E n̂ s E n̂ 50 kV Clean air b z r a r b q b a E Dirty air qin 0 + - Electricity Electric Fields Electric Charge •Electric forces affect only objects with charge •Charge is measured in Coulombs (C). A Coulomb is a lot of charge •Charge comes in both positive and negative amounts •Charge is conserved – it can neither be created nor destroyed •Charge is usually denoted by q or Q •There is a fundamental charge, called e Particle •All elementary particles have charges that Proton are simple multiples of e Neutron Electron e 1.602 1019 C Oxygen nuc. Red dashed line means you should be able to ++ use this on a test, but you needn’t memorize it q e 0 -e 8e 2e Charge Can Be Spread Out Charge may be at a point, on a line, on a surface, or throughout a volume •Linear charge density units C/m •Multiply by length •Surface charge density units C/m2 •Multiply by area •Charge density units C/m3 •Multiply by volume Concept Question V lwh 4 cm3 2 cm A box of dimensions 2 cm 2 cm 1 cm has charge density = 5.0 C/cm3 throughout and linear charge density = – 3.0 C/cm along one long diagonal. What is the total charge? A) 2 C B) 5 C C) 11 C D) 29 C E) None of the above 5.0 C/cm3 2 cm 2 2 2 L l 2 w2 h2 2 2 1 cm 3 cm q V L 5 4 3 3 C 11 C The Nature of Matter •Matter consists of positive and negative charges in very large quantities •There are nuclei with positive charges •Surrounded by a “sea” of negatively + + + + charged electrons + + + + •To charge an object, you can add some charge to the object, or remove some charge + + + + •But normally only a very small fraction •10-12 of the total charge, or less + + + + •Electric forces are what hold things together •But complicated by quantum mechanics •Some materials let charges move long distances, others do not •Normally it is electrons that do the moving Insulators only let their charges move a very short distance Conductors allow their charges to move a very long distance Some ways to charge objects •By rubbing them together •Not well understood •By chemical reactions •This is how batteries work •By moving conductors in a magnetic field •Get to this in March •By connecting them to conductors that have charge already •That’s how outlets work •Charging by induction •Bring a charge near an extended conductor •Charges move in response •Separate the conductors – + •Remove the charge – + – – + – + + –– –– + + – + + + + Coulomb’s Law •Like charges repel, and unlike charges attract •The force is proportional to the charges r •It depends on distance q1 ke q1q2 9 2 2 ˆ k 8.988 10 N m / C F2 r e 2 r ke q1q2 F2 r2 q2 Other ways of writing this formula •The r-hat just tells you the direction of the force •When working with components, often helps to rewrite the r-hat •Sometimes this formula is written in terms of a ke q1q2 F2 r 3 quantity0 called the permittivity of free space r F2 q1q2 rˆ 2 4 0 r 0 1 8.854 1012 C2 /N/m 2 4 ke Concept Question 5.0 cm What is the direction of the force +2.0 C 5.0 cm 5.0 cm on the purple charge? –2.0 C A) Up B) Down C) Left D) Right E) None of the above –2.0 C •The separation between the purple charge and each of the L 5 cm 2 5 cm 2 7.1 cm other charges is identical •The magnitude of those forces is 2 9 2 2 6 8.988 10 N m / C 2 10 C identical ke q1q2 F 7.2 N 2 2 r 0.071 m •The blue charge creates a repulsive force at 45 down and left •The green charge creates an attractive force at 45 up and left •The sum of these two vectors points straight left Ftot 7.2 N 2 10.2 N angle 180 Sample Problem Three charges are distributed as shown at right. Where can we place a fourth charge of magnitude 3.0 mC such that the total force on the 1.0 mC vanishes? 1.0 mC 2.0 mC 1.0 m 2.0 m ke q1q2 ke 2 mC 1 mC ˆ ˆ ke mC2 F1 rˆ i 2i 2 2 2 r m 1 m 3.0 mC ? 2 ke 4 mC 1 mC ˆ k mC -4.0 mC ke q1q2 e ˆ ˆ j j F2 r 2 2 2 m r 2 m 2 2 k mC ke mC e ˆ ˆ F 2.236 F3 2i j , 0 F1 F2 F3 , 3 2 2 m m ke q1q2 3ke mC2 F3 2 r2 r tan 1 12 27 3 2.236 2 r 1.16 m r m2 Angle 360 333 Forces From Continuous Charges •If you have a spread out charge, it is tempting to q start by calculating the total charge r •Generally not the way to go dl •The charge of the line is easy to find, Q = L •But the distance and direction is hard to find •To deal with this problem, you have to divide it up into little segments of length dl •Then calculate the charge dQ = dl for each little piece ke dl F q rˆ •Find the separation r for each little piece 2 r •Add them up – integrate •For a 2D object, it becomes a double integral ke dA F q rˆ •For a 3D object, it becomes a triple integral 2 r ke dV F q rˆ 2 r The Electric Field •Suppose we have some distribution of charges •We are about to put a small charge q0 at a point r •What will be the force on the charge at r? •Every term in the force is proportional to q0 •The answer will be proportional to q0 •Call the proportionality constant E, the electric field F0 q0E F0 The units for electric E q0 field are N/C F qE q0 r •It is assumed that the test charge q0 is small enough that the other charges don’t move in response •The electric field E is a function of r, the position •It is a vector field, it has a direction in space everywhere •The electric field is assumed to exist even if there is no test charge q0 present Electric Field From a Point Charge F0 E q0 q r q0 •From a single point charge, the electric field is easy to find •It points away from positive charges •It points towards negative charges + ke qq0 F0 2 rˆ r ke q E 2 rˆ r - Electric Field from Two Charges •Electric field is a vector • We must add the vector components of the contributions of multiple charges ke qi E 2 rˆi ri i + + + - Electric Fields From Continuous Charges P r •If you have a spread out charge, we can add up the dl contribution to the electric field from each part •To deal with this problem, you have to divide it up into little segments of length dl •Then calculate the charge dQ = dl for each little piece •Find the separation r and the direction r-hat for each ke dl little piece E rˆ 2 r •Add them up – integrate •For a 2D object, it becomes a double integral ke dA •For a 3D object, it becomes a triple integral E rˆ 2 r ke dV E rˆ 2 r Sample Problem •Divide the charge into little segments dl – Because it is on the y-axis, dl = dy •The vector r points from the source of the electric field to the point of measurement ry – It’s magnitude is r = y – It’s direction is the minus-y direction rˆ ˆj •Substitute into the integral – Limits of integral are y= a and y = ke dy Qy 2 dy ˆj k Qˆj E 2 2 e 2 2 2 y y a y a a a ke dl E rˆ 2 r What is the electric field at the origin for a line of charge on r the y-axis with linear charge density (y) = Qy2/(y2+a2) stretching from y = a to y = ? y dy (y) y=a P x •Pull constants out of the integral •Look up the integral kQ k Q kQ y E ˆj e tan 1 ˆj e tan 1 tan 1 1 ˆj e jˆ keQ a 2 4 a a aa 4a •Substitute limits Sample Problem •Divide the line charge into little segments •Find the charge dQ = dx for each piece •Find the separation r for each little piece r xˆi aˆj P x r x2 a2 •Add them up – integrate b c c xˆ i aˆj dx ke dx r dx E 2 rˆ ke 3 ke 2 2 3/2 r r b b b x a c •Look up c xdx dx ˆj ke ˆi a integrals 2 2 3/2 2 2 3/2 b x a b x a c ˆ ˆ i x a j ˆi c a ˆj ˆi b a ˆj ke ke 2 2 x2 a2 c2 a2 b a b c r a dx c What is the electric field at the point P for a line with constant linear charge density and the geometry sketched above? Electric Field Lines •Electric field lines are a good way to visualize how electric fields work •They are continuous oriented lines showing the direction of the electric field •They start on positive charges and end on negative charges (or infinity) + - •They never cross •Where they are close together, the field is strong •The bigger the charge, the more field lines come out Sample Problem Sketch the field lines coming from the charges below, if q is positive •Let’s have four lines for each unit of q •Eight lines coming from red, eight going into green, four coming from blue •Most of the “source” lines from red and blue will “sink” into green •Remaining lines must go to infinity +2q -2q +q Acceleration in a Constant Electric Field •If a charged particle is in a constant electric field, it is easy to figure out what happens qE F qE F ma a m •We can then use all standard formulas for constant acceleration A proton accelerates from rest in a constant electric field of 100 N/C. How far must it accelerate to reach escape velocity from the Earth (11.186 km/s)? •Look up the mass and charge of a proton m 1.676 1027 kg •Find the acceleration 19 q e 1.602 10 C 19 qE 1.602 10 C 100 N/C 9 2 9.558 10 m/s a m 1.676 1027 kg 2 2 v v •Use PHY 113 formulas to get the distance f i 2ad 2 4 2 •Solve for the distance 1.1186 10 m/s vf 0.006545 m 0.65 cm d 9 2 2a 2 9.5585 10 m/s