Electric Potential for Point Charges So far, we have been talking about the energy of charges in an electric field. But electric fields can be related back to point charges. In this section we introduce the electric potential for point charges. In the last lecture we saw that E = -  V /  x. In reality, this is a derivative, and because E is actually a vector, this is really a 3 dimensional vector derivative.

We can use calculus (see next page) to integrate to get the expression for the potential for a point charge, which is OSE:V Q = kQ/r = Q/4  0 r. We have taken V=0 at r=  from the point charge Q. Note the 1/r dependence (for F and E the dependence was 1/r 2 ). V is a scalar and F and E are vectors, so working with V should be much easier. As you might expect, the collection due to a sum of point charges is equal to the sum of potentials. I’ll make this an OSE. OSE:V net =  V i. As with all equations involving charges, the sign on Q is important.

noooo…. correctly including vector nature We can use calculus to derive the expression (previous page) for the potential for a point charge…

for a point charge remember, the r in the integral is a “dummy” variable in this derivation, I switch from using Q to q for our point charge—no particular reason

For a continuous distribution of charges, replace the sum by an integral.

Example: What minimum work is required by an external force to bring a charge q=3.00  C from a great distance away (take r =  ) to a point m from a charge Q = 20.0  C? This problem can be solved without a diagram, although you may make one if it helps you. OSE:W i  f = q  V i  f W i  f = q (V f - V i ) W i  f = q (kQ/r f - kQ/r i )

W i  f = kqQ (1/r f - 1/r i ) W i  f = (9x10 9 )(3x10 -6 )(20x10 -6 ) / (0.5) 0, because r i =  W i  f = kqQ /r f W i  f = 1.08 J

Example: Calculate the electric potential at point A in the figure below due to the two charges shown. x y Q 2 =+50  C Q 1 =-50  C 52 cm 60 cm 30 cm  =30º A OSE: V net =  V i. V A = V 1 + V 2.

OSE:V Q = kQ/r = Q/4  0 r V A = V 1 + V 2 V A = kQ 1 /r 1 + kQ 2 /r 2 x y Q 2 =+50  C Q 1 =-50  C 52 cm 60 cm 30 cm  =30º A r 1 = r 2 =

All the numbers are in SI units, making the calculation easy. V A = (9x10 9 )[(-50x10 -6 /0.6) + (+50x10 -6 /0.3)] V A = kQ 1 /r 1 + kQ 2 /r 2 Which would you rather work: this example, or the example from lecture 22, which calculated the electric field, used vectors, and took 6 slides? V A = 7.5x10 5 V.

Conceptual example. All charges in the figure have the same magnitude (i) (ii) (iii) Which set has a positive potential energy? Which set has the most negative potential energy? Which set requires the most work to separate the charges to  ?

Electric Dipoles An electric dipole is two charges +Q and –Q separated by a distance l. The figure shows electric field lines and equipotential lines for an electric dipole.

Electric dipoles appear frequently in physics, chemistry, and biology. Potential due to a dipole. l r r rr  P -Q +Q sorry, klunky figure, but I don’t feel like re-doing it

l r r rr  P -Q +Q If P is far from the charges, so that r>>l, then  r = l cos  and r >>  r. becomes klunky script lowercase letter l, but I don’t feel like fixing it right now

l r r rr  P -Q +Q The product Ql is called the dipole moment of the dipole.

C+C+ O -2 Example: dipole moment of C==O at point P (see text for numbers). P The potential at P is much greater if you remove one of the charges (makes sense; charges almost “cancel”).

Storing Electrical Energy: Capacitance A capacitor is basically two parallel conducting plates with insulating material in between. The capacitor doesn’t have to look like metal plates. Capacitor for use in high-performance audio systems. When a capacitor is connected to an external potential, charges flow onto the plates and create a potential difference between the plates. + - Capacitor plates build up charge. - - The battery in this circuit has some voltage V. We haven’t defined what that means yet.

+ - If the external potential is disconnected, charges remain on the plates, so capacitors are good for storing charge (and energy). Capacitors are also very good at releasing their stored charge all at once. The capacitors in your TV are so good at storing energy that touching the two terminals at the same time can be fatal, even though the TV may not have been used for months. High-voltage capacitors like these are supposed to have “bleeder resistors” that drain the charge away after the circuit is turned off. I wouldn’t bet my life on it. + - Graphic from

assortment of capacitors The charge acquired by each plate of a capacitor is Q=CV where C is the capacitance of the capacitor. OSE: Q = CV. The unit of C is the farad but most capacitors have values of C ranging from picofarads to microfarads (pF to  F). micro  10 -6, nano  10 -9, pico 

The capacitance of an object depends only on the materials it is made of and its geometry. For a parallel plate capacitor with plates of area A separated by a distance d, the capacitance is C=  0 A/d. d area = A The material in between the plates in this case is “air.” C=  0 A/d.*  0 is the permittivity of free space (and approximately of air). *Not an OSE yet—not completely general.

If an insulating sheet (“dielectric”) is placed between the plates, the capacitance increases by a factor K, which depends on the material in the sheet. K is called the dielectric constant of the material. dielectric Thus C = K  0 A / d is true in general (OSE) because K is 1 for a vacuum, and approximately 1 for air. You can also define  = K  0 and write C =  A / d (we won’t). OSE: C = K  0 A / d

Example (a) Calculate the capacitance of a capacitor whose plates are 20 x 3 cm and are separated by a 1.0 mm air gap. d = area = 0.2 x 0.03 OSE: C = K  0 A / d C = 1(8.85x )(0.2x0.03) / C =53x F C = 53 pF If you keep everything in SI (mks) units, the result is automatically in SI units.

(b) What is the charge on each plate of the capacitor is connected to a 12 volt* battery? 0 V +12 V *Remember, it’s the potential difference that matters. OSE: Q = CV Q = (53x )(12) Q = 6.4x C If you keep everything in SI (mks) units, the result is automatically in SI units. V= 12

(c) What is the electric field between the plates? 0 V +12 V If you keep everything in SI (mks) units, the result is automatically in SI units. d = E V= 12

Anybody confused by this symbol “V” I’ve been using? Maybe you should be! V is the symbol for electrical potential, also called potential. The units of V are volts, abbreviated V. V is also the voltage of a battery, or the voltage in an electrical circuit. Actually, the V of a battery is really the potential difference, measured in volts, between the terminals of a battery. Nowhere have I called V an energy. The symbol V is often used for potential energy, but I will not do that in this course. I count 4 different meanings for V. You have to be aware of the context!

Dielectrics The dielectric is the thin insulating sheet in between the plates of a capacitor. dielectric Any reasons to use a dielectric (other than to make your life more complicated)?  Lets you apply higher voltages (so more charge).  Lets you place the plates closer together (make d smaller).  Increases the value of C because K>1. OSE: C = K  0 A / d OSE: Q = CV

Visit howstuffworks to read about capacitors and learn their advantages/disadvantages compared to batteries!howstuffworks Example A capacitor connected as shown acquires a charge Q. V While the capacitor is still connected to the battery, a dielectric material is inserted. Will Q increase, decrease, or stay the same? Why? V V=0

Storage of Electric Energy The electrical energy stored in a capacitor is OSE: U capacitor = QV/2 = CV 2 /2 = Q 2 /2C It is no accident that we use the symbol U for the energy stored. This is another kind of potential energy. Use it in your energy conservation equations just like any other form of energy! The derivation follows, for those who love calculus.

work to move charge dq through potential V (from last lecture) from Q=CV work to put charge Q on capacitor C is constant the other forms follow from definitions

Example: A camera flash unit stores energy in a 150  F capacitor at 200 V. How much electric energy can be stored? U capacitor = CV 2 /2 U capacitor = (150x10 -6 )(200) 2 / 2 U capacitor = 3.0 J

Big concepts from this chapter: ● We defined electric potential. This lets us calculate electric potential energies. A new component to add to your already- existing conservation of energy toolbox. ● Electric field and potential are related. A new component to add to your already-existing electric field toolbox. ● Capacitance. Yet another conservation of energy variation. ● Electron volt, electric dipoles—important, but applications of fundamental concepts.

Official Starting Equations: V a = (PE) a /q W i  f = q  V i  f  PE i  f = q  V i  f E f – E i = (W other ) i  f A big idea (and OSE) from mechanics:

V Q = kQ/r = Q/4  0 r. V net =  V i. Q = CV C = K  0 A / d U capacitor = QV/2 = CV 2 /2 = Q 2 /2C