Dipole Moment-running man20130526

The .bination of two equal point charges of opposite sign separated by a small distance (l) is called an electric dipole or simply dipole and the product (Q.l) is known as the electric dipole moment. When these two charges of opposite sign are separated by even very small distance say l there exist a finite resultant field but when the two charges are superimposed the resultant filed would be zero. It is because a molecule in the normal state is neutral electrically. When the molecule is situated or brought to an electric filed the positive atomic nuclei and the negative electrons are pulled in opposite directions as a result, their centers are displaced in the direction of the field. Such molecule is said to be polarized. This leads to the concept of dipole which means a pair of equal and opposite charges with very small distance between them. When the two charges +Q and ""Q are superimposed, the resultant field is zero and the normal state of molecule is achieved. Calculating dipole moment In the simplest case of dipole moment i.e., when two point charges with +Q and ""Q are separated with l distance then the dipole moment is given as P = q d Where p is dipole moment or displacement vector pointing form ""Q to +Q. The potential due to electric dipole is given as = constant distance factor Dipole moment Angle factor For continuous distribution of charge then the dipole moment is given as P( r ) = "vp( r0) (r0 "" r) d3 r0 In order to find out the electric field due to a dipole we use gradient concept E = -"V and taking gradient of potential equation which is shown in above we can get the electric field which is vary inversely as cube of the distance r whereas it was varying inversely as square of the distance in case of a point charge. An arrangement of two equal and opposite point charges infinitesimal distance apart constitutes an electric dipole. Consider an electric dipole (or j ust dipole) as a pair of equal and opposite point charges q and -q (so that ‘Lq = 0) separated by a very small fixed distance * 2a. The strength of a dipole is called the dipole moment p. It is a vector quantity in contrast to a charge, which is a scalar. Its magnitude is the product of the charge q and the separation (2a) or the length (2a). p = (2a)q and its direction is from the negative charge -q to the positive charge q. Dipole moment = p = (charge q) x (vector from negative to positive charge) Dipole moment is measured in coulomb metre (Cm). If both charge q and separation 2a are not equal to zero and are finite, then the dipole has a finite size, a location, a direction and a strength. But what happens to a dipole in which the charge q gets larger and larger, while the separation (2a) gets smaller and smaller, keeping the product I P I = 2aq constant ? In the limit we get what we may call an ideal dipole. We do not give q and (2a) separately for it, but only the dipole moment vector p with a definite magnitude and direction. In other words for a point dipole, only its location and dipole moment vector are relevant as it has no size. "We write la instead of a to make easier some of the mathematical manipulations. DIPOLE FIELD AND DIPOLE BEHAVIOUR IN A UNIFORM TWO DIMENSIONAL ELECTRIC FIELD The electric field produced by a dipole is called the dipole field. Let us consider an electric dipole consisting of two charges, +q and -q with equal magnitudes and opposite signs (so that Zq = 0), separated by a fixed small distance 2a. Let us find the field strength due to such a dipole at a general point A (x, y, o) in space at a distance r a from the origin 0. Let the two electric charges, +q and -q, be situated at points r, and r2 in space (other than r) from the field point A. The geometry of the dipole is shown in Fig. (115) where O is the mid-point of the dipole and also it is the origin of the coordinate axes. The dipole is along the y-axis. For the x-.ponent of the field strength at the field point A (x, y), we have Ex – ^ cos – cos 2 4n0 r,2 r Since cos 8X = cos 02 = y- and r2 = x2 + (y- a)2 and r*=x2 + (y + a)2" so E – q x____-_ ^lx2+iy-a)T [x2+(y+a)2f. Similarly, ^or the y-.ponent, we obtain E – q fsin 01 _ sin 02 or Ey = S-_(‘y~a)___(y + a)_’ H[x(y-a)T [x(y+a)T. Nowr y , so we have [x2+(ya)2]’ = [xW+JiTafl* = r"3 1 + a r2 . According to Binomial Theorem 2! 3! Thus, (l|)"a= ………. 2 8 Therefore, for r 2a, based on binomial expansion and retaining terms By: Aninda – Online math problem solver can be a device or a man. There are a various number of sites that give this application. These online math problem solver sites have certain devices which can be utilized to take care of particular math questions together with problems. 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