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Electric Potential Equipotentials and Energy Today: Mini-Quiz + Hints for HWK

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Electric Potential Equipotentials and Energy Today: Mini-quiz + hints for HWK U = qV Should lightening rods have a small or large radius of curvature ? V is proportional to 1/R. If you want a high voltage to pass through the rod then use a small radius of curvature. Air is normally an insulator, however for large E fields (E>3 x 10 6 V/m) it starts conducting. Empire State Building, NYC Electrical Potential Review: Wa → b = work done by force in going from a to b along path. b b r r b r r Wa →b = F • ld = qE • ld ∫a ∫a F b r r θ ∆U = Ub −U a = −Wa →b = −∫ qE • ld a a dl U = potential energy b r r ∆U Ub −U a Wa →b ∆V = Vb −Va = = = − = −∫ E • ld q q q a • Potential difference is minus the work done per unit charge by the electric field as the charge moves from a to b. • Only changes in V are important; can choose the zero at any point. Let Va = 0 at a = infinity and Vb → V, then: r r r V = electric potential V = −∫ E • ld ∞ allows us to calculate V everywhere if we know E Potential from charged spherical shell • E-field (from Gauss' Law) V q q • : 4πεπεπε 000 R 4πεπεπε 000 r r < R Er = 0 1 q • r >R: E = R r r 4πε 2 0r R • Potential R • r > R: r === r r r r 1 Qq V r( ) === −−− ∫∫∫ E ••• ld === −−− ∫∫∫ E r ( dr ) === r === ∞∞∞ ∞∞∞ 4πεπεπε 0 r • r < R: r===r r r r aR r 1 Qq V )r( === −−− ∫∫∫ E••• ld === −−−∫∫∫ Er dr( )=== −−−∫∫∫ Er dr( )−−−∫∫∫ Er dr( ) === +++0 4πεπεπε 0 aR r===∞∞∞ ∞∞∞ ∞∞∞ Ra ELECTRIC POTENTIAL for Charged Sphere (Y&F, ex.23.8) Suppose we have a charged q metal sphere with charge q. What is the electric potential as a function of radius r? N.B. V is continuous but E is not. Clicker exercise 1 A point charge Q is fixed at the a center of an uncharged conducting Q spherical shell of inner radius a and outer radius b. b – What is the value of the potential Va at the inner surface of the spherical shell? 1 Q 1 Q V = V = (a) Va = 0 (b) a (c) a 4 πε 0 a 4πε 0 b E Clicker exercise out 1 A point charge Q is fixed at the center of an uncharged conducting a spherical shell of inner radius a Q and outer radius b. – What is the value of the potential at Va b the inner surface of the spherical shell? 1 Q 1 Q V a = V = (a) Va = 0 (b) 4πε a (c) a 0 4πε 0 b • How to start?? The only thing we know about the potential is its definition: a r r Va ≡Va −V∞ = −∫ E • ld ∞ • To calculate Va, we need to know the electric field E • Outside the spherical shell: r Q rˆ E = • Apply Gauss’ Law to sphere: 2 4πε 0 r • Inside the spherical shell: E = 0 b r r a r r 1 Q 1 Q V = V = − ∫ E • ld − ∫ E • ld Va = + 0 a a 4πε 0 b ∞ b 4πε 0 b Preflight 6: Two spherical conductors are separated by a large distance. They each carry the same positive charge Q. Conductor A has a larger radius than conductor B. A B 2) Compare the potential at the surface of conductor A with the potential at the surface of conductor B. a) VA > V B b) VA = V B c) VA < V B Electrical Potential Two ways to find V at any point in space: r r r • Use electric field: V = − E • ld ∫∞ • Sum or Integrate over charges: q 1 r P 1 q 1 = i q2 r2 V ∑ i 4πε 0 ri r3 q 3 P r 1 dq dq V = ∫ 4πε 0 r Examples of integrating over a distribution of charge: • line of charge (review this one) • ring of charge • disk of charge You should be able to do these. Infinite line charge or conducting cylinder. r Linear charge density λ λ E = 2πε 0r b λ λ rb Va−= V b ∫ Edr = ∫ = ln( ) a 2πε0r 2 πε 0 r a Suppose we set rb to infinity, Instead, set ra=r and rb=r 0 at potential is infinite some fixed radius r 0. Cover homework hints Potential from a charged sphere 1 q V( r ) = 4πε 0 r E (where V ( ∞ ) ≡ 0 ) r Equipotential • The electric field of the charged sphere has spherical symmetry. • The potential depends only on the distance from the center of the sphere, as is expected from spherical symmetry. • Therefore, the potential is constant on a sphere which is concentric with the charged sphere. These surfaces are called equipotentials. • Notice that the electric field is perpendicular to the equipotential surface at all points . Equipotentials Defined as: The locus of points with the same potential. • Example: for a point charge, the equipotentials are spheres centered on the charge. The electric field is always perpendicular to an equipotential surface! B r r Why?? VB −VA = −∫ E • ld A Along the surface, there is NO change in V (it’s an equipotential!) B r Therefore, r − ∫ E • ld = ∆V = 0 A r r We can conclude then, that E • ld is zero. If the dot product of the field vector and the displacement vector is zero, then these two vectors are perpendicular, or the electric field is always perpendicular to the equipotential surface. EXAMPLES of Equipotential Lines Conductors + + + + + + + + + + + + + + • Claim The surface of a conductor is always an equipotential surface (in fact, the entire conductor is an equipotential). • Why?? If surface were not equipotential, there would be an electric field component parallel to the surface and the charges would move!! Preflight 6: A B 3) The two conductors are now connected by a wire. How do the potentials at the conductor surfaces compare now ? a) VA > V B b) VA = V B c) VA < V B 4) What happens to the charge on conductor A after it is connected to conductor B ? a) QA increases b) Q A decreases c) Q A doesn’t change Charge on Conductors? • How is charge distributed on the surface of a conductor? – KEY: Must produce E=0 inside the conductor and E normal to the surface . Spherical example (with little off-center charge): E=0 inside conducting shell. + + + + - - + - - - charge density induced on + - + inner surface non-uniform. - +q - + - - + - - + charge density induced on + - - outer surface uniform + - + + + + E outside has spherical symmetry centered on spherical conducting shell. Equipotential Example • Field lines more closely spaced near end with most curvature – higher E-field • Field lines ⊥⊥⊥ to surface near the surface (since surface is equipotential). • Near the surface, equipotentials have similar shape as surface. • Equipotentials will look more circular (spherical) at large r. Conservation of Energy • The Coulomb force is a CONSERVATIVE force (i.e., the work done by it on a particle which moves around a closed path returning to its initial position is ZERO.) • Therefore, a particle moving under the influence of the Coulomb force is said to have an electric potential energy defined by: U = qV this “q” is the “test charge” in other examples... • The total energy (kinetic + electric potential) is then conserved for a charged particle moving under the influence of the Coulomb force. Lecture 6, ACT 3 Two test charges are brought 3A separately to the vicinity of a positive charge Q. Q r – charge +q is brought to pt A, a q distance r from Q. A – charge +2 q is brought to pt B, a distance 2r from Q. Q 2r B – Compare the potential energy of q 2q (UA) to that of 2q (UB): (a) UA < UB (b) UA = UB (c) UA > UB 3B • Suppose charge 2q has mass m and is released from rest from the above position (a distance 2r from Q). What is its velocity vf as it approaches r = ∞? 1 Qq v = 1 Qq (c) (a) f (b) v f = v f = 0 4πε 0 mr 2πε 0 mr Clicker Problem • Two test charges are brought 3A separately to the vicinity of positive charge Q. r Q q – charge +q is brought to pt A, a A distance r from Q. Q 2q – charge +2 q is brought to pt B, a 2r distance 2r from Q. B – Compare the potential energy of q (UA) to that of 2q (UB): (a) UA < UB (b) UA = UB (c) UA > UB • The potential energy of q is proportional to Qq /r. • The potential energy of 2q is proportional to Q(2q)/(2 r). • Therefore, the potential energies UA and UB are EQUAL!!! Clicker Problem 3B • Suppose charge 2q has mass m and is released from rest from the above position (a distance 2r from Q). What is its velocity vf as it approaches r = ∞? 1 Qq 1 Qq (a) v f = (b) v f = (c) v f = 0 4πε 0 mr 2πε 0 mr • What we have here is a little combination of 170 and 272. • The principle at work here is CONSERVATION OF ENERGY. • Initially: • The charge has no kinetic energy since it is at rest. • The charge does have potential energy (electric) = UB. • Finally: • The charge has no potential energy ( U ∝ 1/ R) • The charge does have kinetic energy = KE 1 Q( 2q ) 1 2 1 Qq = mv 2 v = U =KE f f B 4πε 0 2r 2 2πε 0 mr How to obtain vector E field from V r r r • Determine V from E: V = − E • ld ∫∞ Example: V due to spherical charge distribution.
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