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Transcript
Electric Dipole PH 203 Professor Lee Carkner Lecture 3 EB PAL # 2 Electric Field q EA q +2q Distance to point P is 5 cm (hypotenuse of a 3-4-5 right triangle) Top angle of triangle, sin q = 3/5, q = 37 deg. EA = kq/r2 = (8.99X109)(5X10-6) / (0.05)2 = 1.8X107 N/C qA = 270 - q = 233 deg EB = (8.99X109)(2)(5X10-6) / (0.05)2 = 3.6X107 N/C qB = q + 90 = 127 deg E PAL # 2 Electric Field +2q EAX = EA cos qA = (1.8X107)(cos 233) = -1.1X107 N/C EAY = EA sin qA = (1.8X107)(sin 233) = -1.4X107 N/C EBX =EB cos qB = (3.6X107)(cos 127) = -2.2X107 N/C EBY = EB sin qB = (3.6X107)(sin 127) = 2.9X107 N/C EX = EAX + EBX = -3.3X107 N/C EY = EAY + EBY = 1.5X107 N/C E2 = EX2 + EY2 E = 3.6X107 N/C q = arctan (EY/EX) = 24 deg above negative x-axis q = 180-22 =156 deg from positive x-axis The Dipole Dipole moment = p = qd z is the distance from the center of the dipole to some point on the dipole axis Dipole Field At a distance z that is large compared to d, the electric field reduces to: E = (1/(2pe0)) (p/z3) Note that: E falls off very rapidly with z Doubling charge is the same as doubling distance between charges Dipole in an External Field Assumptions: The dipole’s structure is rigid and unchangeable Since the two charges are equal and opposite, the two forces are equal and opposite However, since the charges are not in the same place, there is a net torque on the dipole A dipole in an external field will have no translation motion, but will have rotational motion Dipole Torque Torque is then Fd sin q = Eqd sin q Remember that p is direction from negative charge to positive charge How will the dipole spin? The dipole wants to align the dipole moment with E Dipole has the most torque when perpendicular to the field (q = 90) Dipole Energy We set the potential energy (U) to be zero when the dipole is at right angles to the field (q= 90) Dipole perpendicular to field: U = 0 Can write potential energy as: The work (done by an external force) to turn a dipole in a field is thus: W = Uf - Ui Charge Distribution The charge density: Linear = l = C/m Surface = Volume = r =C/m3 For example: dE = (1/(4pe0)) (lds/r2) Rings For a uniform charged ring: E = qz / (4pe0(z2+R2)3/2) Disks For a uniform charged disk: E = (s/2e0)(1 –[z/(z2+R2)½] The field depends not on the total charge but the charge density Next Time Read 23.1-23.4 Problems: Ch 22, P: 42, 52, 53, Ch 23, P: 4, 5 What is true about the magnitude and direction of the fields from charges A and B at point P? A) They are equal in magnitude and point in the same direction B) They are equal in magnitude and point towards charges A and B C) They are unequal in magnitude and point away from charges A and B D) They are unequal in magnitude and 180 apart in direction E) The net field at P is zero A B What is true about the x and y components of the fields from charges A and B at point P? A) They both add B) They both cancel C) The x components add and the y components cancel D) The x components cancel and the y components add E) We can’t tell with out knowing the magnitude of q A B The above electric field, A) B) C) D) E) increases to the right increases to the left increases up increases down is uniform An electron placed at A, A) Would move left and feel twice the force as an electron at B B) Would move right and feel twice the force as an electron at B C) Would move left and feel half the force as an electron at B D) Would move right and feel half the force as an electron at B E) Would move right and feel the same force as an electron at B