Electric Potential

Q Q 4pe0r 4pe0R a a Electric Potential V Definitions Examples R r C B r B A r q A path independence

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!! Text Reference: Chapter 24.1- 4 Examples: 24.1,2 and 4-6

Conservation of Energy of a particle from phys 111 • Kinetic Energy (K) • non-relativistic • Potential Energy (U) • determined by force law • 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)

Example: Gravitational Force is conservative(and attractive) U(r) pt 1 pt 2 0 U(r1) More potential energy Less potential energy U(r2) • Consider a comet in a highly elliptical orbit • At point 1, particle has a lot of potential energy, but little kinetic energy Total energy = K + U is constant! • At point 2, particle has little potential energy, but a lot of kinetic energy

+ - Electric forces are conservative, too • 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 • Consider a charged particle traveling through a region of static electric field: • 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)<- it’s a vector field • What is the potential energy, U(x,y,z) of the charge in this field? • Must define where in spaceU(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 takeQofrom whereUis zero to point(x,y,z) • Define V(x,y,z)=U(x,y,z) / Qo (U = QV) • U depends on what Qo is, 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

Felec Fwe supply = -Felec q E 0 A B Electric potential difference • Suppose charge q0 is moved from pt A to pt B through a region of space described by electric field E. • 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≡WAB will have to be done to accomplish this task.

\ Þ 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

B A ´ ´ x -1mC (c)ΔVAB> 0 (b)ΔVAB= 0 (a)ΔVAB< 0 • The simplest way to get the sign of the potential difference is to imagine placing a positive charge at pointAand determining which way it would move. Remember that it will always “fall” to lower potential. • A positive charge atAwould be attracted to the -1mC charge; therefore NEGATIVE work would be done to move the charge from A toB. • You can also determine the sign directly from the definition: Since , ΔVAB <0 !! Lecture 5, ACT 1 • 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 betweenAand B? (ΔVABºVB - VA )

E q 0 A B ΔVAB is Independent of Path • 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. • ΔVABis the same for any path chosen to move from A to B (because electric forces are conservative).

B C q h E r dl A Does it really work? • Consider case of constant field: • Direct:A - B • 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  see Appendix.

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? 2A A (a) WAB < 0 B (c) WAB > 0 (b) WAB = 0 Lecture 5, ACT 2 • 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!!

line integral 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. • agood reference point is infinity ... we often setV = 0 • the electric potential is then defined as: • for a point charge at origin, integrate in from infinity along some axis, e.g., the x-axis • here “r” is distance to origin

Q Q (b) VA= VB (c)VA> VB (a)VA< VB Lecture 5, ACT 2 Two test charges are brought separately to the vicinity of positive charge Q. 2B r q + A 2q 2r • charge+qis brought to ptA, a distancerfrom Q. + B • THE POTENTIAL IS A FUNCTION OF THE SPACE !!!!! • The Potential does not depend on the “test” charge at all. • Like ACT 1, except here a positive source charge creates the field. • charge +2qis brought to ptB, a distance2r fromQ. • Compare the potential at pointA(VA) to that atB(VB): • A positive “test charge” would move (“fall”) from point A towards pointB. • Therefore,VA > VB • In fact, since pointBis twice as far from the charge as point A, we calculate thatVA = 2VB!!

Potential from charged spherical shell V Q Q • E-field (from Gauss' Law) 4pe0a 4pe0r • r< a: Er = 0 a r 1 Q • r >a: Er = 2 4 pe r 0 a a • Potential • r> a: • r < a:

V Q Q 4pe0a 4pe0r a r a R a What does the result mean? Er • This is the plot of the radial component of the electric field of a charged spherical shell: Notice that inside the shell, the electric field is zero. Outside the shell, the electric field falls off as1/r2. The potential forr>ais given by the integral ofEr. This integral is simply the area underneath theErcurve. R a r

r1 x q1 r2 r3 q2 q3 Þ Potential from N charges The potential from a collection of N charges is just the algebraic sum of the potential due to each charge separately.

Lecture 5, ACT 3 • Which of the following charge distributions produces V(x) = 0 for all points on thex-axis? (we are defining V(x) º 0 at x = ¥ ) +2mC +2mC +1mC +2mC +1mC -2mC x x x -1mC -2mC -2mC -1mC +1mC -1mC (c) (a) (b) The key here is to realize that to calculate the total potential at a point, we must only make an ALGEBRAIC sum of the individual contributions. Therefore, to makeV(x)=0for allx, we must have the+Qand-Qcontributions cancel, which means that any point on the x-axis must be equidistant from+2mCand -2mCand also from+1mC and-1mC. This condition is met only in case (a)!

Dielectric Breakdown Insulator Conductor Arc discharge equalizes the potential Sparks • High electric fields can ionize nonconducting materials (“dielectrics”) • Breakdown can occur when the field is greater than the “dielectric strength” of the material. • E.g., in air, What is ΔV? Ex. Note: High humidity can also bleed the charge off  reduce ΔV.

Lecture 5, ACT 4 r2 r1 Ball 1 Ball 2 (a) Ball 1 (b) Ball 2 (c) Same Time \ Smaller r higherE closer to breakdown Ex. High Voltage Terminals must be big! Two charged balls are each at the same potential V. Ball 2 is twice as large as ball 1. • As V is increased, which ball will induce breakdown first?

Lightning! _ _ _ + + + Collisions produce charged particles. The heavier particles (-) sit near the bottom of the cloud; the lighter particles (+) near the top. Stepped Leader Negatively charged electrons begin zigzagging downward. Attraction As the stepped leader nears the ground, it draws a streamer of positive charge upward. Flowing Charge As the leader and the streamer come together, powerful electric current begins flowing Contact! Intense wave of positive charge, a “return stroke,” travels upward at 108 m/s Factoids:

Reading Assignment: Appendix B; Chapter 24.3,5 and 25.1 examples: 24.7,11,13,15 and 25.1 Summary • If we know the electric field E, • allows us to calculate the potential function V everywhere (define VA = 0 above) • Potential due to n charges: • Equipotential surfaces are surfaces where the potential is constant.

Appendix A: Independent of Path? B r2 A r1 q • Consider any path from A to B as being made up of a succession of arc plus radial parts as above. The work along the arcs will always be 0, leaving just the sum of the radial parts. All inner sums will cancel, leaving just the initial and final radii as above... Therefore it's general!

Appendix A: Independent of Path? E B r B A r q A • We want to evaluate potential difference from A to B • What path should we choose to evaluate the integral? • If we choose straight line, the integral is difficult to evaluate. • Magnitude different at each pt along line. • Angle betweenEand path is different at each pt along line. E C • If we choose path ACB as shown, our calculation is much easier! • From A to C, E is perpendicular to the path. i.e • From C to B, E is parallel to the path. i.e B r B A r q A

Appendix A: Independent of Path? • Evaluate potential difference from A to B along path ACB. E C by definition: B Evaluate the integral: r B A r q A

C Appendix A: Independent of Path? B r B A r q A B r2 A r1 q This is the same result as above!! The straight line path is better approximated by Increasing the number of arcs and radial pieces. • How general is this result? • Consider the approximation to the straight path fromA->B(white arrow) = 2 arcs (radii = r1 and r2) plus the 3 connecting radial pieces. • For the 2 arcs + 3 radials path:

I II III IV r Appendix B Calculate the potentialV(r)at the point shown (r<a) uncharged conductor c b a solid sphere with total chargeQ

I II III uncharged conductor IV c r b a solid sphere with total chargeQ Calculate the potential V(r)at the point shown (r < a) Calculating Electric Potentials • Where do we know the potential, and where do we need to know it? V=0 atr = ... we need r < a... • Determine E(r) for all regions in between these two points • Determine DV for each region by integration ... and so on ... • Check the sign of each potential differenceDV DV > 0means we went “uphill”DV< 0means we went“downhill” (from the point of viewof a positivecharge)

I II III IV c r b a Calculate the potentialV(r)at the point shown (r < a) Calculating Electric Potentials • Look at first term: • Line integral from infinity to c has to be positive, pushing against a force: Line integral is going “in” which is just the opposite of what usually is done - controlled by limits • What’s left?

I II III IV r Calculate the potential V(r)at the point shown (r < a) Calculating Electric Potentials • Look at third term: c b a • Line integral from b to a, again has to be positive, pushing against a force: Line integral is going “in” which is just the opposite of what usually is done - controlled by limits • What’s left? Previous slide we have calculated this already

I II III IV c r b a Calculate the potential V(r)at the point shown (r < a) Calculating Electric Potentials • Look at last term: • Line integral from a to r, again has to be positive, pushing against a force. • But this time the force doesn’t vary the same way, since “r ’” determines the amount of source charge This is the charge that is inside “r” and sources field • What’s left to do? • ADD THEM ALL UP! • Sum the potentials

I II III IV c r b a Potential increase from moving into the sphere Calculate the potential V(r)at the point shown (r < a) Calculating Electric Potentials • Add up the terms from I, III and IV: III IV I An adjustment to account for the fact that the conductor is an equipotential, DV= 0 from c →b The potential difference from infinity to a if the conducting shell weren’t there

I II III IV c r b a Calculating Electric Potentials Summary The potential as a function of r for all 4 regions is: Ir > c: II b < r < c: III a < r < b: IV r < a:

I II III IV c r b a Let’s try some numbers Q = 6m C a = 5cmb = 8cm c = 10cm Ir > c: V(r = 12cm) = 449.5 kV II b < r < c: V(r =9cm) = 539.4 kV III a < r < b: V(r = 7cm) = 635.7 kV IV r < a: V(r = 3cm) = 961.2 kV

Electric Potential