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Dipole in an Electric Field

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Dipole in an Electric Field Physics 2113 Isaac Newton (1642–1727) Physics 2113 Lecture 08: FRI 12 OCT CH22: Electric Fields Michael Faraday (1791–1867) Electric field due to a charged disk Idea: superpose several rings, of E P infinitesimal width dR q Charge per unit area σ = 2 a πb Charge of ring of radius R and width dR dq = 2π RdR σ q Electric field due to ring at point P R dR a dq dE = 2 2 3/ 2 b 4πε 0 (R + a ) Integrating b b aσ 2π R dR aσ ⎡ 1 ⎤ σ ⎡ a ⎤ E = = − = 1− ∫ 2 2 3/ 2 ⎢ 2 2 ⎥ ⎢ 2 2 ⎥ 0 4πε 0 (R + a ) 2ε0 ⎣ R + a ⎦0 2ε0 ⎣ b + a ⎦ Electric Dipole Moment The direction of dipole moment is taken from negative to positive charge. Using expression for the electric field of the dipole, we find qd p E = 3 = 3 2πε 0 x 2πε 0 x Dipole in an electric field A point charge feels a force proportional to the electric field ! ! F = q E Assume the field is constant throughout the size of the dipole. Forces are equal and opposite: no net force! Forces act at different points, therefore they have a nontrivial torque. τ = F d sinθ = q E d sinθ ! ! The torque acting on the dipole can be written as τ = P ×E The cross product takes care of the sin θ in the magnitude and the direction is given by the right hand rule (clockwise in the figure). The torque felt by a dipole in a field implies the ability to do work (for instance, turn the sha of a motor). The poten>al energy is given by It is maximum when p and E point in opposite direc>ons, and minimum when they are parallel (note the minus sign). The dipole is “wound up” when it is opposing the field, ready to do work, and cannot do any more work when it is aligned with the field. Microwave ovens operate by generang alternang fields and making the dipoles present in the water molecules align and realign, creang fric>on among the molecules. Maximum torque is at 90 degrees. Example : Arc of Charge y • Figure shows a uniformly charged rod of charge -Q bent into a circular arc of radius R, centered at (0,0). • What is the direction of the electric field at the origin? x (a) Field is 0. (b) Along +y • Choose symmetric elements (c) Along -y • x components cancel Example : Arc of Charge y • Figure shows a uniformly charged rod of charge -Q bent into a circular arc of radius R, centered at (0,0). • What is the direction of the electric field at the origin? x (a) Field is 0. (b) Along +y • Choose symmetric elements (c) Along -y • x components cancel Arc of Charge: details y • Figure shows a uniformly charged rod of charge -Q bent into a circular arc of radius R, centered at (0,0). 450 • Compute the direction & x magnitude of E at the origin. kdQ y dQ = λRdθ dEx = dE cosθ = cosθ R2 dθ / 2 / 2 π k(λRdθ )cosθ kλ π Ex = = cosθdθ θ ∫ R2 R ∫ 0 0 x kλ kλ kλ E = E = E = 2 λ = 2Q/(πR) x R y R R Generalizing the previous result: From last class: Summary22 Summary : Defini'on of Electric Field Field due to an Electric Dipole • The electric field at any point • The magnitude of the electric field set up by the dipole at a distant point on the dipole axis is Eq. 22-1 Electric Field Lines Eq. 22-9 • provide a means for visualizing the direc>ons and the magnitudes of Field due to a Charged Disk electric fields • The electric field magnitude at a point on the central axis through a Field due to a Point Charge uniformly charged disk is given by • The magnitude of the electric field E set up by a point charge q at a distance r from the charge is Eq. 22-26 Eq. 22-3 Summary22 Summary : Force on a Point Charge in an Dipole in an Electric Field Electric Field • The electric field exerts a torque on a • When a point charge q is placed in an dipole external electric field E Eq. 22-34 Eq. 22-28 • The dipole has a poten>al energy U associated with its orientaon in the field Eq. 22-38 .
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