272238
A unit charge moves on an equipotential surface from a point \(A\) to point \(B\), then
1 \(V_A-V_B=+v e\)
2 \(V_A-V_B=0\)
3 \(V_A-V_B=-v e\)
4 it is stationary
Explanation:
(b)
NCERT Pige- 60 / N-54
Electrostatic Potentials and Capacitance
272241
Which of the following is NOT the property of equipotential surface?
1 They do not cross each other.
2 The rate of change of potential with distance on them is zero.
3 For a uniform electric field they are concentric spheres.
4 They can be imaginary spheres.
Explanation:
(c) As all other statements are correct. In uniform electric field equipotential surfaces are never concentric spheres but are planes \(\perp\) to Electric field lines.
NCERT Page-60 / N-54|CBSE Sample 2021-2022
Electrostatic Potentials and Capacitance
272243
The potential at a point \(x\) (measured in \(m\) \} due to some charges situated on the \(x\)-axis is given by \(V(x)=\) \(20 /\left(x^2-4\right)\) volt. The electric field E at \(x=4 \mu \mathrm{~m}\) is given by
1 (10/9)volt/ \(/ \mu \mathrm{m}\) and in the +vex direction
2 (5/3) volt/ / m and in the -vex direction
3 (5/3)volt \(/ \mu \mathrm{m}\) and in the \(+v e x\) direction
4 (10/9) volt\(/ \mu \mathrm{m}\) and in the -ve \(x\) direction
Explanation:
(a) Here, \(V(x)=\frac{20}{x^2-4}\) volt
We know that \(E=-\frac{d V}{d x}=-\frac{d}{d x}\left(\frac{20}{x^2-4}\right)\)
or, \(E=+\frac{40 x}{\left(x^2-4\right)^2}\)
At \(x=4 \mu \mathrm{~m}\),
\(\mathrm{E}=+\frac{40 \times 4}{\left(4^2-4\right)^2}=+\frac{160}{144}=+\frac{10}{9}\) volt \(/ \mu \mathrm{m}\).
Positive sign indicates that \(\overrightarrow{\mathrm{E}}\) is in +ve x-direction.
NCERT Page-61/N-55
Electrostatic Potentials and Capacitance
272244
The expression \(E=-\frac{d v}{d T}\) implies, that electric field is in that direction in which
1 increase in potential is steepest.
2 decrease in potential is steepest.
3 change in potential is minimum.
4 None of these
Explanation:
(b) As we move towards a positive charge distribution \(V\) increases i.e., \(\frac{d v}{d r}\) is positive. The increase in potential is steepest when we move exactly towards charge distribution. But Eis in a direction exactly away from charge distribution, therefore E is in exactly opposite direction in which increase in potential is steepest. Hence \(E=-\frac{d V}{d r}\).
NCERT Page-61/N-55
Electrostatic Potentials and Capacitance
272245
From a point charge, there is a fixed point A. At A, there is an electric field of \(500 \mathrm{~V} / \mathrm{m}\) and potential difference of 3000 V . Distance between point charge and A will be
1 6 m
2 12 m
3 16 m
4 24 m
Explanation:
\(\{\mathrm{a}\} \mathrm{E}=500 \mathrm{~V} / \mathrm{m}\)
\(\mathrm{V}=3000 \mathrm{~V}\).
We know that electric field \((E)=500=\frac{V}{d}\)
or \(d=\frac{3000}{500}=6 \mathrm{~m}\)
272238
A unit charge moves on an equipotential surface from a point \(A\) to point \(B\), then
1 \(V_A-V_B=+v e\)
2 \(V_A-V_B=0\)
3 \(V_A-V_B=-v e\)
4 it is stationary
Explanation:
(b)
NCERT Pige- 60 / N-54
Electrostatic Potentials and Capacitance
272241
Which of the following is NOT the property of equipotential surface?
1 They do not cross each other.
2 The rate of change of potential with distance on them is zero.
3 For a uniform electric field they are concentric spheres.
4 They can be imaginary spheres.
Explanation:
(c) As all other statements are correct. In uniform electric field equipotential surfaces are never concentric spheres but are planes \(\perp\) to Electric field lines.
NCERT Page-60 / N-54|CBSE Sample 2021-2022
Electrostatic Potentials and Capacitance
272243
The potential at a point \(x\) (measured in \(m\) \} due to some charges situated on the \(x\)-axis is given by \(V(x)=\) \(20 /\left(x^2-4\right)\) volt. The electric field E at \(x=4 \mu \mathrm{~m}\) is given by
1 (10/9)volt/ \(/ \mu \mathrm{m}\) and in the +vex direction
2 (5/3) volt/ / m and in the -vex direction
3 (5/3)volt \(/ \mu \mathrm{m}\) and in the \(+v e x\) direction
4 (10/9) volt\(/ \mu \mathrm{m}\) and in the -ve \(x\) direction
Explanation:
(a) Here, \(V(x)=\frac{20}{x^2-4}\) volt
We know that \(E=-\frac{d V}{d x}=-\frac{d}{d x}\left(\frac{20}{x^2-4}\right)\)
or, \(E=+\frac{40 x}{\left(x^2-4\right)^2}\)
At \(x=4 \mu \mathrm{~m}\),
\(\mathrm{E}=+\frac{40 \times 4}{\left(4^2-4\right)^2}=+\frac{160}{144}=+\frac{10}{9}\) volt \(/ \mu \mathrm{m}\).
Positive sign indicates that \(\overrightarrow{\mathrm{E}}\) is in +ve x-direction.
NCERT Page-61/N-55
Electrostatic Potentials and Capacitance
272244
The expression \(E=-\frac{d v}{d T}\) implies, that electric field is in that direction in which
1 increase in potential is steepest.
2 decrease in potential is steepest.
3 change in potential is minimum.
4 None of these
Explanation:
(b) As we move towards a positive charge distribution \(V\) increases i.e., \(\frac{d v}{d r}\) is positive. The increase in potential is steepest when we move exactly towards charge distribution. But Eis in a direction exactly away from charge distribution, therefore E is in exactly opposite direction in which increase in potential is steepest. Hence \(E=-\frac{d V}{d r}\).
NCERT Page-61/N-55
Electrostatic Potentials and Capacitance
272245
From a point charge, there is a fixed point A. At A, there is an electric field of \(500 \mathrm{~V} / \mathrm{m}\) and potential difference of 3000 V . Distance between point charge and A will be
1 6 m
2 12 m
3 16 m
4 24 m
Explanation:
\(\{\mathrm{a}\} \mathrm{E}=500 \mathrm{~V} / \mathrm{m}\)
\(\mathrm{V}=3000 \mathrm{~V}\).
We know that electric field \((E)=500=\frac{V}{d}\)
or \(d=\frac{3000}{500}=6 \mathrm{~m}\)
272238
A unit charge moves on an equipotential surface from a point \(A\) to point \(B\), then
1 \(V_A-V_B=+v e\)
2 \(V_A-V_B=0\)
3 \(V_A-V_B=-v e\)
4 it is stationary
Explanation:
(b)
NCERT Pige- 60 / N-54
Electrostatic Potentials and Capacitance
272241
Which of the following is NOT the property of equipotential surface?
1 They do not cross each other.
2 The rate of change of potential with distance on them is zero.
3 For a uniform electric field they are concentric spheres.
4 They can be imaginary spheres.
Explanation:
(c) As all other statements are correct. In uniform electric field equipotential surfaces are never concentric spheres but are planes \(\perp\) to Electric field lines.
NCERT Page-60 / N-54|CBSE Sample 2021-2022
Electrostatic Potentials and Capacitance
272243
The potential at a point \(x\) (measured in \(m\) \} due to some charges situated on the \(x\)-axis is given by \(V(x)=\) \(20 /\left(x^2-4\right)\) volt. The electric field E at \(x=4 \mu \mathrm{~m}\) is given by
1 (10/9)volt/ \(/ \mu \mathrm{m}\) and in the +vex direction
2 (5/3) volt/ / m and in the -vex direction
3 (5/3)volt \(/ \mu \mathrm{m}\) and in the \(+v e x\) direction
4 (10/9) volt\(/ \mu \mathrm{m}\) and in the -ve \(x\) direction
Explanation:
(a) Here, \(V(x)=\frac{20}{x^2-4}\) volt
We know that \(E=-\frac{d V}{d x}=-\frac{d}{d x}\left(\frac{20}{x^2-4}\right)\)
or, \(E=+\frac{40 x}{\left(x^2-4\right)^2}\)
At \(x=4 \mu \mathrm{~m}\),
\(\mathrm{E}=+\frac{40 \times 4}{\left(4^2-4\right)^2}=+\frac{160}{144}=+\frac{10}{9}\) volt \(/ \mu \mathrm{m}\).
Positive sign indicates that \(\overrightarrow{\mathrm{E}}\) is in +ve x-direction.
NCERT Page-61/N-55
Electrostatic Potentials and Capacitance
272244
The expression \(E=-\frac{d v}{d T}\) implies, that electric field is in that direction in which
1 increase in potential is steepest.
2 decrease in potential is steepest.
3 change in potential is minimum.
4 None of these
Explanation:
(b) As we move towards a positive charge distribution \(V\) increases i.e., \(\frac{d v}{d r}\) is positive. The increase in potential is steepest when we move exactly towards charge distribution. But Eis in a direction exactly away from charge distribution, therefore E is in exactly opposite direction in which increase in potential is steepest. Hence \(E=-\frac{d V}{d r}\).
NCERT Page-61/N-55
Electrostatic Potentials and Capacitance
272245
From a point charge, there is a fixed point A. At A, there is an electric field of \(500 \mathrm{~V} / \mathrm{m}\) and potential difference of 3000 V . Distance between point charge and A will be
1 6 m
2 12 m
3 16 m
4 24 m
Explanation:
\(\{\mathrm{a}\} \mathrm{E}=500 \mathrm{~V} / \mathrm{m}\)
\(\mathrm{V}=3000 \mathrm{~V}\).
We know that electric field \((E)=500=\frac{V}{d}\)
or \(d=\frac{3000}{500}=6 \mathrm{~m}\)
272238
A unit charge moves on an equipotential surface from a point \(A\) to point \(B\), then
1 \(V_A-V_B=+v e\)
2 \(V_A-V_B=0\)
3 \(V_A-V_B=-v e\)
4 it is stationary
Explanation:
(b)
NCERT Pige- 60 / N-54
Electrostatic Potentials and Capacitance
272241
Which of the following is NOT the property of equipotential surface?
1 They do not cross each other.
2 The rate of change of potential with distance on them is zero.
3 For a uniform electric field they are concentric spheres.
4 They can be imaginary spheres.
Explanation:
(c) As all other statements are correct. In uniform electric field equipotential surfaces are never concentric spheres but are planes \(\perp\) to Electric field lines.
NCERT Page-60 / N-54|CBSE Sample 2021-2022
Electrostatic Potentials and Capacitance
272243
The potential at a point \(x\) (measured in \(m\) \} due to some charges situated on the \(x\)-axis is given by \(V(x)=\) \(20 /\left(x^2-4\right)\) volt. The electric field E at \(x=4 \mu \mathrm{~m}\) is given by
1 (10/9)volt/ \(/ \mu \mathrm{m}\) and in the +vex direction
2 (5/3) volt/ / m and in the -vex direction
3 (5/3)volt \(/ \mu \mathrm{m}\) and in the \(+v e x\) direction
4 (10/9) volt\(/ \mu \mathrm{m}\) and in the -ve \(x\) direction
Explanation:
(a) Here, \(V(x)=\frac{20}{x^2-4}\) volt
We know that \(E=-\frac{d V}{d x}=-\frac{d}{d x}\left(\frac{20}{x^2-4}\right)\)
or, \(E=+\frac{40 x}{\left(x^2-4\right)^2}\)
At \(x=4 \mu \mathrm{~m}\),
\(\mathrm{E}=+\frac{40 \times 4}{\left(4^2-4\right)^2}=+\frac{160}{144}=+\frac{10}{9}\) volt \(/ \mu \mathrm{m}\).
Positive sign indicates that \(\overrightarrow{\mathrm{E}}\) is in +ve x-direction.
NCERT Page-61/N-55
Electrostatic Potentials and Capacitance
272244
The expression \(E=-\frac{d v}{d T}\) implies, that electric field is in that direction in which
1 increase in potential is steepest.
2 decrease in potential is steepest.
3 change in potential is minimum.
4 None of these
Explanation:
(b) As we move towards a positive charge distribution \(V\) increases i.e., \(\frac{d v}{d r}\) is positive. The increase in potential is steepest when we move exactly towards charge distribution. But Eis in a direction exactly away from charge distribution, therefore E is in exactly opposite direction in which increase in potential is steepest. Hence \(E=-\frac{d V}{d r}\).
NCERT Page-61/N-55
Electrostatic Potentials and Capacitance
272245
From a point charge, there is a fixed point A. At A, there is an electric field of \(500 \mathrm{~V} / \mathrm{m}\) and potential difference of 3000 V . Distance between point charge and A will be
1 6 m
2 12 m
3 16 m
4 24 m
Explanation:
\(\{\mathrm{a}\} \mathrm{E}=500 \mathrm{~V} / \mathrm{m}\)
\(\mathrm{V}=3000 \mathrm{~V}\).
We know that electric field \((E)=500=\frac{V}{d}\)
or \(d=\frac{3000}{500}=6 \mathrm{~m}\)
272238
A unit charge moves on an equipotential surface from a point \(A\) to point \(B\), then
1 \(V_A-V_B=+v e\)
2 \(V_A-V_B=0\)
3 \(V_A-V_B=-v e\)
4 it is stationary
Explanation:
(b)
NCERT Pige- 60 / N-54
Electrostatic Potentials and Capacitance
272241
Which of the following is NOT the property of equipotential surface?
1 They do not cross each other.
2 The rate of change of potential with distance on them is zero.
3 For a uniform electric field they are concentric spheres.
4 They can be imaginary spheres.
Explanation:
(c) As all other statements are correct. In uniform electric field equipotential surfaces are never concentric spheres but are planes \(\perp\) to Electric field lines.
NCERT Page-60 / N-54|CBSE Sample 2021-2022
Electrostatic Potentials and Capacitance
272243
The potential at a point \(x\) (measured in \(m\) \} due to some charges situated on the \(x\)-axis is given by \(V(x)=\) \(20 /\left(x^2-4\right)\) volt. The electric field E at \(x=4 \mu \mathrm{~m}\) is given by
1 (10/9)volt/ \(/ \mu \mathrm{m}\) and in the +vex direction
2 (5/3) volt/ / m and in the -vex direction
3 (5/3)volt \(/ \mu \mathrm{m}\) and in the \(+v e x\) direction
4 (10/9) volt\(/ \mu \mathrm{m}\) and in the -ve \(x\) direction
Explanation:
(a) Here, \(V(x)=\frac{20}{x^2-4}\) volt
We know that \(E=-\frac{d V}{d x}=-\frac{d}{d x}\left(\frac{20}{x^2-4}\right)\)
or, \(E=+\frac{40 x}{\left(x^2-4\right)^2}\)
At \(x=4 \mu \mathrm{~m}\),
\(\mathrm{E}=+\frac{40 \times 4}{\left(4^2-4\right)^2}=+\frac{160}{144}=+\frac{10}{9}\) volt \(/ \mu \mathrm{m}\).
Positive sign indicates that \(\overrightarrow{\mathrm{E}}\) is in +ve x-direction.
NCERT Page-61/N-55
Electrostatic Potentials and Capacitance
272244
The expression \(E=-\frac{d v}{d T}\) implies, that electric field is in that direction in which
1 increase in potential is steepest.
2 decrease in potential is steepest.
3 change in potential is minimum.
4 None of these
Explanation:
(b) As we move towards a positive charge distribution \(V\) increases i.e., \(\frac{d v}{d r}\) is positive. The increase in potential is steepest when we move exactly towards charge distribution. But Eis in a direction exactly away from charge distribution, therefore E is in exactly opposite direction in which increase in potential is steepest. Hence \(E=-\frac{d V}{d r}\).
NCERT Page-61/N-55
Electrostatic Potentials and Capacitance
272245
From a point charge, there is a fixed point A. At A, there is an electric field of \(500 \mathrm{~V} / \mathrm{m}\) and potential difference of 3000 V . Distance between point charge and A will be
1 6 m
2 12 m
3 16 m
4 24 m
Explanation:
\(\{\mathrm{a}\} \mathrm{E}=500 \mathrm{~V} / \mathrm{m}\)
\(\mathrm{V}=3000 \mathrm{~V}\).
We know that electric field \((E)=500=\frac{V}{d}\)
or \(d=\frac{3000}{500}=6 \mathrm{~m}\)
