359545
The variation of electric potential with distance from a fixed point is shown in the figure. What is the value of electric field \({(V / m)}\) at \({x=2 m}\)
1 Zero
2 \({6 / 2}\)
3 \({6 / 1}\)
4 \({6 / 3}\)
Explanation:
from \({x=0}\) to \({x=1 m}\) \(E=\dfrac{(6-2)}{(1-0)}=4 V / m\) from \({x=1 m}\) to \({x=2 m}\), \(E=0 V / m\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359546
The potential of the electric field produced by point charge at any point (\(x\), \(y\), \(z\)) is given by \( - 2{x^2}\) where \(x\), \(y\) are in metre and \(V\) is in volts. The intensity of the electric field at \(( - 2,\,1,\,0)\) is :
1 \( - 17V{m^{ - 1}}\)
2 \( + 17V{m^{ - 1}}\)
3 \( - 8V{m^{ - 1}}\)
4 \( + 12V{m^{ - 1}}\)
Explanation:
Given that \(V = - 2{x^2}\) Intensity of the electric field \({E_x} = \frac{{ - dV}}{{dx}} = 4x = 4\left( { - 2} \right)\) \((E\,at\,x = - 2)\) \( = - 8V{m^{ - 1}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359547
A uniform electric field having a magnitude \({E_0}\) and direction along the positive \(X\)-axis exists . If the potential \(V\) is zero at \(x = 0\) , then its value at \(X = + x\) will be:
359548
The ionisation potential of mercury is \(10.39\;V\). How far an electron must travel in an electric field of \(1.5 \times {10^6}\;V/m\) to gain sufficient energy to ionise mercury?
Ionisation potential \((V)\) of mercury is the energy required to strip it of an electron. The electric field strength is given by, \(E=\dfrac{V}{d}\) where, \(d\) is distance between plates creating electric field. Given, \(V=10.39 V\) \(E = 1.5 \times {10^6}V{m^{ - 1}}\) \(\therefore d = \frac{V}{E} = \frac{{10.39}}{{1.5 \times {{10}^6}}}\;m\) Hence, distance travelled by electron to gain ionisation energy is \(=\dfrac{10.39}{1.5 \times 10^{6}} m\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359549
The electric potential in a region is represented as \({V=2 x+3 y-z}\). Obtain expression for the electric field strength.
359545
The variation of electric potential with distance from a fixed point is shown in the figure. What is the value of electric field \({(V / m)}\) at \({x=2 m}\)
1 Zero
2 \({6 / 2}\)
3 \({6 / 1}\)
4 \({6 / 3}\)
Explanation:
from \({x=0}\) to \({x=1 m}\) \(E=\dfrac{(6-2)}{(1-0)}=4 V / m\) from \({x=1 m}\) to \({x=2 m}\), \(E=0 V / m\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359546
The potential of the electric field produced by point charge at any point (\(x\), \(y\), \(z\)) is given by \( - 2{x^2}\) where \(x\), \(y\) are in metre and \(V\) is in volts. The intensity of the electric field at \(( - 2,\,1,\,0)\) is :
1 \( - 17V{m^{ - 1}}\)
2 \( + 17V{m^{ - 1}}\)
3 \( - 8V{m^{ - 1}}\)
4 \( + 12V{m^{ - 1}}\)
Explanation:
Given that \(V = - 2{x^2}\) Intensity of the electric field \({E_x} = \frac{{ - dV}}{{dx}} = 4x = 4\left( { - 2} \right)\) \((E\,at\,x = - 2)\) \( = - 8V{m^{ - 1}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359547
A uniform electric field having a magnitude \({E_0}\) and direction along the positive \(X\)-axis exists . If the potential \(V\) is zero at \(x = 0\) , then its value at \(X = + x\) will be:
359548
The ionisation potential of mercury is \(10.39\;V\). How far an electron must travel in an electric field of \(1.5 \times {10^6}\;V/m\) to gain sufficient energy to ionise mercury?
Ionisation potential \((V)\) of mercury is the energy required to strip it of an electron. The electric field strength is given by, \(E=\dfrac{V}{d}\) where, \(d\) is distance between plates creating electric field. Given, \(V=10.39 V\) \(E = 1.5 \times {10^6}V{m^{ - 1}}\) \(\therefore d = \frac{V}{E} = \frac{{10.39}}{{1.5 \times {{10}^6}}}\;m\) Hence, distance travelled by electron to gain ionisation energy is \(=\dfrac{10.39}{1.5 \times 10^{6}} m\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359549
The electric potential in a region is represented as \({V=2 x+3 y-z}\). Obtain expression for the electric field strength.
359545
The variation of electric potential with distance from a fixed point is shown in the figure. What is the value of electric field \({(V / m)}\) at \({x=2 m}\)
1 Zero
2 \({6 / 2}\)
3 \({6 / 1}\)
4 \({6 / 3}\)
Explanation:
from \({x=0}\) to \({x=1 m}\) \(E=\dfrac{(6-2)}{(1-0)}=4 V / m\) from \({x=1 m}\) to \({x=2 m}\), \(E=0 V / m\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359546
The potential of the electric field produced by point charge at any point (\(x\), \(y\), \(z\)) is given by \( - 2{x^2}\) where \(x\), \(y\) are in metre and \(V\) is in volts. The intensity of the electric field at \(( - 2,\,1,\,0)\) is :
1 \( - 17V{m^{ - 1}}\)
2 \( + 17V{m^{ - 1}}\)
3 \( - 8V{m^{ - 1}}\)
4 \( + 12V{m^{ - 1}}\)
Explanation:
Given that \(V = - 2{x^2}\) Intensity of the electric field \({E_x} = \frac{{ - dV}}{{dx}} = 4x = 4\left( { - 2} \right)\) \((E\,at\,x = - 2)\) \( = - 8V{m^{ - 1}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359547
A uniform electric field having a magnitude \({E_0}\) and direction along the positive \(X\)-axis exists . If the potential \(V\) is zero at \(x = 0\) , then its value at \(X = + x\) will be:
359548
The ionisation potential of mercury is \(10.39\;V\). How far an electron must travel in an electric field of \(1.5 \times {10^6}\;V/m\) to gain sufficient energy to ionise mercury?
Ionisation potential \((V)\) of mercury is the energy required to strip it of an electron. The electric field strength is given by, \(E=\dfrac{V}{d}\) where, \(d\) is distance between plates creating electric field. Given, \(V=10.39 V\) \(E = 1.5 \times {10^6}V{m^{ - 1}}\) \(\therefore d = \frac{V}{E} = \frac{{10.39}}{{1.5 \times {{10}^6}}}\;m\) Hence, distance travelled by electron to gain ionisation energy is \(=\dfrac{10.39}{1.5 \times 10^{6}} m\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359549
The electric potential in a region is represented as \({V=2 x+3 y-z}\). Obtain expression for the electric field strength.
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359545
The variation of electric potential with distance from a fixed point is shown in the figure. What is the value of electric field \({(V / m)}\) at \({x=2 m}\)
1 Zero
2 \({6 / 2}\)
3 \({6 / 1}\)
4 \({6 / 3}\)
Explanation:
from \({x=0}\) to \({x=1 m}\) \(E=\dfrac{(6-2)}{(1-0)}=4 V / m\) from \({x=1 m}\) to \({x=2 m}\), \(E=0 V / m\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359546
The potential of the electric field produced by point charge at any point (\(x\), \(y\), \(z\)) is given by \( - 2{x^2}\) where \(x\), \(y\) are in metre and \(V\) is in volts. The intensity of the electric field at \(( - 2,\,1,\,0)\) is :
1 \( - 17V{m^{ - 1}}\)
2 \( + 17V{m^{ - 1}}\)
3 \( - 8V{m^{ - 1}}\)
4 \( + 12V{m^{ - 1}}\)
Explanation:
Given that \(V = - 2{x^2}\) Intensity of the electric field \({E_x} = \frac{{ - dV}}{{dx}} = 4x = 4\left( { - 2} \right)\) \((E\,at\,x = - 2)\) \( = - 8V{m^{ - 1}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359547
A uniform electric field having a magnitude \({E_0}\) and direction along the positive \(X\)-axis exists . If the potential \(V\) is zero at \(x = 0\) , then its value at \(X = + x\) will be:
359548
The ionisation potential of mercury is \(10.39\;V\). How far an electron must travel in an electric field of \(1.5 \times {10^6}\;V/m\) to gain sufficient energy to ionise mercury?
Ionisation potential \((V)\) of mercury is the energy required to strip it of an electron. The electric field strength is given by, \(E=\dfrac{V}{d}\) where, \(d\) is distance between plates creating electric field. Given, \(V=10.39 V\) \(E = 1.5 \times {10^6}V{m^{ - 1}}\) \(\therefore d = \frac{V}{E} = \frac{{10.39}}{{1.5 \times {{10}^6}}}\;m\) Hence, distance travelled by electron to gain ionisation energy is \(=\dfrac{10.39}{1.5 \times 10^{6}} m\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359549
The electric potential in a region is represented as \({V=2 x+3 y-z}\). Obtain expression for the electric field strength.
359545
The variation of electric potential with distance from a fixed point is shown in the figure. What is the value of electric field \({(V / m)}\) at \({x=2 m}\)
1 Zero
2 \({6 / 2}\)
3 \({6 / 1}\)
4 \({6 / 3}\)
Explanation:
from \({x=0}\) to \({x=1 m}\) \(E=\dfrac{(6-2)}{(1-0)}=4 V / m\) from \({x=1 m}\) to \({x=2 m}\), \(E=0 V / m\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359546
The potential of the electric field produced by point charge at any point (\(x\), \(y\), \(z\)) is given by \( - 2{x^2}\) where \(x\), \(y\) are in metre and \(V\) is in volts. The intensity of the electric field at \(( - 2,\,1,\,0)\) is :
1 \( - 17V{m^{ - 1}}\)
2 \( + 17V{m^{ - 1}}\)
3 \( - 8V{m^{ - 1}}\)
4 \( + 12V{m^{ - 1}}\)
Explanation:
Given that \(V = - 2{x^2}\) Intensity of the electric field \({E_x} = \frac{{ - dV}}{{dx}} = 4x = 4\left( { - 2} \right)\) \((E\,at\,x = - 2)\) \( = - 8V{m^{ - 1}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359547
A uniform electric field having a magnitude \({E_0}\) and direction along the positive \(X\)-axis exists . If the potential \(V\) is zero at \(x = 0\) , then its value at \(X = + x\) will be:
359548
The ionisation potential of mercury is \(10.39\;V\). How far an electron must travel in an electric field of \(1.5 \times {10^6}\;V/m\) to gain sufficient energy to ionise mercury?
Ionisation potential \((V)\) of mercury is the energy required to strip it of an electron. The electric field strength is given by, \(E=\dfrac{V}{d}\) where, \(d\) is distance between plates creating electric field. Given, \(V=10.39 V\) \(E = 1.5 \times {10^6}V{m^{ - 1}}\) \(\therefore d = \frac{V}{E} = \frac{{10.39}}{{1.5 \times {{10}^6}}}\;m\) Hence, distance travelled by electron to gain ionisation energy is \(=\dfrac{10.39}{1.5 \times 10^{6}} m\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359549
The electric potential in a region is represented as \({V=2 x+3 y-z}\). Obtain expression for the electric field strength.
