359550
The electric field in a region is given by the vector \(\overrightarrow E = (4\hat i - 1\hat j)\left( {\frac{N}{C}} \right).\)The potential will be constant along
1 The line \(3y = 4x\)
2 \(Y\) - axis
3 The line \(y = 4x\)
4 \(X\) - axis
Explanation:
The electric field vector is \({\mathop{\rm Tan}\nolimits} \alpha = \frac{4}{1}\) \(y = 4x\) is the line along which the potential is constant
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359551
A field of \(100\,V{m^{ - 1}}\) is directed at \({30^o}\) to positive \(x\)-axis. Find \({V_A} - {V_B}\) if \(OA = 2\,m\) and \(OB = 4\,m\)
1 \(100(\sqrt 3 - 2)V\)
2 \(200(2 + \sqrt 3 )V\)
3 \(100(2 + \sqrt 3 )V\)
4 \(100(2 - \sqrt 3 )V\)
Explanation:
We know that \({V_2} - {V_1} = - \int_1^2 {\overline E \,.d\overline r } \) Given that \(\overline E = 50\sqrt 3 \,\hat i + 50\,\hat j\) \({V_A} - {V_0} = - \int\limits_{{\rm{x}} = 0}^{{\rm{x}} = 2} {\overline E \,.d\overline r } = - {E_{\rm{x}}}\,\Delta {\rm{x}} = - 100\sqrt 3 \) \({V_B} - {V_0} = - \int\limits_{y = 0}^{y = 4} {\overline E \,.d\overline r } = - {E_y}\,\Delta y = - 200\) \({V_B} - {V_A} = 100\left( {\sqrt 3 - 2} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359552
In a region, the potential is represented by \(V(x,y,z) = 6x - 8xy - 8y + 6yz,\) where \(V\) is in volts and \(x,y,z\) are in metres. The electric forces experienced by a charge of 2\(C\) situated at point (1,1,1) is
1 \(24\,N\)
2 \(28\,N\)
3 \(6\sqrt 5 \,N\)
4 \(4\sqrt {35} N\)
Explanation:
As we know that \({E_x} = \frac{{ - \partial V}}{{\partial x}},{E_y} = \frac{{ - \partial V}}{{\partial y}}\& {E_z} = \frac{{ - \partial V}}{{\partial z}}\) As \(V = 6x - 8xy - 8y + 6yz\) \(\bar E = - [(6 - 8y)\hat i + ( - 8x - 8 + 6z)\hat j + 6y\hat k]\) The value of E at coordinate (1,1,1) \(E = - [ - 2\hat i - 10\hat j + 6\hat k]\) So,\({E_{net}} = \sqrt {{{(2)}^2} + {{(10)}^2} + {{( - 6)}^2}} = 2\sqrt {35} \,NC\) and force on charge q due to \({E_{net}}\) is given by \(F = q\;{E_{net}} = 2 \times 2\sqrt {35} = 4\sqrt {35} \,N\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359553
The potential of the electric field produced by a point charge at any point (x, y, z) is given by \(V = 3{x^2} + 5\), where x, y, z are in metres and V is in volts. The intensity of the electric field at \(( - 2,1,0)\) is
1 \( + 17\,V\,{m^{ - 1}}\)
2 \( - 17\,V\,{m^{ - 1}}\)
3 \( + 12\,V\,{m^{ - 1}}\)
4 \( - 12\,V\,{m^{ - 1}}\)
Explanation:
As potential is independent of \(y\) and \(z\), electric field components in those two directions are zero. Since \(V = 3{x^2} + 5\) \(E = - \,\,\frac{{dV}}{{dx}} = - \frac{d}{{dx}}(3{x^2} + 5) = - 6x\) Intensity of the electic field at \(x = - 2\) \( = - 6( - 2) = + 12\,V{m^{ - 1}}\)
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359550
The electric field in a region is given by the vector \(\overrightarrow E = (4\hat i - 1\hat j)\left( {\frac{N}{C}} \right).\)The potential will be constant along
1 The line \(3y = 4x\)
2 \(Y\) - axis
3 The line \(y = 4x\)
4 \(X\) - axis
Explanation:
The electric field vector is \({\mathop{\rm Tan}\nolimits} \alpha = \frac{4}{1}\) \(y = 4x\) is the line along which the potential is constant
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359551
A field of \(100\,V{m^{ - 1}}\) is directed at \({30^o}\) to positive \(x\)-axis. Find \({V_A} - {V_B}\) if \(OA = 2\,m\) and \(OB = 4\,m\)
1 \(100(\sqrt 3 - 2)V\)
2 \(200(2 + \sqrt 3 )V\)
3 \(100(2 + \sqrt 3 )V\)
4 \(100(2 - \sqrt 3 )V\)
Explanation:
We know that \({V_2} - {V_1} = - \int_1^2 {\overline E \,.d\overline r } \) Given that \(\overline E = 50\sqrt 3 \,\hat i + 50\,\hat j\) \({V_A} - {V_0} = - \int\limits_{{\rm{x}} = 0}^{{\rm{x}} = 2} {\overline E \,.d\overline r } = - {E_{\rm{x}}}\,\Delta {\rm{x}} = - 100\sqrt 3 \) \({V_B} - {V_0} = - \int\limits_{y = 0}^{y = 4} {\overline E \,.d\overline r } = - {E_y}\,\Delta y = - 200\) \({V_B} - {V_A} = 100\left( {\sqrt 3 - 2} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359552
In a region, the potential is represented by \(V(x,y,z) = 6x - 8xy - 8y + 6yz,\) where \(V\) is in volts and \(x,y,z\) are in metres. The electric forces experienced by a charge of 2\(C\) situated at point (1,1,1) is
1 \(24\,N\)
2 \(28\,N\)
3 \(6\sqrt 5 \,N\)
4 \(4\sqrt {35} N\)
Explanation:
As we know that \({E_x} = \frac{{ - \partial V}}{{\partial x}},{E_y} = \frac{{ - \partial V}}{{\partial y}}\& {E_z} = \frac{{ - \partial V}}{{\partial z}}\) As \(V = 6x - 8xy - 8y + 6yz\) \(\bar E = - [(6 - 8y)\hat i + ( - 8x - 8 + 6z)\hat j + 6y\hat k]\) The value of E at coordinate (1,1,1) \(E = - [ - 2\hat i - 10\hat j + 6\hat k]\) So,\({E_{net}} = \sqrt {{{(2)}^2} + {{(10)}^2} + {{( - 6)}^2}} = 2\sqrt {35} \,NC\) and force on charge q due to \({E_{net}}\) is given by \(F = q\;{E_{net}} = 2 \times 2\sqrt {35} = 4\sqrt {35} \,N\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359553
The potential of the electric field produced by a point charge at any point (x, y, z) is given by \(V = 3{x^2} + 5\), where x, y, z are in metres and V is in volts. The intensity of the electric field at \(( - 2,1,0)\) is
1 \( + 17\,V\,{m^{ - 1}}\)
2 \( - 17\,V\,{m^{ - 1}}\)
3 \( + 12\,V\,{m^{ - 1}}\)
4 \( - 12\,V\,{m^{ - 1}}\)
Explanation:
As potential is independent of \(y\) and \(z\), electric field components in those two directions are zero. Since \(V = 3{x^2} + 5\) \(E = - \,\,\frac{{dV}}{{dx}} = - \frac{d}{{dx}}(3{x^2} + 5) = - 6x\) Intensity of the electic field at \(x = - 2\) \( = - 6( - 2) = + 12\,V{m^{ - 1}}\)
359550
The electric field in a region is given by the vector \(\overrightarrow E = (4\hat i - 1\hat j)\left( {\frac{N}{C}} \right).\)The potential will be constant along
1 The line \(3y = 4x\)
2 \(Y\) - axis
3 The line \(y = 4x\)
4 \(X\) - axis
Explanation:
The electric field vector is \({\mathop{\rm Tan}\nolimits} \alpha = \frac{4}{1}\) \(y = 4x\) is the line along which the potential is constant
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359551
A field of \(100\,V{m^{ - 1}}\) is directed at \({30^o}\) to positive \(x\)-axis. Find \({V_A} - {V_B}\) if \(OA = 2\,m\) and \(OB = 4\,m\)
1 \(100(\sqrt 3 - 2)V\)
2 \(200(2 + \sqrt 3 )V\)
3 \(100(2 + \sqrt 3 )V\)
4 \(100(2 - \sqrt 3 )V\)
Explanation:
We know that \({V_2} - {V_1} = - \int_1^2 {\overline E \,.d\overline r } \) Given that \(\overline E = 50\sqrt 3 \,\hat i + 50\,\hat j\) \({V_A} - {V_0} = - \int\limits_{{\rm{x}} = 0}^{{\rm{x}} = 2} {\overline E \,.d\overline r } = - {E_{\rm{x}}}\,\Delta {\rm{x}} = - 100\sqrt 3 \) \({V_B} - {V_0} = - \int\limits_{y = 0}^{y = 4} {\overline E \,.d\overline r } = - {E_y}\,\Delta y = - 200\) \({V_B} - {V_A} = 100\left( {\sqrt 3 - 2} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359552
In a region, the potential is represented by \(V(x,y,z) = 6x - 8xy - 8y + 6yz,\) where \(V\) is in volts and \(x,y,z\) are in metres. The electric forces experienced by a charge of 2\(C\) situated at point (1,1,1) is
1 \(24\,N\)
2 \(28\,N\)
3 \(6\sqrt 5 \,N\)
4 \(4\sqrt {35} N\)
Explanation:
As we know that \({E_x} = \frac{{ - \partial V}}{{\partial x}},{E_y} = \frac{{ - \partial V}}{{\partial y}}\& {E_z} = \frac{{ - \partial V}}{{\partial z}}\) As \(V = 6x - 8xy - 8y + 6yz\) \(\bar E = - [(6 - 8y)\hat i + ( - 8x - 8 + 6z)\hat j + 6y\hat k]\) The value of E at coordinate (1,1,1) \(E = - [ - 2\hat i - 10\hat j + 6\hat k]\) So,\({E_{net}} = \sqrt {{{(2)}^2} + {{(10)}^2} + {{( - 6)}^2}} = 2\sqrt {35} \,NC\) and force on charge q due to \({E_{net}}\) is given by \(F = q\;{E_{net}} = 2 \times 2\sqrt {35} = 4\sqrt {35} \,N\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359553
The potential of the electric field produced by a point charge at any point (x, y, z) is given by \(V = 3{x^2} + 5\), where x, y, z are in metres and V is in volts. The intensity of the electric field at \(( - 2,1,0)\) is
1 \( + 17\,V\,{m^{ - 1}}\)
2 \( - 17\,V\,{m^{ - 1}}\)
3 \( + 12\,V\,{m^{ - 1}}\)
4 \( - 12\,V\,{m^{ - 1}}\)
Explanation:
As potential is independent of \(y\) and \(z\), electric field components in those two directions are zero. Since \(V = 3{x^2} + 5\) \(E = - \,\,\frac{{dV}}{{dx}} = - \frac{d}{{dx}}(3{x^2} + 5) = - 6x\) Intensity of the electic field at \(x = - 2\) \( = - 6( - 2) = + 12\,V{m^{ - 1}}\)
359550
The electric field in a region is given by the vector \(\overrightarrow E = (4\hat i - 1\hat j)\left( {\frac{N}{C}} \right).\)The potential will be constant along
1 The line \(3y = 4x\)
2 \(Y\) - axis
3 The line \(y = 4x\)
4 \(X\) - axis
Explanation:
The electric field vector is \({\mathop{\rm Tan}\nolimits} \alpha = \frac{4}{1}\) \(y = 4x\) is the line along which the potential is constant
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359551
A field of \(100\,V{m^{ - 1}}\) is directed at \({30^o}\) to positive \(x\)-axis. Find \({V_A} - {V_B}\) if \(OA = 2\,m\) and \(OB = 4\,m\)
1 \(100(\sqrt 3 - 2)V\)
2 \(200(2 + \sqrt 3 )V\)
3 \(100(2 + \sqrt 3 )V\)
4 \(100(2 - \sqrt 3 )V\)
Explanation:
We know that \({V_2} - {V_1} = - \int_1^2 {\overline E \,.d\overline r } \) Given that \(\overline E = 50\sqrt 3 \,\hat i + 50\,\hat j\) \({V_A} - {V_0} = - \int\limits_{{\rm{x}} = 0}^{{\rm{x}} = 2} {\overline E \,.d\overline r } = - {E_{\rm{x}}}\,\Delta {\rm{x}} = - 100\sqrt 3 \) \({V_B} - {V_0} = - \int\limits_{y = 0}^{y = 4} {\overline E \,.d\overline r } = - {E_y}\,\Delta y = - 200\) \({V_B} - {V_A} = 100\left( {\sqrt 3 - 2} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359552
In a region, the potential is represented by \(V(x,y,z) = 6x - 8xy - 8y + 6yz,\) where \(V\) is in volts and \(x,y,z\) are in metres. The electric forces experienced by a charge of 2\(C\) situated at point (1,1,1) is
1 \(24\,N\)
2 \(28\,N\)
3 \(6\sqrt 5 \,N\)
4 \(4\sqrt {35} N\)
Explanation:
As we know that \({E_x} = \frac{{ - \partial V}}{{\partial x}},{E_y} = \frac{{ - \partial V}}{{\partial y}}\& {E_z} = \frac{{ - \partial V}}{{\partial z}}\) As \(V = 6x - 8xy - 8y + 6yz\) \(\bar E = - [(6 - 8y)\hat i + ( - 8x - 8 + 6z)\hat j + 6y\hat k]\) The value of E at coordinate (1,1,1) \(E = - [ - 2\hat i - 10\hat j + 6\hat k]\) So,\({E_{net}} = \sqrt {{{(2)}^2} + {{(10)}^2} + {{( - 6)}^2}} = 2\sqrt {35} \,NC\) and force on charge q due to \({E_{net}}\) is given by \(F = q\;{E_{net}} = 2 \times 2\sqrt {35} = 4\sqrt {35} \,N\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359553
The potential of the electric field produced by a point charge at any point (x, y, z) is given by \(V = 3{x^2} + 5\), where x, y, z are in metres and V is in volts. The intensity of the electric field at \(( - 2,1,0)\) is
1 \( + 17\,V\,{m^{ - 1}}\)
2 \( - 17\,V\,{m^{ - 1}}\)
3 \( + 12\,V\,{m^{ - 1}}\)
4 \( - 12\,V\,{m^{ - 1}}\)
Explanation:
As potential is independent of \(y\) and \(z\), electric field components in those two directions are zero. Since \(V = 3{x^2} + 5\) \(E = - \,\,\frac{{dV}}{{dx}} = - \frac{d}{{dx}}(3{x^2} + 5) = - 6x\) Intensity of the electic field at \(x = - 2\) \( = - 6( - 2) = + 12\,V{m^{ - 1}}\)
