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Transcript
Electric Potential Q V 4pe0 R Q 4pe0 r R r C B r a B q r A A path independence a Today… • Conservative Forces and Energy Conservation – Total energy is constant and is sum of kinetic and potential • Introduce Concept of Electric Potential – A property of the space and sources as is the Electric Field – Potential differences drive all biological & chemical reactions, as well as all electric circuits • Calculating Electric Potentials put V(infinity)=0 – Charged Spherical Shell – N point charges – Example: electric potential of a charged sphere • Electrical Breakdown – Sparks – Lightning!! Conservation of Energy of a particle from phys 1301 • Kinetic Energy (K) – non-relativistic • Potential Energy (U) 1 2 K mv 2 – determined by force law U ( x, y, z ) • for Conservative Forces: K+U is constant – total energy is always constant • examples of conservative forces – gravity; gravitational potential energy – springs; coiled spring energy (Hooke’s Law): U(x)=kx2 – electric; electric potential energy (today!) • examples of non-conservative forces (heat) – friction – viscous damping (terminal velocity) – electrical resistance Example: Gravitational Force is conservative (and attractive) • Consider a comet in a highly elliptical orbit U(r) pt 2 pt 1 0 U(r1) GMm U (r ) r • At point 1, particle has a lot of potential energy, but little kinetic energy • At point 2, particle has little potential energy, but a lot of kinetic energy More potential energy U(r2) Less potential energy Total energy = K + U is constant! Electric forces are conservative, too • Consider a charged particle traveling through a region of static electric field: + • A negative charge is attracted to the fixed positive charge • negative charge has more potential energy and less kinetic energy far from the fixed positive charge, and… • more kinetic energy and less potential energy near the fixed positive charge. • But, the total energy is conserved • We will now discuss electric potential energy and the electrostatic potential…. Electric potential and potential energy • Imagine a positive test charge, Qo, in an external electric field, E x, y, z (a vector field) • What is the potential energy, U(x,y,z) of the charge in this field? – Must define where in space U(x,y,z) is zero, perhaps at infinity (for charge distributions that are finite) – U(x,y,z) is equal to the work you have to do to take Qo from where U is zero to point (x,y,z) Electric potential and potential energy • Define V x, y, z U x, y, z Q0 (U = QV) • U depends on Qo , but V is independent of Qo (which can be + or -) • V(x,y,z) is the electric potential in volts associated with E x, y, z . (1V = 1 J/c) – V(x,y,z) is a scalar field Electric potential difference • Suppose charge q0 is moved from pt A to pt B through a region of space described by electric field E. Fwe supply = -Felec Felec q0 A E B • To move a charge in an E-field, we must supply a force just equal and opposite to that experienced by the charge due to the E-field. • Since there will be a force on the charge due to E, a certain amount of work WAB≡WAB will have to be done to accomplish this task. Electric potential difference, cont. • Remember: work is force times distance \ • To get a positive test charge from the lower potential to the higher potential you need to invest energy - you need to do work • The overall sign of this: A positive charge would “fall” from a higher potential to a lower one • If a positive charge moves from high to low potential, it can do work on you; you do “negative work” on the charge Question 1 A E C B 2) Points A, B, and C lie in a uniform electric field. What is the potential difference between points A and B and A and C? ΔVAB = VB - VA and VAC = VC - VA a) ΔVAB > 0 and VAC > 0 b) ΔVAB = 0 and VAC < 0 c) VAB = 0 and VAC > 0 d) ΔVAB < 0 and VAC < 0 Question 1 A E C B 2) Points A, B, and C lie in a uniform electric field. What is the potential difference between points A and B and A and C? ΔVAB = VB - VA and VAC = VC - VA a) ΔVAB > 0 and VAC > 0 b) ΔVAB = 0 and VAC < 0 c) VAB = 0 and VAC > 0 d) ΔVAB < 0 and VAC < 0 Motion in a region of electric field •A positive charge is released from rest in a region of electric field •The charge moves towards a region of lower potential energy, i.e lower electrical potential –Away from a positive charge (+ve electric field), you have to do work to push them together –Towards a negative charge (-ve electric field), you have to do work to pull them apart •To move without changing the electrical potential energy –Must move perpendicular to the field ( F dl 0 ) Question 2 • A single charge ( Q = -1mC) is fixed at the origin. Define point A at x = + 5m and point B at x = +2m. – What is the sign of the potential difference between A and B? (Δ VAB VB - VA ) (a) ΔVAB < 0 (b) Δ VAB = 0 B -1mC (c) Δ VAB > 0 A x Question 2 • A single charge ( Q = -1mC) is fixed at the origin. Define point A at x = + 5m and point B at x = +2m. B – What is the sign of the potential difference between A and B? (Δ VAB VB - VA ) (a) ΔVAB < 0 (b) Δ VAB = 0 A -1mC (c) Δ VAB > 0 • Imagine placing a positive charge at point A and determining which way it would move. Remember that it will always “fall” to lower potential. •A positive charge at A would be attracted to the -1mC charge; therefore NEGATIVE work would be done to move the charge from A to B. •You can also determine the sign directly from the definition: Since , ΔVAB <0 !! x Δ VAB is Independent of Path q0 E A B • The integral is the sum of the tangential (to the path) component of the electric field along a path from A to B. • This integral does not depend upon the exact path chosen to move from A to B. • Δ VAB is the same for any path chosen to move from A to B (because electric forces are conservative). Does it really work? • Consider case of constant field: – Direct: A - B B h A q C r dl • Long way round: A - C – B • So here we have at least one example of a case in which the integral is the same for BOTH paths. • In fact, it works for all paths (proof next time) E Question 3 A positive charge Q is moved from A to B along the path shown. What is the sign of the work done to move the charge from A to B? (a) WAB < 0 (b) WAB = 0 (c) WAB > 0 A B Question 3 A positive charge Q is moved from A to B along the path shown. What is the sign of the work done to move the charge from A to B? (a) WAB < 0 • (b) WAB = 0 A B (c) WAB > 0 A direct calculation of the work done to move a positive charge from point A to point B is not easy. • Neither the magnitude nor the direction of the field is constant along the straight line from A to B. • But, you DO NOT have to do the direct calculation. • • Remember: potential difference is INDEPENDENT OF THE PATH!! Therefore we can take any path we wish. Choose a path along the arc of a circle centered at the charge. Along this path at every point!! Electric Potential: where is it zero? • So far we have only considered potential differences. • Define the electric potential of a point in space as the potential difference between that point and a reference point. • a good reference point is infinity ... we often set V = 0 • the electric potential is then defined as: • for a point charge at origin, integrate in from infinity along a line r • here “r” is distance to origin Vr V dl E r q qdl Vr V 2 4 pe l 4 pe l 0 0 r line integral Potential from charged spherical shell • E-field (from Gauss' Law) • r < a: Er = 0 • r >a: • Potential • r > a: a Radius = a a 1 Q Er = 4pe 0r 2 V Q 4pe0 r Q 4pe0 a a • r < a: r What does the result mean? • This is the plot of the radial component of the electric field of a charged spherical shell: Er Notice that inside the shell, the electric field is zero. Outside the shell, the electric field falls off as 1/r2. The potential for r>a is given by the integral of Er. This integral is simply the area underneath the Er curve. a R r V Q 4pe0 a Q 4pe0 r a a R a r