Equipotentials and Energy

1. Equipotentials and Energy Electric Potential

2. Last time: Potential energy and potential • Potential energy stored in a static charge distribution • work we do to assemble the charges • Electric potential energy of a charge in the presence of a set of source charges • potential energy of the test charge equals the potential from the sources times the test charge: U = qV • Electrostatic potential at any point in space • sources give rise toV(r): allows us to calculate the potential Veverywhere if we know the electric field E (define VA = 0 somewhere)

3. Potential from charged spherical shell (See Appendix A for a more complex example) 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:

4. 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

5. a Q b 1 • A point charge Q is fixed at the center of an uncharged conducting spherical shell of inner radius a and outer radius b. • What is the value of the potential Vaat the inner surface of the spherical shell? (b) (a) (c)

6. Eout • How to start?? The only thing we know about the potential is its definition: • To calculate Va, we need to know the electric field E • Outside the spherical shell: • Apply Gauss’ Law to sphere: • Inside the spherical shell: E = 0 1 • A point charge Q is fixed at the center of an uncharged conducting spherical shell of inner radius a and outer radius b. • What is the value of the potential Va at the inner surface of the spherical shell? a Q b (b) (a) (c)

7. 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 > VBb) VA = VBc) VA < VB

8. Equipotential Potential from a charged sphere Last time… Er (where ) • 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 along a sphere which is concentric with the point charge. These surfaces are called equipotentials. • Notice that the electric field is perpendicular to the equipotential surface at all points.

9. 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! Why?? Along the surface, there is NO change inV(it’s an equipotential!) Therefore, We can conclude then, that 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.

10. Electric Dipole Equipotentials

11. Some fish have the ability to produce & detect electric fields • Navigation, object detection, communication with other electric fish • “Strongly electric fish” (eels) can stun their prey Electric Fish Black ghost knife fish Dipole-like equipotentials More info: Prof. Mark Nelson, Beckman Institute, UIUC -Electric current flows down the voltage gradient -An object brought close to the fish alters the pattern of current flow

12. + + + + + + + + + + + + + + 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!!

13. Preflight 6: B A 3) The two conductors are now connected by a wire. How do the potentials at the conductor surfaces compare now ? a) VA > VB b) VA = VB c) VA< VB 4) What happens to the charge on conductor A after it is connected to conductor B ? a) QAincreases b) QAdecreases c) QAdoesn’t change

14. + + + E=0 inside conducting shell. + + - - - - - + - + - - +q charge density induced on inner surface non-uniform. + - - + - + - + - - - + charge density induced on outer surface uniform + + + + Eoutside has spherical symmetry centered on spherical conducting shell. 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):

15. -q An uncharged spherical conductor has a weirdly shaped cavity carved out of it. Inside the cavity is a charge -q. How much charge is on the cavity wall? 2A (a) Less than< q(b) Exactly q(c) More than q 2B How is the charge distributed on the cavity wall? (a) Uniformly (b) More charge closer to –q (c) Less charge closer to -q 2C How is the charge distributed on the outside of the sphere? (a) Uniformly (b) More charge near the cavity (c) Less charge near the cavity

16. -q An uncharged spherical conductor has a weirdly shaped cavity carved out of it. Inside the cavity is a charge -q. How much charge is on the cavity wall? 2A (a) Less than< q(b) Exactly q(c) More than q By Gauss’ Law, since E=0 inside the conductor, the total charge on the inner wall must be +q (and therefore -q must be on the outside surface of the conductor, since it has no net charge).

17. -q The induced charge will distribute itself nonuniformly to exactly cancel everywhere in the conductor. The surface charge density will be higher near the -q charge. 2B How is the charge distributed on the cavity wall? (a) Uniformly (b) More charge closer to -q (c) Less charge closer to -q

18. -q 2C How is the charge distributed on the outside of the sphere? (a) Uniformly (b) More charge near the cavity (c) Less charge near the cavity As in the previous example, the charge will be uniformly distributed (because the outer surface is symmetric). Outside the conductor the E field always points directly to the center of the sphere, regardless of the cavity or charge. Note: this is why your radio, cell phone, etc. won’t work inside a metal building!

19. ConductorsversusInsulatorsCharges move to Charges cannotcancel electric field move at allin the conductor E=0  equipotential Charge distribution surface on insulatorunaffected byexternal fieldsAll charge on surface Charge can sit “inside”(Appendix B describes method of “images” to find the surface charge distribution on a conductor [only for your reading pleasure!])

20. r L But: r S Smaller sphere has the larger surface charge density ! Þ Charge on Conductor Demo • How is the charge distributed on a non-spherical conductor?? Claim largest charge density at smallest radius of curvature. • 2 spheres, connected by a wire, “far” apart • Both at same potential

21. 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.

22. 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.

23. r2 r1 Ball 1 Ball 2 Smallerr higherE closer to breakdown (a)Ball 1 (b)Ball 2 (c)Same Time Ex. High Voltage Terminals must be big! Followup Question: 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?

24. Therefore, a particle moving under the influence • of the Coulomb force is said to have an • electric potential energy defined by: this “q” is the “test charge” in other examples... 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.) • The total energy (kinetic + electric potential) is then conserved for a charged particle moving under the influence of the Coulomb force.

25. Preflight 6: A E C B 6) If a negative charge is moved from point A to point B, its electric potential energy a) increases b) decreases c) doesn’t change

26. r q A Q • charge +q is brought to pt A, a distance r from Q. • charge +2q is brought to pt B, a distance 2r from Q. Q B 2r (b)UA= UB (c)UA> UB (a)UA< UB • Compare the potential energy of q (UA) to that of 2q(UB): 2q 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 =¥? (c) (a) (b) Two test charges are brought separately to the vicinity of a positive charge Q. 3A

27. r Q q A Q 2q 2r B (b)UA= UB (c)UA> UB (a)UA< UB • Two test charges are brought separately to the vicinity of positive charge Q. • charge +q is brought to pt A, a distance rfrom Q. • charge +2q is brought to pt B, a distance 2r from Q. • Compare the potential energy of q (UA) to that of 2q (UB): 3A • The potential energy of q is proportional to Qq/r. • The potential energy of 2q is proportional to Q(2q)/(2r). • Therefore, the potential energies UA and UB are EQUAL!!!

28. 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 =¥? (c) (a) (b) • What we have here is a little combination of 111 and 112. • 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 (Uµ 1/R) • The charge does have kinetic energy = KE

29. Accelerators • Electrostatic: Van de Graaff • electrons ®100 keV ( 105 eV) • Electromagnetic: Fermilab • protons ®1TeV ( 1012 eV) Energy Units MKS: U = QV1 coul-volt = 1 joule for particles (e, p, ...) 1 eV = 1.6x10-19 joules

30. Summary • If we know the electric field everywhere, we can calculate the potential, e.g., • The place where V=0 is “arbitrary” (often at infinity) • Physically, DV is what counts • Equipotential surfaces are surfaces where the potential is constant • Conductors are equipotentials • “Breakdown” can occur if the electric field exceeds the “dielectric strength” • Next time  capacitors

31. I II III IV r Appendix A: Electrical potential examples Calculate the potentialV(r)at the point shown (r<a) uncharged conductor c b a solid sphere with total chargeQ

32. 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)

33. 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?

34. 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

35. 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

36. 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

37. 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:

38. 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

39. - + - + - + - + + - - - - - - + + + + + + + + + + Appendix B: FYI: Induced charge distribution on conductorvia“method of images” • Consider a source charge brought close to a conductor: + • Charge distribution “induced” on conductor by source charge: • Induced charge distribution is “real” and sources E-field so that the total is zero inside conductor! • resulting E-field is sum of field from source charge and induced charge distribution • E-field is locally perpendicular to surface • With enough symmetry, can solve fors on conductor • how? Gauss’ Law

40. - - - - - Appendix B: (FYI) Induced charge distribution on conductorvia“method of images” • Consider a source charge brought close to a planar conductor: • Charge distribution “induced” on conductor by source charge • conductor is equipotential • E-field is normal to surface • this is just like a dipole + - • Method of Images for a charge (distribution) near a flat conducting plane: • reflect the point charge through the surface and put a charge of opposite sign there • do this for all source charges • E-field at plane of symmetry - the conductor surface determines s.

41. What does grounding do?1. Acts as an “infinite” source or sink of charge.2. The charges arrange themselves in such a way as to minimize the global energy (e.g., E0 at infinity, V0 at infinity).3. Typically we assign V = 0 to ground.