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Download Example 16-7 Field of an Electric Dipole
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Example 16-7 Field of an Electric Dipole A combination of two point charges of the same magnitude but opposite signs is called an electric dipole. Figure 16-13 shows an electric dipole made up of a point charge +q and a point charge 2q separated by a distance 2d. Derive expressions for the magnitude and direction of the net electric field due to these two charges at a point P a distance y along the midline of the dipole. E+ P E– r Figure 16-13 An electric dipole The field produced by an electric dipole at any point is the –q +q vector sum of the fields Es + and Es - caused by the positive charge +q and the negative charge 2q, respectively. Set Up The net field is the vector sum of the field s + due to the charge +q (which points E s - due to the away from +q) and the field E charge 2q (which points toward 2q). Note that this problem is very similar to Example 16-4 in Section 16-4, in which we used vector addition to find the net electric force exerted by two charges (one positive and one negative) on a third charge; here we use vector addition to find the net electric field due to the positive and negative charge. As in Example 16-4, we’ll choose the positive x direction to be to the right and the positive y axis to be upward and add the two vectors using components. Solve Use Equation 16-4 to find the magnitudes of s + and E s - at P. the fields E 2d Magnitude of the electric field due to a point charge Q: E = k Q r 2 y y E+ (16-4) E– Total electric field: s = E s1 + E s 2 E (16-5) +q O O –q O x The distance from +q to point P is the same as the distance from 2q to P. Call this distance r: r = 2y 2 + d 2 Since +q and 2q have the same magnitude (q, which is positive) and are the same distance from P, Equation 16-4 tells us that the fields that the two charges produce at P have the same magnitude: E + = E- = kq r 2 = kq 2 y + d2 s + and E s-, Find the x and y components of E and use these to calculate the components of the net field at P. s + and E s - are The components of E E+, x = E+ cos u E+, y = E+ sin u E2, x = E2 cos u E2, y = 2E2 sin u Since the magnitudes E+ and E2 are equal, the components of the net field at P are E x = E + , x + E -, x = E + cos u + E - cos u 2kq cos u = 2 y + d2 E y = E + , y + E -, y = E + sin u + 1 -E - sin u2 kq kq = 2 sin u - 2 sin u = 0 2 y + d y + d2 From the figure, cos u = d d = 2 r 2y + d 2 So the components of the net electric field are 2kq Ex = 2 d 2 1y + d 2 2y + d Ey = 0 Reflect We can check our result by substituting y = 0, so that the point P is directly between the two charges and a distance d from each s - both point to the s + and E charge. Then E right, and the magnitude of the net electric field should be equal to the sum of the mags-. s + and E nitudes of E Note that at points very far from the dipole, so that y is much greater than d, the magnitude of the field is inversely proportional to the cube of y: At double the distance, the field of a dipole is 11>22 3 = 1>8 as great. This is a much more rapid decrease with distance than the field of a single point charge, for which E is inversely proportional to the square of the distance: At double the distance, the field of a point charge is 11>22 2 = 1>4 as great. The dipole field decreases much more rapidly because the fields of +q and 2q partially cancel each other. 2 2 = 2kqd 2 1y + d 2 2 3>2 The net electric field at point P is to the right and has magnitude E = 2kqd> 1y 2 + d 2 2 3>2. At y = 0, the net electric field has magnitude E = 2kqd 10 + d 2 2 3>2 = 2kqd d3 = 2kq d2 = 2a kq d2 b This is just twice the magnitude of the field due to each individual charge: E + = E- = kq d2 If y is much greater than d, y2 + d2 is approximately equal to y2. Then the magnitude of the net electric field due to the dipole is approximately E net = 2kqd 2 3>2 1y 2 = 2kqd y3