1 All the charges whether inside or outside thegaussian surface contribute to the electric flux.
2 Electric flux depends upon the geometry of thegaussian surface.
3 Gauss theoremcan be applied to non-uniform electric field.
4 The electric field over thegaussian surface remains continuous and uniform at every point.
Explanation:
(d) All other statements except (iv) are in correct.
The electric field over the Gaussian surface remains continuous and uniform at every point.
NCERT Page-33 / N-29 | CBSE Sample 2021-2022
Electric Charges and Fields
272201
The total electric flux emanating from a closed surface enclosing an $\alpha $-particle is (e-electronic charge)
1 $\frac{2e}{{{\varepsilon }_{0}}}$
2 $\frac{e}{{{\varepsilon }_{0}}}$
3 $e{{\varepsilon }_{0}}$
4 $\frac{{{\varepsilon }_{0}}e}{4}$
Explanation:
(a) According to Gauss's law total electric flux through a closed surface is $\frac{1}{{{\varepsilon }_{0}}}$ times the total charge inside that surface.
Electric flux, ${{\phi }_{E}}=\frac{q}{{{\varepsilon }_{0}}}$
Charge on $\alpha $-particle $=2e~\therefore {{\phi }_{E}}=\frac{2e}{{{\varepsilon }_{0}}}$
NCERT Page-34 / N-30
Electric Charges and Fields
272202
For a given surface the Gauss's law is stated as $\oint ~\vec{E}\cdot d\vec{A}=0$. From this we can conclude that
1 $E$ is necessarily zero on the surface
2 $E$ is perpendicular to the surface at every point
3 the total flux through the surface is zero
4 the flux is only going out of the surface
Explanation:
(c) $\oint ~\vec{E}\cdot d\vec{A}=0$, represents charge inside close surface is zero. Electric field as any point on the surface may be zero.
NCERT Page-33 / N-30
Electric Charges and Fields
272203
The electric field inside a spherical shell of uniform surface charge density is
1 zero
2 constant different from zero
3 proportional to the distance from the curve
4 None of the above
Explanation:
(a) Electric charge resides only on the surface of a spherical shell. According to Gauss's theorem the total electric flux over a closed surface is equal to the $\frac{1}{{{\varepsilon }_{0}}}$ times the total charge enclosed by the closed surface.
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Electric Charges and Fields
272200
Which statement is true for Gauss law-
1 All the charges whether inside or outside thegaussian surface contribute to the electric flux.
2 Electric flux depends upon the geometry of thegaussian surface.
3 Gauss theoremcan be applied to non-uniform electric field.
4 The electric field over thegaussian surface remains continuous and uniform at every point.
Explanation:
(d) All other statements except (iv) are in correct.
The electric field over the Gaussian surface remains continuous and uniform at every point.
NCERT Page-33 / N-29 | CBSE Sample 2021-2022
Electric Charges and Fields
272201
The total electric flux emanating from a closed surface enclosing an $\alpha $-particle is (e-electronic charge)
1 $\frac{2e}{{{\varepsilon }_{0}}}$
2 $\frac{e}{{{\varepsilon }_{0}}}$
3 $e{{\varepsilon }_{0}}$
4 $\frac{{{\varepsilon }_{0}}e}{4}$
Explanation:
(a) According to Gauss's law total electric flux through a closed surface is $\frac{1}{{{\varepsilon }_{0}}}$ times the total charge inside that surface.
Electric flux, ${{\phi }_{E}}=\frac{q}{{{\varepsilon }_{0}}}$
Charge on $\alpha $-particle $=2e~\therefore {{\phi }_{E}}=\frac{2e}{{{\varepsilon }_{0}}}$
NCERT Page-34 / N-30
Electric Charges and Fields
272202
For a given surface the Gauss's law is stated as $\oint ~\vec{E}\cdot d\vec{A}=0$. From this we can conclude that
1 $E$ is necessarily zero on the surface
2 $E$ is perpendicular to the surface at every point
3 the total flux through the surface is zero
4 the flux is only going out of the surface
Explanation:
(c) $\oint ~\vec{E}\cdot d\vec{A}=0$, represents charge inside close surface is zero. Electric field as any point on the surface may be zero.
NCERT Page-33 / N-30
Electric Charges and Fields
272203
The electric field inside a spherical shell of uniform surface charge density is
1 zero
2 constant different from zero
3 proportional to the distance from the curve
4 None of the above
Explanation:
(a) Electric charge resides only on the surface of a spherical shell. According to Gauss's theorem the total electric flux over a closed surface is equal to the $\frac{1}{{{\varepsilon }_{0}}}$ times the total charge enclosed by the closed surface.
1 All the charges whether inside or outside thegaussian surface contribute to the electric flux.
2 Electric flux depends upon the geometry of thegaussian surface.
3 Gauss theoremcan be applied to non-uniform electric field.
4 The electric field over thegaussian surface remains continuous and uniform at every point.
Explanation:
(d) All other statements except (iv) are in correct.
The electric field over the Gaussian surface remains continuous and uniform at every point.
NCERT Page-33 / N-29 | CBSE Sample 2021-2022
Electric Charges and Fields
272201
The total electric flux emanating from a closed surface enclosing an $\alpha $-particle is (e-electronic charge)
1 $\frac{2e}{{{\varepsilon }_{0}}}$
2 $\frac{e}{{{\varepsilon }_{0}}}$
3 $e{{\varepsilon }_{0}}$
4 $\frac{{{\varepsilon }_{0}}e}{4}$
Explanation:
(a) According to Gauss's law total electric flux through a closed surface is $\frac{1}{{{\varepsilon }_{0}}}$ times the total charge inside that surface.
Electric flux, ${{\phi }_{E}}=\frac{q}{{{\varepsilon }_{0}}}$
Charge on $\alpha $-particle $=2e~\therefore {{\phi }_{E}}=\frac{2e}{{{\varepsilon }_{0}}}$
NCERT Page-34 / N-30
Electric Charges and Fields
272202
For a given surface the Gauss's law is stated as $\oint ~\vec{E}\cdot d\vec{A}=0$. From this we can conclude that
1 $E$ is necessarily zero on the surface
2 $E$ is perpendicular to the surface at every point
3 the total flux through the surface is zero
4 the flux is only going out of the surface
Explanation:
(c) $\oint ~\vec{E}\cdot d\vec{A}=0$, represents charge inside close surface is zero. Electric field as any point on the surface may be zero.
NCERT Page-33 / N-30
Electric Charges and Fields
272203
The electric field inside a spherical shell of uniform surface charge density is
1 zero
2 constant different from zero
3 proportional to the distance from the curve
4 None of the above
Explanation:
(a) Electric charge resides only on the surface of a spherical shell. According to Gauss's theorem the total electric flux over a closed surface is equal to the $\frac{1}{{{\varepsilon }_{0}}}$ times the total charge enclosed by the closed surface.
1 All the charges whether inside or outside thegaussian surface contribute to the electric flux.
2 Electric flux depends upon the geometry of thegaussian surface.
3 Gauss theoremcan be applied to non-uniform electric field.
4 The electric field over thegaussian surface remains continuous and uniform at every point.
Explanation:
(d) All other statements except (iv) are in correct.
The electric field over the Gaussian surface remains continuous and uniform at every point.
NCERT Page-33 / N-29 | CBSE Sample 2021-2022
Electric Charges and Fields
272201
The total electric flux emanating from a closed surface enclosing an $\alpha $-particle is (e-electronic charge)
1 $\frac{2e}{{{\varepsilon }_{0}}}$
2 $\frac{e}{{{\varepsilon }_{0}}}$
3 $e{{\varepsilon }_{0}}$
4 $\frac{{{\varepsilon }_{0}}e}{4}$
Explanation:
(a) According to Gauss's law total electric flux through a closed surface is $\frac{1}{{{\varepsilon }_{0}}}$ times the total charge inside that surface.
Electric flux, ${{\phi }_{E}}=\frac{q}{{{\varepsilon }_{0}}}$
Charge on $\alpha $-particle $=2e~\therefore {{\phi }_{E}}=\frac{2e}{{{\varepsilon }_{0}}}$
NCERT Page-34 / N-30
Electric Charges and Fields
272202
For a given surface the Gauss's law is stated as $\oint ~\vec{E}\cdot d\vec{A}=0$. From this we can conclude that
1 $E$ is necessarily zero on the surface
2 $E$ is perpendicular to the surface at every point
3 the total flux through the surface is zero
4 the flux is only going out of the surface
Explanation:
(c) $\oint ~\vec{E}\cdot d\vec{A}=0$, represents charge inside close surface is zero. Electric field as any point on the surface may be zero.
NCERT Page-33 / N-30
Electric Charges and Fields
272203
The electric field inside a spherical shell of uniform surface charge density is
1 zero
2 constant different from zero
3 proportional to the distance from the curve
4 None of the above
Explanation:
(a) Electric charge resides only on the surface of a spherical shell. According to Gauss's theorem the total electric flux over a closed surface is equal to the $\frac{1}{{{\varepsilon }_{0}}}$ times the total charge enclosed by the closed surface.