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Electric Potential (III) Text sections 25.5, 25.6 Fields, potential, and conductors Practice: Chapter 25, problems 29, 33, 35, 39, 57, 61 Read page 710, Van de Graaff generator We can use two completely different methods: 1. 2. Find from Gauss’s Law, then… 1 O +++++++++++++++++ L b Total charge Q, uniform linear charge density (So ) Find: V at point O y dq b r x x dx Charge/unit length: 2 dq + + dq + + + + + + + + + Total charge Q, uniform Find: V at centre C R + + + C (Homework exercise: review the calculation for the electric field E, which is harder.) + + + Total charge Q, uniform linear density Find: V at centre C R + C (Homework exercise: review the calculation for the electric field E) + + At the center of the semicircle, the potential is: A) less than kQ/R B) equal to kQ/R C) greater than kQ/R 3 dq + + + + + + + Total charge Q, uniform r R + + + C Find: V at centre C Solution: dV (at C) due to dq: r=R is a constant (same for each dq) Example: Spherical Charges 1) Find the electric field as a function of r using Gauss’s Law. 2) Imagine pushing a “test charge” in from infinity along a radial line: the potential change with each small change dr in distance is dV = - E(r) dr. 3) Integrate from R to infinity to find V(R) (relative to infinity) at any position R. 4 Solid Conducting Sphere,radius R E R r R r V Solid Conducting Sphere,radius R E R r R r V 5 A charge +Q is placed on a spherical conducting shell. What is the potential (relative to infinity) at the centre? A) B) C) D) keQ/R1 keQ/R2 keQ/ (R1 - R2) zero +Q R1 R2 Fields > 3 x 106 V/m will cause a spark in dry air. What is the maximum potential to which an isolated metal sphere of radius 1 cm can be charged, without causing a spark? A) 300V B) 3kV C) 30kV D) 300 kV E) 3MV 6 Fields 3 x 106 V/m will cause a spark in dry air. Find the maximum potential on a metal sphere of radius… a) 1 mm b) 1 m 1) inside conductor is an equipotential. 3) surface (just outside.) 4) Excess charge is on the surface; and _______________________________________ 4) Empty cavity inside a conductor, as well. 5) 7 + Large - - Weak - - + + + + - + Metal foil wrapper $ 500 Computer chip 8 A conducting sphere of radius R1, carrying charge Q, is surrounded by a thick conducting shell with no net charge. What is the potential of the inner sphere, relative to infinity? A) B) C) D) +Q V = zero 0 < V < keQ/R1 V = keQ/R1 V > keQ/R1 The dashed green line represents a spherical gaussian surface inside the conducting material. The total electric flux through this surface (in units of Q/ ε0) is -3Q R3 R1 +Q R2 A) 0 B) - Q/ε0 C) +Q/ε0 D) -2Q/ε0 E) 3Q/ε0 9 So, the charge on the inner surface of the outer shell is A) 0 B) - Q C) +Q D) -2Q E) +3Q R3 R1 +Q R2 -3Q ? + + + r + a Radius a, total charge Q r Uniform volume charge density Find: V(r) for r<a + + + + answer: 10 -Q +Q R1 R2 R3 Find: Potential difference E R1 V V1 R2 R3 r Find: V2 R1 R2 R3 r 11 Solution (cont’d): + + + + + E=0 + R V = constant = Vo + + + Charge Q + + + Outside (r > R): (Just outside) Vo (on sphere) = V (just outside) = So… Quiz: A large conducting sphere has a net charge Q. A second, smaller conducting sphere with no net charge is now connected to it by a conducting wire. When the system comes to equilibrium, which of the following are true? A) the charges on the spheres will be equal B) the surface charge densities (charge per unit area) on the surfaces are equal C) the potentials on the spheres are equal D) the electric fields just outside the spheres are equal. 12 Solution (Cont’d): Thus… Therefore… i) Point source: (choose V0 as r ) or ii) Several point sources: (Scalar) iii) Continuous distribution: OR … I. Find from Gauss’s Law (if possible) II. Integrate, (a “line integral”) 13 + Total charge Q, uniform r r + + + Find: V at centre C C Solution: dV (at C) due to dq: r is a constant (same for each dq) r + a + + + + + + + + + + + + + dq Radius a, total charge Q r Uniform volume charge density Find: V(r) 14 Solution (cont’d): Inside (r > a): Now, Potential: i.e. So for r > a : Same as for point charge!! What about r < a? Can we write Solution (cont’d): For r < a: Note: This is NOT equal to 15