1. Electric Potential Chapter 25

2. ELECTRIC POTENTIAL DIFFERENCE The fundamental definition of the electric potential V is given in terms of the electric field:  V AB = -  A B E · dl  V AB = Electric potential difference between the points A and B = V B -V A. This is not the way we will usually calculate electric potentials, but we will explore this in a couple of simple examples to understand it better. A B

3. Constant electric field  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 easy in this case because E is constant and the angle between E and dl is constant. L dl

4. A B E C d This line integral is the same for any path connecting the same endpoints. For example, try the two-step path A to 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    V AB = E L cos    L  V AB = -  A B E · dl Constant electric field Notice: the electric field points downhill

5. Equipotential Surfaces (lines) For a constant field E all of the points along the vertical line A are at the same potential. E A

6. Equipotential Surfaces (lines) For a constant field E all of the points along the vertical line A are at the same potential. Pf:  V bc =-∫E·dl=0 because E  dl. We can say line A is at potential V A. E A b c

7. Equipotential Surfaces (lines) For a constant field E all of the points along the vertical line A are at the same potential. Pf:  V bc =-∫E·dl=0 because E  dl. We can say line A is at potential V A. E A x The same is true for any vertical line: all points along it are at the same potential. For example, all points on the dotted line a distance x from A are at the same potential V x, where  V Ax = E x

8. Equipotential Surfaces (lines) For a constant field E all of the points along the vertical line A are at the same potential. Pf:  V bc =-∫E·dl=0 because E  dl. We can say line A is at potential V A. E A x A line (or surface in 3D) of constant potential is known as an Equipotential The same is true for any vertical line: all points along it are at the same potential. For example, all points on the dotted line a distance x from A are at the same potential V x, where  V Ax = E x

9. 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

10. 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)

11. Equipotential Surfaces Equipotential plots are like contour maps of hills and valleys. The electric field is the local slope, and points downhill. Lines of constant E Lines of constant V (perpendicular to E) It is sometimes useful to draw pictures of equipotentials rather than electric field lines:

12. Equipotential Surfaces 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. The electric field is the local slope, and points downhill.

13. 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

14. 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

15. 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. c b a q Electric Potential of a Point Charge

16. The Electric Potential This is the most important thing to know about electric potential: the potential of a point charge q. q Remember: this is the electric potential with respect to infinity – we chose V(∞) to be zero. Electric Potential of a Point Charge Never do this derivation again. Instead, know this simple result by heart: r

17. Potential Due to a Group of Charges The second most important thing to know about electric potential is how to calculate it given more than one charge For isolated point charges just add the potentials created by each charge (superposition) For a continuous distribution of charge …

18. Potential Produced by a Continuous Distribution of Charge dq A dV A = k dq / r r V A =  dV A =  k dq / r Remember: k=1/(4  0 ) In the case of a continuous charge distribution, divide the distribution up into small pieces and then sum (integrate) the contribution from each bit:

19. A charge density per unit length  stretches a length L. Find the electric potential at a point d from one end. Example: a line of charge Break the charge into little bits: say a length dx at position x. x L d x r = d+L-x dq = dx P

20. A charge density per unit length  stretches a length L. Find the electric potential at a point d from one end. 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: x L d x r = d+L-x dq = dx P

21. A charge density per unit length  stretches a length L. Find the electric potential at a point d from one end. 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: x L d x r = d+L-x dq = dx P so

22. A charge density per unit length  stretches a length L. Find the electric potential at a point d from one end. 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: x L d x r = d+L-x dq = dx P so

23. A charge density per unit length  stretches a length L. Find the electric potential at a point d from one end. 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: x L d x r = d+L-x dq = dx P so

24. 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

25. Field and Electric Potential Remember from calculus that integrals are antiderivatives. By the fundamental theorem of calculus you can “undo” the integral: Given

26. Field and Electric Potential Very similarly you can get E(r) from derivatives of V(r). Remember from calculus that integrals are antiderivatives. By the fundamental theorem of calculus you can “undo” the integral: Given

27. Field and Electric Potential Very similarly you can get E(r) from derivatives of V(r). Remember from calculus that integrals are antiderivatives. By the fundamental theorem of calculus you can “undo” the integral: Given Choose V(r 0 )=0. Then

28. Field and Electric Potential Very similarly you can get E(r) from derivatives of V(r).  is the gradient operator Remember from calculus that integrals are antiderivatives. By the fundamental theorem of calculus you can “undo” the integral: The third most important thing to know about potentials. Given Choose V(r 0 )=0. Then 

29. Force and Potential Energy This is entirely analogous to the relationship between a conservative force and its potential energy. can be inverted: In a very similar way the electric potential and field are related by: can be inverted: The reason is that V is simply potential energy per unit charge.

30. Example: a disk of charge Suppose the disk has radius R and a charge per unit area . Find the potential and electric field at a point up the z axis. Divide the object into small elements of charge and find the potential dV at P due to each bit. So here let a bit be a small ring of charge width dw and radius w. dw P r R w z dq =  2  wdw

31. This is easier than integrating over the components of vectors. Here we integrate over a scalar and then take partial derivatives. Example: a disk of charge dw P r R w z By symmetry one sees that E x =E y =0 at P. Find E z from

32. Example: point charge Put a point charge q at the origin. q Find V(r): here this is easy: r

33. Example: point charge Put a point charge q at the origin. q Find V(r): here this is easy: r Then find E(r) from the derivatives:

34. Example: point charge Put a point charge q at the origin. q Find V(r): here this is easy: r Then find E(r) from the derivatives: Derivative:

35. Example: point charge Put a point charge q at the origin. q Find V(r): here this is easy: r Then find E(r) from the derivatives: Derivative: So:

36. Energy of a Charge Distribution How much energy (  work) is required to assemble a charge distribution ?. CASE I: Two Charges Bringing the first charge does not require energy (  work)

37. Energy of a Charge Distribution How much energy (  work) is required to assemble a charge distribution ?. CASE I: Two Charges Bringing the first charge does not require energy (  work) Bringing the second charge requires to perform work against the field of the first charge. r Q1Q1 Q2Q2

38. Energy of a Charge Distribution CASE I: Two Charges Bringing the second charge requires to perform work against the field of the first charge. r Q1Q1 Q2Q2 W = Q 2 V 1 with V 1 = (1/4  0 ) (Q 1 /r)  W = (1/4  0 ) (Q 1 Q 2 /r) = U U = (1/4  0 ) (Q 1 Q 2 /r) U = potential energy of two point charges

39. Energy of a Charge Distribution CASE II: Several Charges U 12 = (1/4  0 ) (Q 1 Q 2 /r) U 12 = potential energy of a pair of point charges a QQ QQ How much energy is stored in this square charge distribution?, or … What is the electrostatic potential energy of the distribution?, or … How much work is needed to assemble this charge distribution? To answer it is necessary to add up the potential energy of each pair of charges  U =  U ij

40. CASE III: Parallel Plate Capacitor Energy of a Charge Distribution -Q +Q fields cancel fields add d E A Electric Field  E =  /  0 = Q /  0 A (  = Q / A) Potential Difference  V = E d = Q d /  0 A

41. CASE III: Parallel Plate Capacitor Energy of a Charge Distribution -Q +Q fields cancel fields add d E A Now, suppose moving an additional very small positive charge dq from the negative to the positive plate. We need to do work. How much work? dW = V dq = (q d /  0 A) dq We can use this expression to calculate the total work needed to charge the plates to Q, -Q

42. CASE III: Parallel Plate Capacitor Energy of a Charge Distribution -Q +Q fields cancel fields add d E A dW = V dq = (q d /  0 A) dq The total work needed to charge the plates to Q, -Q, is given by: W =  dW =  (q d /  0 A) dq = (d /  0 A)  q dq W = (d /  0 A) [Q 2 / 2] = d Q 2 / 2  0 A

43. CASE III: Parallel Plate Capacitor Energy of a Charge Distribution -Q +Q fields cancel fields add d E A W = U = d Q 2 / 2  0 A The work done in charging the plates ends up as stored potential energy of the final charge distribution Where is the energy stored ? The energy is stored in the electric field

44. CASE III: Parallel Plate Capacitor Energy of a Charge Distribution -Q +Q fields cancel fields add d E A U = d Q 2 / 2  0 A = (1/2)  0 E 2 A d The energy U is stored in the field, in the region between the plates. E = Q / (  0 A) The volume of this region is Vol = A d, so we can define the energy density u E as: u E = U / A d = (1/2)  0 E 2

45. Energy of a Charge Distribution. u E = U / A d = (1/2)  0 E 2 Electric Energy Density Although we derived this expression for the uniform field of a parallel plate capacitor, this is a universal expression valid for any electric field. When we have an arbitrary charge distribution, we can use u E to calculate the stored energy U dU = u E d(Vol) = (1/2)  0 E 2 d(Vol)  U = (1/2)  0  E 2 d(Vol) CASE IV: Arbitrary Charge Distribution [The integral covers the entire region in which the field E exists]

46. A Shrinking Sphere A sphere of radius R 1 carries a total charge Q distributed evenly over its surface. How much work does it take to shrink the sphere to a smaller radius R 2 ?.