Consider two point charges of equal magnitude q but opposite sign, separated by a dis- tance d (this is called an electric dipole). 1. Calculate the electric field set up by these two charges, for points in the xz-plane (the field will look the same in any plane containing the z-axis, because of rota- tional symmetry). Work in Cartesian coordinates, with the z-axis passing through the two charges, and the origin halfway between them; thus let the z-coordinate of the positive charge be d/2 and the z-coordinate of the negative charge -d/2. Show that the field is (x,0, z – d/2) (x,0, z + d/2) E(x,0, 2) = 4T€, ((x² + (z – d/2)²)³/2 (x² + (z + d/2)²)³/2, 2. Calculate the far-field approximation for points on the x-axis, i.e. approximate the field at points with z = 0 and with x> d. 3. Do the same for points on the z-axis, i.e. for x = 0 and z » d. Take care that your approximation is not too crude (a 0 answer won't do). To this purpose, bring the two terms of the solution on common denominator and simplify before you make any approximations! In both cases, the answer should depend on the quantity d q, and the field strength should decrease with thee third power of the distance to the origin
Consider two point charges of equal magnitude q but opposite sign, separated by a dis- tance d (this is called an electric dipole). 1. Calculate the electric field set up by these two charges, for points in the xz-plane (the field will look the same in any plane containing the z-axis, because of rota- tional symmetry). Work in Cartesian coordinates, with the z-axis passing through the two charges, and the origin halfway between them; thus let the z-coordinate of the positive charge be d/2 and the z-coordinate of the negative charge -d/2. Show that the field is (x,0, z – d/2) (x,0, z + d/2) E(x,0, 2) = 4T€, ((x² + (z – d/2)²)³/2 (x² + (z + d/2)²)³/2, 2. Calculate the far-field approximation for points on the x-axis, i.e. approximate the field at points with z = 0 and with x> d. 3. Do the same for points on the z-axis, i.e. for x = 0 and z » d. Take care that your approximation is not too crude (a 0 answer won't do). To this purpose, bring the two terms of the solution on common denominator and simplify before you make any approximations! In both cases, the answer should depend on the quantity d q, and the field strength should decrease with thee third power of the distance to the origin
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Transcribed Image Text:Consider two point charges of equal magnitude q but opposite sign, separated by a dis-
tance d (this is called an electric dipole).
1. Calculate the electric field set up by these two charges, for points in the xz-plane
(the field will look the same in any plane containing the z-axis, because of rota-
tional symmetry). Work in Cartesian coordinates, with the z-axis passing through
the two charges, and the origin halfway between them; thus let the z-coordinate of
the positive charge be d/2 and the z-coordinate of the negative charge -d/2. Show
that the field is
(x,0, z – d/2)
(x,0, z + d/2)
E(x,0, 2) =
4T€, ((x² + (z – d/2)²)³/2 (x² + (z + d/2)²)³/2,
2. Calculate the far-field approximation for points on the x-axis, i.e. approximate the
field at points with z = 0 and with x> d.
3. Do the same for points on the z-axis, i.e. for x = 0 and z » d. Take care that your
approximation is not too crude (a 0 answer won't do). To this purpose, bring the
two terms of the solution on common denominator and simplify before you make
any approximations!
In both cases, the answer should depend on the quantity d q, and the field strength should
decrease with the third power of the distance to the origin.
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