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Transcript
Lecture 3: May 22nd 2009 Physics for Scientists and Engineers II Physics for Scientists and Engineers II , Summer Semester 2009 Gauss’s Law (Karl Friedrich Gauss 1777-1855) • • Example of surface of a cube in a constant electric field resulted in a net flux of zero through that closed surface. Today: We will look at a spherical closed surface (shell) concentric around a point charge q and calculate the electric flux through that shell of radius r. dA E E parallel to d A everywhere on surface E also has the same magnitude everywhere on the q surface of the shell : E k e 2 r d E Ed A EdA q r E Ed A EdA E dA shell shell E 4r 2 k e E 1 4o 4q shell q 4r 2 ke 4q 2 r q o fE is independent of r and proportional to q. Physics for Scientists and Engineers II , Summer Semester 2009 Generalization For non-spherical closed surfaces surrounding q, the net electric flux is the same as for the spherical concentric shell (just harder to calculate). Example: dA E r2 q r1 E Ed A EdA EdA surface left half righthalf Eleft 2r2 Eright 2r1 k e 2 E Physics for Scientists and Engineers II , Summer Semester 2009 1 4o 4q 2 q o q q 2 2 2 r k 2r1 ke 4q 2 e 2 2 r2 r1 Generalization The electric flux through a closed surface surrounding a point charge q equals E What if multiple charges are inside a surface? E d A E 1 E 2 E 3 E 4 ... d A (superposi tion principle of E ) E d A E1 d A E 2 d A E 3 d A ... q1 0 q2 0 q3 0 .... Physics for Scientists and Engineers II , Summer Semester 2009 qinside 0 q o Gauss’s Law E Ed A Physics for Scientists and Engineers II , Summer Semester 2009 qin o Example: Electric Field due to an Infinitely Long Uniformly Charged Rod…..doing it the hard way y dE a dq = l dx P r x x l Contributi on to E from dq : l dx sin iˆ 2 r l dx x ke 2 iˆ r r l x dx ke 3 iˆ r l x dx ˆ ke i 3 2 2 2 a x d E ke Physics for Scientists and Engineers II , Summer Semester 2009 l dx cos ˆj 2 r l dx a ˆ ke 2 j r r l a dx ke 3 ˆj r l a dx ˆ ke j 3 2 2 2 a x ke y dq = l dx P dE a r x x l d E ke Contributi on to E from dq : a l x dx 2 x2 3 2 iˆ ke a l a dx 2 x2 3 ˆj 2 l x dx ˆ l a dx E ke i k e 3 3 2 2 2 2 2 2 a x a x Total electric field at point P : E x ke l a x x 2 2 3 dx 2 Physics for Scientists and Engineers II , Summer Semester 2009 E y ke l a a 1 2 x 2 3 dx 2 ˆj ….solving the integrals E x ke l a x x 2 2 3 2 Substituti on : x a tan E y ke l 90 90 a dx a 2 E y ke l dx 0 a 2 tan 2 3 2 a d cos 2 a d cos 2 90 a a 2 x 2 3 dx 2 (I wouldn' t expect you to know that) 90 ke l 1 a 90 1 tan 2 90 kl ke l 1 1 e d d cos a 90 1 3 2 cos 2 a 90 2 cos kl kl 2k l 90 e sin 90 e sin 90 sin 90 e a a a Physics for Scientists and Engineers II , Summer Semester 2009 3 2 1 d cos 2 Example: Electric Field due to an Infinitely Long Uniformly Charged Rod…..using Gauss’s Law a L By symmetry : E x 0 and Eradial const. at any point that is radially a distance of " a" away from the line charge. E is perpendicu lar to the cylindrica l round surface everywhere . No flux goes through t he flat end surfaces of the cylinder. Physics for Scientists and Engineers II , Summer Semester 2009 The charge enclosed in the cylinder is : q inside l L Gauss' s Law applied : E Ed A E dA cylinder surface E 2aL E E 2aL qinside 0 lL 0 lL 0 lL l l 2k e 2aL 0 2a 0 a Question: Why would Gauss’s law be unpractical to use if we picked a spherical surface in this particular problem? Answer: The electric field vector is not constant along the spherical surface. Physics for Scientists and Engineers II , Summer Semester 2009 Rules for Using Gauss’s Law to Evaluate the Electric Field E can be argued by symmetry t o be constant in magnitude and parallel to d A over a part or all of the surface. Ed A EdA All other surfaces have zero flux going through t hem ( E d A 0). This condition is fulfilled if either E 0, or if E d A on those surfaces. Physics for Scientists and Engineers II , Summer Semester 2009 Example: Spherical Shell with Charge on it. Determine E(r) Constant volume charge density. Total charge = Q Region 1: r < a E Ed A qinside 0 0 E 0 Region 2: a < r < b a b E qinside 0 E qinside qinside k e 0 4r 2 r2 4 4 qinside Vinside r 3 a 3 3 3 Q r 3 a3 4 3 4 3 r a Q 3 4 4 3 b a3 3 33 b a 3 3 Region 3: r > b E Ed A E Ed A qinside 0 Q 0 Q Q k e 4r 2 0 r2 Physics for Scientists and Engineers II , Summer Semester 2009 Q r 3 a3 Q a3 r 2 E ke 2 3 ke 3 r b a3 b a 3 r Conductors in Electrostatic Equilibrium • • Electrostatic equilibrium in a conductor: The net motion of charges within the conductor is zero. Properties of conductors in electrostatic equilibrium: E = 0 inside the conductor (hollow or solid). Charged conductors: Charge is on the surface. Just outside the surface of the conductor: E surface E , where is local surface charge density. 0 Surface charge density is greatest where surface has smallest radius of curvature. (…and therefore E is largest just outside those surface areas.) Physics for Scientists and Engineers II , Summer Semester 2009 Conductors in Electrostatic Equilibrium E = 0 inside the conductor (hollow or solid). How do we know? Consider an uncharged conductor (or charged, it doesn’t matter) …positive charge remains on the right side as negative charge accumulates on the left… …and place it into an electric field… Fe - - E free electrons experience a force (here, to the left side) E extra + E Charges keep separating and creating extra electric field. ….until extra field cancels the external electric field…. ….and the separation of charges stops (electrons on the inside no longer experience a net force.) Einside=0 Physics for Scientists and Engineers II , Summer Semester 2009 Conductors in Electrostatic Equilibrium Charged conductors: Charge is on the surface. How do we know? Consider an charged conductor of some shape We know : E 0 on the inside Imagine a Gaussian closed surface just beneath the actual surface. Since E inside 0 E 0 through this surface no net charge inside of Gaussian surface 0 entire net charge must reside on the surface Physics for Scientists and Engineers II , Summer Semester 2009 Conductors in Electrostatic Equilibrium E surface How do we know? Suppose E was not perpendicular to surface E Then E would have a component parallel to the surface Motion of charges would occur along the surface (not yet in electrostatic equilibrium). In electrostatic equilibrium there cannot be a component of E parallel to the surface. Physics for Scientists and Engineers II , Summer Semester 2009 Conductors in Electrostatic Equilibrium E , where is the local surface charge density. 0 How do we know? Consider a surface element of area A, small enough to be “flat” E Consider a Gaussian surface (e.g., cylinder with end caps parallel to surface, one end cap On the inside, one just outside) Calculate flux through the cylinder E EA Physics for Scientists and Engineers II , Summer Semester 2009 qinside 0 A 0 E 0 Chapter 25: Work Done by Electric Force – Small Displacement • Imagine a charge q0 is moved in an electric field by some small displacement: E Fe ds + q0 Work done by electric force : WFe F e d s q0 E d s Physics for Scientists and Engineers II , Summer Semester 2009 Chapter 25: Work Done by Electric Force – Along a Path • Imagine a charge q0 is moved in an electric field along some path ds E Fe + q Work done by electric force : WFe F path e d s q0 E ds path “path integral” or “line integral” Physics for Scientists and Engineers II , Summer Semester 2009 Work Done on Charge by Electric Force Along a Closed Path • Imagine a charge in an electric field is moved by external force as shown in a closed path. E x1 x2 x3 + Fe x5 x4 a WFe F e d s F e x1 F e x 2 F e x 3 F e x 4 F e x 5 Fe a Fe a 0 path (Work done by Fe 0 actually for any closed path) • In other words: Electrostatic force is a conservative force. An electric potential (a scalar quantity) can be defined. (just like we did with the gravitational force / potential) Physics for Scientists and Engineers II , Summer Semester 2009