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We will continue the analysis of the electric dipole starting from part d of problem 1, but in the ong-distance limit. It will be convenient to use the distance, r, from the origin, r = √x² + z² and the ngle between r and the z axis: z = r cos0, x= r sin 0. The long-distance limit corresponds to r >> d. In the denominator of your result from problem 1 part d, ou will have terms that look like √x² + (z±d/2)². Rewrite those in terms of r and r cos and factor -ut an r2 from each of the terms inside the radical. Then pull that r² out of the square root leaving an xpression inside the quadratic that is amenable to a binomial approximation However we will only

We will continue the analysis of the electric dipole starting from part d of problem 1, but in the ong-distance limit. It will be convenient to use the distance, r, from the origin, r = √x² + z² and the ngle between r and the z axis: z = r cos0, x= r sin 0. The long-distance limit corresponds to r >> d. In the denominator of your result from problem 1 part d, ou will have terms that look like √x² + (z±d/2)². Rewrite those in terms of r and r cos and factor -ut an r2 from each of the terms inside the radical. Then pull that r² out of the square root leaving an xpression inside the quadratic that is amenable to a binomial approximation However we will only

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We will continue the analysis of the electric dipole starting from part d of problem 1, but in the long-distance limit. It will be convenient to use the distance, r, from the origin, r = angle 0 between and the z axis: Vx2 + z2 and the z = r cos 0, x = r sin 0. The long-distance limit corresponds to r >> d. In the denominator of your result from problem 1 part d, you will have terms that look like Vx2 + (z ± d/2)². Rewrite those in terms of r and r cos e and factor out an r2 from each of the terms inside the radical. Then pull that r² out of the square root leaving an expression inside the quadratic that is amenable to a binomial approximation. However, we will only keep terms to first power in d/r so you will want to drop a term inside each square root before applying the binomial approximation. Apply the binomial approximation to the two terms you obtained from problem 1d and combine them. Show that the electric field in the long-distance limit can be written 1 qd (1) 4m€0 r3 (3 cos Of – k), 4περ r3 where, since we are still evaluating the field for y = 0, 7 = xi + zk. ||
Transcribed Image Text:We will continue the analysis of the electric dipole starting from part d of problem 1, but in the long-distance limit. It will be convenient to use the distance, r, from the origin, r = angle 0 between and the z axis: Vx2 + z2 and the z = r cos 0, x = r sin 0. The long-distance limit corresponds to r >> d. In the denominator of your result from problem 1 part d, you will have terms that look like Vx2 + (z ± d/2)². Rewrite those in terms of r and r cos e and factor out an r2 from each of the terms inside the radical. Then pull that r² out of the square root leaving an expression inside the quadratic that is amenable to a binomial approximation. However, we will only keep terms to first power in d/r so you will want to drop a term inside each square root before applying the binomial approximation. Apply the binomial approximation to the two terms you obtained from problem 1d and combine them. Show that the electric field in the long-distance limit can be written 1 qd (1) 4m€0 r3 (3 cos Of – k), 4περ r3 where, since we are still evaluating the field for y = 0, 7 = xi + zk. ||
But, once we have put the field in the form of Eq. 1, it can be trivially generalized by relaxing the y = 0 requirement so that 7 = r_ Îl + zk. The expression in Eq. 1 doesn't need to change. It is common to express the result in Eq. 1 in terms of an "electric dipole moment", p, which for our geometry is defined p = qdk. The magnitude is the product of the charge, q and the distance between the two charges in the dipole (thus the positions at ±d/2). The direction of the dipole moment vector is in the direction of the displacement between the charges. Show that the long-distance field can be written 1 3 (f p) î – p 4TE0 p3
Transcribed Image Text:But, once we have put the field in the form of Eq. 1, it can be trivially generalized by relaxing the y = 0 requirement so that 7 = r_ Îl + zk. The expression in Eq. 1 doesn't need to change. It is common to express the result in Eq. 1 in terms of an "electric dipole moment", p, which for our geometry is defined p = qdk. The magnitude is the product of the charge, q and the distance between the two charges in the dipole (thus the positions at ±d/2). The direction of the dipole moment vector is in the direction of the displacement between the charges. Show that the long-distance field can be written 1 3 (f p) î – p 4TE0 p3
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