2. Electric Potential q A C B r A B r path independence a a Rr VQ 4   r Q 4   R

3. 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!!

4. Conservation of Energy of a particle from phys 1301 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 )= kx 2 –electric;electric potential energy (today!) examples of non-conservative forces (heat) –friction –viscous damping (terminal velocity) –electrical resistance

5. Example: Gravitational Force is conservative (and attractive) 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 More potential energy Less potential energy pt 1 pt 2 0 U(r)U(r) U(r1)U(r1) U(r2)U(r2)

6. 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….

7. Electric potential and potential energy Imagine a positive test charge, Q o, in an external electric field, (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 Q o from where U is zero to point ( x,y,z )

8. Define (U = QV) U depends on Q o, but V is independent of Q o (which can be + or -) V ( x,y,z ) is the electric potential in volts associated with. (1V = 1 J/c) –V ( x,y,z ) is a scalar field Electric potential and potential energy

9. Electric potential difference Suppose charge q 0 is moved from pt A to pt B through a region of space described by electric field E. A B q 0 E F elec F we supply = - F elec 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 W AB ≡ W A  B will have to be done to accomplish this task.

10. 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 Electric potential difference, cont.

11. 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? ΔV AB = V B - V A and  V AC = V C - V A a) ΔV AB > 0 and  V AC > 0 b) ΔV AB = 0 and  V AC < 0 c)  V AB = 0 and  V AC > 0 d) ΔV AB < 0 and  V AC < 0 E A B C Question 1

12. 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? ΔV AB = V B - V A and  V AC = V C - V A a) ΔV AB > 0 and  V AC > 0 b) ΔV AB = 0 and  V AC < 0 c)  V AB = 0 and  V AC > 0 d) ΔV AB < 0 and  V AC < 0 E A B C Question 1

13. 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 ( ) Motion in a region of electric field

14. Question 2 A single charge ( Q = -1  C) 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 V AB  V B - V A ) x -1  C   A B (a) Δ V AB <  (b) Δ V AB =  (c) Δ V AB > 

15. Question 2 A single charge ( Q = -1  C) 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 V AB  V B - V A ) x -1  C   A B (a) Δ V AB <  (b) Δ V AB =  (c) Δ V AB >  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 -1  C 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, Δ V AB <0 !!

16. Δ V AB 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. Δ V AB is the same for any path chosen to move from A to B (because electric forces are conservative). A B q 0 E

17. 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 (proof next time) A C B Eh r  dldl

18. 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 B (a) W AB < 0 (b) W AB = 0 (c) W AB > 0

19. Choose a path along the arc of a circle centered at the charge. Along this path at every point!! Question 3 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. 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 B (a) W AB < 0 (b) W AB = 0 (c) W AB > 0

20. 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 here “ r ” is distance to origin line integral

21. Potential from charged spherical shell a a Q 4   a ar V Q 4   r Potential r > a : r < a : E -field (from Gauss' Law) E r = 0 r < a : E r = 1 4  0 Q r 2  r > a : Radius = a

22. What does the result mean? 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 as 1/ r 2. The potential for r > a is given by the integral of E r. This integral is simply the area underneath the E r curve. a a ar ErEr R Q 4   a ar V Q 4   r R