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Two point charges +q and -q are held fixed at (-d,0) and (d,0) respectively of a x-y coordinate system. Then
- a)The electric field E at all points on x-axis has the same direction.
- b)Electric field at all point on y-axis is along x-axis.
- c)Work has to be done in bringing a test charge from ∞ the origin.
- d)The dipole moment is 2qd along the x-axis.
Correct answer is option 'B'. Can you explain this answer?
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Two point charges +q and -q are held fixed at (-d,0) and (d,0) respect...
When you draw the electric field lines for these 2 charges they will starts from the positive charge and end at the negative charge. And when they pass through the Y axis they will be perpendicular to the Y axis, since both the charges are at equal distance and equal in magnitude. The direction of electric field is found by the tangents to the electric field at that point, since the line is parallel to x axis it's slope will also be zero, So the direction of the electric field line will be along the x axis.
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Two point charges +q and -q are held fixed at (-d,0) and (d,0) respect...
Infinity to a point on the x-axis between the two charges.
a) The electric field at all points on the x-axis will be in the same direction (either towards q or towards -q), but the magnitude of the electric field will vary depending on the distance from each charge. At a point equidistant from both charges (i.e. at x=0), the electric field due to q and -q will cancel out, resulting in zero net electric field. However, at any other point on the x-axis, the electric field will have a non-zero magnitude and will be in the direction towards the closer charge.
b) The electric field at all points on the y-axis will be along the x-axis, as the charges are symmetrically placed with respect to the y-axis. The electric field due to q and -q will have equal magnitude at any point on the y-axis, but opposite direction, resulting in zero net electric field along the y-axis.
c) Work has to be done in bringing a test charge from infinity to a point on the x-axis between the two charges because the test charge will experience a force due to the electric fields of both q and -q. As the test charge moves closer to one of the charges, the force due to that charge will increase, while the force due to the other charge will decrease. Therefore, work has to be done to overcome the repulsive force between the test charge and the charges, and to bring the test charge to a point on the x-axis where the net force is zero. The amount of work required will depend on the distance of the test charge from each charge and the magnitude of the charges.
a) The electric field at all points on the x-axis will be in the same direction (either towards q or towards -q), but the magnitude of the electric field will vary depending on the distance from each charge. At a point equidistant from both charges (i.e. at x=0), the electric field due to q and -q will cancel out, resulting in zero net electric field. However, at any other point on the x-axis, the electric field will have a non-zero magnitude and will be in the direction towards the closer charge.
b) The electric field at all points on the y-axis will be along the x-axis, as the charges are symmetrically placed with respect to the y-axis. The electric field due to q and -q will have equal magnitude at any point on the y-axis, but opposite direction, resulting in zero net electric field along the y-axis.
c) Work has to be done in bringing a test charge from infinity to a point on the x-axis between the two charges because the test charge will experience a force due to the electric fields of both q and -q. As the test charge moves closer to one of the charges, the force due to that charge will increase, while the force due to the other charge will decrease. Therefore, work has to be done to overcome the repulsive force between the test charge and the charges, and to bring the test charge to a point on the x-axis where the net force is zero. The amount of work required will depend on the distance of the test charge from each charge and the magnitude of the charges.
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Two point charges +q and -q are held fixed at (-d,0) and (d,0) respect...
When you draw the electric field lines for these 2 charges they will starts from the positive charge and end at the negative charge. And when they pass through the Y axis they will be perpendicular to the Y axis, since both the charges are at equal distance and equal in magnitude. The direction of electric field is found by the tangents to the electric field at that point, since the line is parallel to x axis it's slope will also be zero, so the direction of the electric field line will be along the x axis.
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Two point charges +q and -q are held fixed at (-d,0) and (d,0) respectively of a x-y coordinate system. Thena)The electric field E at all points on x-axis has the same direction.b)Electric field at all point on y-axis is along x-axis.c)Work has to be done in bringing a test charge from ∞ the origin.d)The dipole moment is 2qd along the x-axis.Correct answer is option 'B'. Can you explain this answer?
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