}=4i\frac{V}{m}$ :
272173
A cylinder of radius $R$ and length $l$ is placed in a uniform electric field $E$ parallel to the axis of the cylinder. The total flux over the curved surface of the cylinder is
1 zero
2 $\pi {{R}^{2}}E$
3 $2\pi {{R}^{2}}E$
4 $E/\pi {{R}^{2}}$
Explanation:
(a) For the curved surface, $\theta ={{90}^{\circ }}$
$\therefore \phi =E$ ds $cos{{90}^{\circ }}=0$.
NCERT Page-26 / N-22
Electric Charges and Fields
272174
Positive electric flux indicates that electric lines of force are directed
1 outwards
2 inwards
3 either (a) or (b)
4 None of these
Explanation:
(a)
NCERT Page-26 / N-22
Electric Charges and Fields
272175
Electric flux over a surface in an electric field may be
1 positive
2 zero
3 negative
4 All of these
Explanation:
(d)
NCERT Page-26 / N-22
Electric Charges and Fields
272176
Consider an electric field $\vec{E}={{E}_{0}}x$ where ${{E}_{0}}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is
272177
A surface has the area vector $\vec{A}=\left( 2i+3j \right){{m}^{2}}$. The flux of an electric field through it if the field is $\vec{E}=4i\frac{V}{m}$ :
272173
A cylinder of radius $R$ and length $l$ is placed in a uniform electric field $E$ parallel to the axis of the cylinder. The total flux over the curved surface of the cylinder is
1 zero
2 $\pi {{R}^{2}}E$
3 $2\pi {{R}^{2}}E$
4 $E/\pi {{R}^{2}}$
Explanation:
(a) For the curved surface, $\theta ={{90}^{\circ }}$
$\therefore \phi =E$ ds $cos{{90}^{\circ }}=0$.
NCERT Page-26 / N-22
Electric Charges and Fields
272174
Positive electric flux indicates that electric lines of force are directed
1 outwards
2 inwards
3 either (a) or (b)
4 None of these
Explanation:
(a)
NCERT Page-26 / N-22
Electric Charges and Fields
272175
Electric flux over a surface in an electric field may be
1 positive
2 zero
3 negative
4 All of these
Explanation:
(d)
NCERT Page-26 / N-22
Electric Charges and Fields
272176
Consider an electric field $\vec{E}={{E}_{0}}x$ where ${{E}_{0}}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is
272177
A surface has the area vector $\vec{A}=\left( 2i+3j \right){{m}^{2}}$. The flux of an electric field through it if the field is $\vec{E}=4i\frac{V}{m}$ :
272173
A cylinder of radius $R$ and length $l$ is placed in a uniform electric field $E$ parallel to the axis of the cylinder. The total flux over the curved surface of the cylinder is
1 zero
2 $\pi {{R}^{2}}E$
3 $2\pi {{R}^{2}}E$
4 $E/\pi {{R}^{2}}$
Explanation:
(a) For the curved surface, $\theta ={{90}^{\circ }}$
$\therefore \phi =E$ ds $cos{{90}^{\circ }}=0$.
NCERT Page-26 / N-22
Electric Charges and Fields
272174
Positive electric flux indicates that electric lines of force are directed
1 outwards
2 inwards
3 either (a) or (b)
4 None of these
Explanation:
(a)
NCERT Page-26 / N-22
Electric Charges and Fields
272175
Electric flux over a surface in an electric field may be
1 positive
2 zero
3 negative
4 All of these
Explanation:
(d)
NCERT Page-26 / N-22
Electric Charges and Fields
272176
Consider an electric field $\vec{E}={{E}_{0}}x$ where ${{E}_{0}}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is
272177
A surface has the area vector $\vec{A}=\left( 2i+3j \right){{m}^{2}}$. The flux of an electric field through it if the field is $\vec{E}=4i\frac{V}{m}$ :
272173
A cylinder of radius $R$ and length $l$ is placed in a uniform electric field $E$ parallel to the axis of the cylinder. The total flux over the curved surface of the cylinder is
1 zero
2 $\pi {{R}^{2}}E$
3 $2\pi {{R}^{2}}E$
4 $E/\pi {{R}^{2}}$
Explanation:
(a) For the curved surface, $\theta ={{90}^{\circ }}$
$\therefore \phi =E$ ds $cos{{90}^{\circ }}=0$.
NCERT Page-26 / N-22
Electric Charges and Fields
272174
Positive electric flux indicates that electric lines of force are directed
1 outwards
2 inwards
3 either (a) or (b)
4 None of these
Explanation:
(a)
NCERT Page-26 / N-22
Electric Charges and Fields
272175
Electric flux over a surface in an electric field may be
1 positive
2 zero
3 negative
4 All of these
Explanation:
(d)
NCERT Page-26 / N-22
Electric Charges and Fields
272176
Consider an electric field $\vec{E}={{E}_{0}}x$ where ${{E}_{0}}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is