359567
The electric potential at any point \(x,y,z\) in metres is given by \(V = 3{x^2}\). The electric field at a point \((2m,0,1\,m)\) is
1 \(12\,V{m^{ - 1}}\)
2 \( - 6\,V{m^{ - 1}}\)
3 \(6\,V{m^{ - 1}}\)
4 \( - 12V{m^{ - 1}}\)
Explanation:
Here, electric potential \(V = 3{x^2}\) Since potential is independent of \(y\) and \(z\), electric field components in those two directions are zero. Hence \(E = - \frac{{dV}}{{dx}} = - \frac{d}{{dx}}(3{x^2}) = - 6x\) \({E_{(2m.0m.1m)}} = - 12\,V\,{m^{ - 1}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359568
A 5 coulomb charge experiences a constant force of \(2000\,N\) when moved between two points, separated by a distance of \(2\,cm\), in a uniform electric field. The potential difference between these two points is
359569
The electric potential at a point in free space due to a charge \(Q\) coulomb is \(Q \times {10^{11}}V.\) The electric field at the point is
1 \(4\pi {\varepsilon _0}Q \times {10^{22}}V/m\)
2 \(12\pi {\varepsilon _0}Q \times {10^{22}}V/m\)
3 \(4\pi {\varepsilon _0}Q \times {10^{20}}V/m\)
4 \(12\pi {\varepsilon _0}Q \times {10^{20}}V/m\)
Explanation:
As potential at a point due to a point charge is \(V = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{r}\) The electric field is \(E = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{{{r^2}}}\) Given that \(V = Q \times {10^{11}}\) From the above equations \(E = 4\pi {\varepsilon _o}Q \times {10^{22}}Vm\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359570
The electric potential at a point \(({\rm{x}},y,z)\) is given by \(V = - {{\rm{x}}^2}y - {\rm{x}}{z^3} + 4\) The electric field \(\overrightarrow E \) at that point is
359567
The electric potential at any point \(x,y,z\) in metres is given by \(V = 3{x^2}\). The electric field at a point \((2m,0,1\,m)\) is
1 \(12\,V{m^{ - 1}}\)
2 \( - 6\,V{m^{ - 1}}\)
3 \(6\,V{m^{ - 1}}\)
4 \( - 12V{m^{ - 1}}\)
Explanation:
Here, electric potential \(V = 3{x^2}\) Since potential is independent of \(y\) and \(z\), electric field components in those two directions are zero. Hence \(E = - \frac{{dV}}{{dx}} = - \frac{d}{{dx}}(3{x^2}) = - 6x\) \({E_{(2m.0m.1m)}} = - 12\,V\,{m^{ - 1}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359568
A 5 coulomb charge experiences a constant force of \(2000\,N\) when moved between two points, separated by a distance of \(2\,cm\), in a uniform electric field. The potential difference between these two points is
359569
The electric potential at a point in free space due to a charge \(Q\) coulomb is \(Q \times {10^{11}}V.\) The electric field at the point is
1 \(4\pi {\varepsilon _0}Q \times {10^{22}}V/m\)
2 \(12\pi {\varepsilon _0}Q \times {10^{22}}V/m\)
3 \(4\pi {\varepsilon _0}Q \times {10^{20}}V/m\)
4 \(12\pi {\varepsilon _0}Q \times {10^{20}}V/m\)
Explanation:
As potential at a point due to a point charge is \(V = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{r}\) The electric field is \(E = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{{{r^2}}}\) Given that \(V = Q \times {10^{11}}\) From the above equations \(E = 4\pi {\varepsilon _o}Q \times {10^{22}}Vm\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359570
The electric potential at a point \(({\rm{x}},y,z)\) is given by \(V = - {{\rm{x}}^2}y - {\rm{x}}{z^3} + 4\) The electric field \(\overrightarrow E \) at that point is
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359567
The electric potential at any point \(x,y,z\) in metres is given by \(V = 3{x^2}\). The electric field at a point \((2m,0,1\,m)\) is
1 \(12\,V{m^{ - 1}}\)
2 \( - 6\,V{m^{ - 1}}\)
3 \(6\,V{m^{ - 1}}\)
4 \( - 12V{m^{ - 1}}\)
Explanation:
Here, electric potential \(V = 3{x^2}\) Since potential is independent of \(y\) and \(z\), electric field components in those two directions are zero. Hence \(E = - \frac{{dV}}{{dx}} = - \frac{d}{{dx}}(3{x^2}) = - 6x\) \({E_{(2m.0m.1m)}} = - 12\,V\,{m^{ - 1}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359568
A 5 coulomb charge experiences a constant force of \(2000\,N\) when moved between two points, separated by a distance of \(2\,cm\), in a uniform electric field. The potential difference between these two points is
359569
The electric potential at a point in free space due to a charge \(Q\) coulomb is \(Q \times {10^{11}}V.\) The electric field at the point is
1 \(4\pi {\varepsilon _0}Q \times {10^{22}}V/m\)
2 \(12\pi {\varepsilon _0}Q \times {10^{22}}V/m\)
3 \(4\pi {\varepsilon _0}Q \times {10^{20}}V/m\)
4 \(12\pi {\varepsilon _0}Q \times {10^{20}}V/m\)
Explanation:
As potential at a point due to a point charge is \(V = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{r}\) The electric field is \(E = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{{{r^2}}}\) Given that \(V = Q \times {10^{11}}\) From the above equations \(E = 4\pi {\varepsilon _o}Q \times {10^{22}}Vm\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359570
The electric potential at a point \(({\rm{x}},y,z)\) is given by \(V = - {{\rm{x}}^2}y - {\rm{x}}{z^3} + 4\) The electric field \(\overrightarrow E \) at that point is
359567
The electric potential at any point \(x,y,z\) in metres is given by \(V = 3{x^2}\). The electric field at a point \((2m,0,1\,m)\) is
1 \(12\,V{m^{ - 1}}\)
2 \( - 6\,V{m^{ - 1}}\)
3 \(6\,V{m^{ - 1}}\)
4 \( - 12V{m^{ - 1}}\)
Explanation:
Here, electric potential \(V = 3{x^2}\) Since potential is independent of \(y\) and \(z\), electric field components in those two directions are zero. Hence \(E = - \frac{{dV}}{{dx}} = - \frac{d}{{dx}}(3{x^2}) = - 6x\) \({E_{(2m.0m.1m)}} = - 12\,V\,{m^{ - 1}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359568
A 5 coulomb charge experiences a constant force of \(2000\,N\) when moved between two points, separated by a distance of \(2\,cm\), in a uniform electric field. The potential difference between these two points is
359569
The electric potential at a point in free space due to a charge \(Q\) coulomb is \(Q \times {10^{11}}V.\) The electric field at the point is
1 \(4\pi {\varepsilon _0}Q \times {10^{22}}V/m\)
2 \(12\pi {\varepsilon _0}Q \times {10^{22}}V/m\)
3 \(4\pi {\varepsilon _0}Q \times {10^{20}}V/m\)
4 \(12\pi {\varepsilon _0}Q \times {10^{20}}V/m\)
Explanation:
As potential at a point due to a point charge is \(V = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{r}\) The electric field is \(E = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{{{r^2}}}\) Given that \(V = Q \times {10^{11}}\) From the above equations \(E = 4\pi {\varepsilon _o}Q \times {10^{22}}Vm\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359570
The electric potential at a point \(({\rm{x}},y,z)\) is given by \(V = - {{\rm{x}}^2}y - {\rm{x}}{z^3} + 4\) The electric field \(\overrightarrow E \) at that point is