PHY2049_05-16-11 - PHY 2049 Physics II We begin with a live...

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5/16/2011 1 PHY 2049: Physics II square6 We begin with a live clicker on Wednesday. square6 Wileyplus homework should be fully operational. square6 Tea and Cookies: We meet on Tuesdays at 5:00PM for tea and cookies in room 2165. quiz square6 The electric field at a distance of 1 m from an isolated point particle with a charge of 2 nC is: square6 A. 1.8N/C B. 180N/C C. 18N/C D. 1800N/C E. none of these Quiz 2 square6 An isolated charged point particle produces an electric field with magnitude E at a point 2m away from the charge. A point at which the field magnitude is 4E is: square6 A. 1m away from the particle square6 B. 0.5m away from the particle square6 C. 2m away from the particle square6 D. 4m away from the particle square6 E. 8m away from the particle PHY 2049: Physics II Last week square6 Coulomb’s law, Electric Field and Gauss’ theorem Today square6 Electric Potential Energy and Electric Potentials square6 Numerous cases Potential Energy and Potential square6 Force => work => change in K=> change in Potential energy square6 Δ U = U f – U i = -W = - Δ K square6 Work done is path independent. K+U = constant. U = k q 1 q 2 /r : interaction energy of two charges. Sign matters PHY 2049: Physics II square6 Electric Potential square6 V = U/q = -W/q square6 Units of Joules/coulomb = volt square6 1 eV = e x 1V = 1.6 x 10 -19 J square6 Also square6 V = kq/r square6 V f –V i = -∫E.ds square6 In case of multiple charges, add as a number
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5/16/2011 2 PHY 2049: Physics II square6 V f -V i = -∫ k q/r 2 dr square6 Choose V i = V (∞)=0 square6 V(r) = kq/r square6 V = kpcosθ/r 2 square6 E = - V/ s = - square6 Uniformly charged disk square6 V = ?? A B C p r Consider the electric dipole shown in the figure We will determine the electric potential created at point P by the two charges of the dipole using superpos V Example : Potential due to an electric dipole ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ition. Point P is at a distance from the center O of the dipole. Line OP makes an angle with the dipole axis 1 4 4 We assume that where is o o r r r q q q V V V r r r r r d d θ πε πε - + + - + - - + - = + = - = ° ( ) ( ) 2 ( ) ( ) 2 2 the charge separation From triangle ABC we have: cos cos 1 cos Also: 4 4 where the electric dipole moment o o r r d q d p r r r V r r p qd θ θ θ πε πε - + - + - = = = (24 - 6) 2 1 cos 4 o p V r θ πε = dq O A Potential created by a line of charge of length L and uniform linear charge density λ at point P. Consider the charge element at point A, a distance from O. From triangle OAP we dq dx x λ = Example : ( 29 ( 29 ( 29 2 2 2 2 2 2 0 2 2 2 2 0 2 2 2 2 have: Here is the distance OP The potential created by at P is: 1 1 4 4 4 ln 4 l l n ln 4 n o o L o L o o r d x d dV dq dq dx dV r d x dx V d x V x d x V L L x dx x d x d d x λ πε πε λ πε λ
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