1. Electric Potential Chapter 25 Electric Potential Energy Electric Potential Equipotential Surfaces

2. Electric Potential: the Bottom Line The electric potential V(r) is an easy way to calculate the electric field – easier than directly using Coulomb’s law. For a collection of charges q i at positions r i the electric potential at r is the scalar sum: For a point charge at the origin: Calculate V and then find the electric field by taking the gradient:

3. Example: two point charges a q q r

4. CONSERVATIVE FORCES A conservative force “gives back” work that has been done against it When the total work done by a force on an object moving around a closed loop is zero, then the force is conservative  F·dr = 0  F is conservative The circle on the integral sign indicates that the integral is taken over a closed path The work done by a conservative force, in moving and object between two points A and B, is independent of the path taken W=  A B F·dr is a function of A and B only – it is NOT a function of the path selected. We can define a potential energy difference as U AB =-W. °

5. POTENTIAL ENERGY Potential energy is defined at each point in space, but it is only the difference in potential energy that matters. Potential energy is measured with respect to a reference point (usually infinity). So we let A be the reference point (i.e, define U A to be zero), and use the above integral as the definition of U at point B. The change  U AB in potential energy due to the electric force is thus  U AB = -q  A B E·dr  U AB = U B – U A = potential energy difference between A and B

6. POTENTIAL ENERGY IN A CONSTANT FIELD E E The potential energy difference between A and B equals the work necessary to move a charge +q from A to B  U AB = U B – U A = -  q E·dl But E = constant, and E.dl = -E dl, so:  U AB = -  q E · dl =  q E dl = q E  dl = q E L  U AB = q E L A B L dl

7. ELECTRIC POTENTIAL DIFFERENCE The potential energy  U depends on the charge being moved. To remove this dependence, we introduce the concept of the electric potential V. This is defined in terms of the difference  V:  V AB =  U AB / q = -  A B E · dl Electrical Potential = Potential Energy per Unit Charge = line integral of -E·dl  V AB = Electric potential difference between the points A and B. Units are Volts (1V = 1 J/C), and so the electric potential is often called the voltage. A positive charge is pushed from regions of high potential to regions of low potential.

8. ELECTRIC POTENTIAL IN A CONSTANT FIELD E The electrical potential difference between A and B equals the work per unit charge necessary to move a charge +q from A to B  V AB = V B – V A = -  E·dl But E = constant, and E.dl = -1 E dl, so:  V AB = -  E·dl =  E dl = E  dl = E L  V AB = E L A B L E dL  V AB =  U AB / q  U AB = q E L

9. Cases in which the electric field E is not aligned with dl  V AB = -  A B E · dl A B E  E. dl = E dl cos    V AB = - E cos   dl = - E L cos  The electric potential difference does not depend on the integration path. So pick a simple path. One possibility is to integrate along the straight line AB. This is convenient in this case because the field E is constant, and the angle  between E and dl is constant. A B

10. Cases in which the electric field E is not aligned with dl A B E C d Another possibility is to choose a path that goes from A to C, and then from C to B  V AB =  V AC +  V CB  V AC = E d  V CB = 0 (E  dL) Thus,  V AB = E d but d = L cos  = - L cos   V AB = - E L cos    L  V AB = -  A B E · dl

11. Equipotential Surfaces (lines)  V AB = E L For a constant field E E L E L B A x  V Ax = E x All the points along the dashed line at x, are at the same potential. The dashed line is an equipotential line Similarly, at a distance x from plate A

12. Equipotential Surfaces (lines) E L x It takes no work to move a charge at right angles to an electric field E  dl   Edl = 0   V = 0 If a surface(line) is perpendicular to the electric field, all the points in the surface (line) are at the same potential. Such surface (line) is called EQUIPOTENTIAL EQUIPOTENTIAL  ELECTRIC FIELD

13. We can make graphical representations of the electric potential in the same way as we have created for the electric field: Equipotential Surfaces Lines of constant E

14. We can make graphical representations of the electric potential in the same way as we have created for the electric field: Equipotential Surfaces Lines of constant E Lines of constant V (perpendicular to E)

15. We can make graphical representations of the electric potential in the same way as we have created for the electric field: Equipotential Surfaces Equipotential plots are like contour maps of hills and valleys. A positive charge would be pushed from hills to valleys. Lines of constant E Lines of constant V (perpendicular to E)

16. Equipotential Surfaces Equipotential plots are like contour maps of hills and valleys. How do the equipotential surfaces look for: (a) A point charge? (b) An electric dipole? + + - E Equipotential plots are like contour maps of hills and valleys. A positive charge would be pushed from hills to valleys.

17. Point Charge q  The Electric Potential b a q What is the electrical potential difference between two points (a and b) in the electric field produced by a point charge q. Electric Potential of a Point Charge

18. Place the point charge q at the origin. The electric field points radially outwards. The Electric Potential c b a q Choose a path a-c-b.  V ab =  V ac +  V cb  V ab = 0 because on this path  V bc = Electric Potential of a Point Charge

19. Place the point charge q at the origin. The electric field points radially outwards. The Electric Potential F=q t E c b a q First find the work done by q’s field when q t is moved from a to b on the path a-c-b. W = W(a to c) + W(c to b) W(a to c) = 0 because on this path hence W W(c to b) = Electric Potential of a Point Charge

20. The Electric Potential F=q t E c b a q  V AB =  U AB / q t And since  V AB = k q [ 1/r b – 1/r a ] Electric Potential of a Point Charge

21. The Electric Potential From this it’s natural to choose the zero of electric potential to be when r a  Letting a be the point at infinity, and dropping the subscript b, we get the electric potential: When the source charge is q, and the electric potential is evaluated at the point r. F=q t E c b a q  V AB = k q [ 1/r b – 1/r a ] V = k q / r Remember: this is the electric potential with respect to infinity Electric Potential of a Point Charge

22. Potential Due to a Group of Charges For isolated point charges just add the potentials created by each charge (superposition) For a continuous distribution of charge …

23. Potential Produced by a Continuous Distribution of Charge In the case of a continuous distribution of charge we first divide the distribution up into small pieces, and then we sum the contribution, to the electric potential, from each piece: dq i

24. Potential Produced by a Continuous Distribution of Charge In the case of a continuous distribution of charge we first divide the distribution up into small pieces, and then we sum the contribution, to the electric potential, from each piece: In the limit of very small pieces, the sum is an integral dq A dV A = k dq / r r V A =  dV A =  k dq / r vol Remember: k=1/(4  0 )

25. Example: a disk of charge Suppose the disk has radius R and a charge per unit area . Find the potential at a point P up the z axis (centered on the disk). Divide the object into small elements of charge and find the potential dV at P due to each bit. For a disk, a bit (differential of area) is a small ring of width dw and radius w. dw P r R w z dq =  2  wdw

26. A charge density per unit length =400 mC/m stretches for 10 cm. Find the electric potential at a point 15 cm from one end. x L d x r = d+L-x dq = dx P Example: a line of charge Break the charge into little bits: say a length dx at position x. The contribution due to this bit at P is:

27. x L d x r = d+L-x dq =ldx P