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## Electric Fields

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**Electric Fields**Electric Flux a R a 2R**y**+ + + + + + + + + R + + + + x + + + + + + + + + • Consider a circular ring with total charge +Q. The charge is spread uniformly around the ring, as shown, so there is λ = Q/2pR charge per unit length. • The electric field at the origin is (b) (a) zero (c) But how would we calculate this??**E(r) = ?**r ++++++++++++++++++++++++++ Electric FieldsfromContinuous Charge Distributions • Examples: • line of charge • charged plates • electron cloud in atoms, … • Principles (Coulomb’s Law + Law of Superposition) remain the same. Only change:**Preflight 3:**B A L 2) A finite line of positive charge is arranged as shown. What is the direction of the Electric field at point A? a) up b) down c) left d) right e) up and left f) up and right 3) What is the direction of the Electric field at point B? a) up b) down c) left d) right e) up and left f) up and right**small piecesof chargedq**total chargeQ • Line of charge: dq = ldx l=charge per unit length • Surface of charge: dq = sdA s=charge per unit area • Volume of Charge: • r=charge per unit volume dq = rdV Charge Densities • How do we represent the charge “Q” on an extended object?**How We Calculate (Uniform) Charge Densities:**• Examples: • 10 coulombsdistributed over a 2-meter rod. Take total charge, divide by “size” 14 pC(pico = 10-12) distributed over a shell of radius 1 μm. 14 pCdistributed over a sphere of radius 1 mm.**E(r) = ?**r ++++++++++++++++++++++++++ q +++++++++++++++++++++++++++++ x Any symmetries ? This may help for easy cancellations Electric field from an infinite line charge Approach: “Add up the electric field contribution from each bit of charge, using superposition of the results to get the final field.” In practice: • Use Coulomb’s Law to find the E-field per segment of charge • Plan to integrate along the line… • x:from-¥to+¥OR q : from -p/2to+p/2**dE**y q r' r ++++++++++++++++ x dx Infinite Line of Charge Charge density = l We need to add up the E-field contributions from all segmentsdx along the line.**dE**y q r' r ++++++++++++++++ x dx Infinite Line of Charge • To find the total field E, we must integrate over all charges along the line. If we integrate overq,we must write r’ and dq in terms of qand dq. • The electric field due to dq is: • Solution: After the appropriate change of variables, we integrate and find: *the calculation is shown in the appendix**dE**y q r' r ++++++++++++++++ x dx Infinite Line of Charge Conclusion: • The Electric Field produced by an infinite line of charge is: - everywhere perpendicular to the line - is proportional to the charge density - decreases as - next lecture: Gauss’ Law makes this trivial!!**Dipole**~ 1/R3 Point Charge ~ 1/R2 Infinite Line of Charge ~ 1/R SummaryElectric Field LinesElectric Field Patterns Coming up: Electric field Flux and Gauss’ Law**The Story Thus Far**Two types of electric charge: opposite charges attract, like charges repel Coulomb’s Law: • Electric Fields • Charges respond to electric fields: • Charges produce electric fields:**The Story Thus Far**• 1. Brute Force: Add up / integrate contribution from each charge. • Often this is pretty difficult. We want to be able to calculate the electric fields from various charge arrangements. Two ways: Ack! Ex: electron cloud around nucleus • 2. Gauss’ Law: The net electric flux through any closed surface is proportional to the charge enclosed by that surface. • In cases of symmetry, this will be MUCH EASIER than the brute force method.**Examine the electric field**• lines produced by the charges • in this figure. • Which statement is true? q1 q2 (a)q1 and q2 have the same sign (b)q1 and q2 have the opposite signs and q1> q2 (c)q1 and q2 have the opposite signs and q1 < q2**Examine the electric field**• lines produced by the charges • in this figure. • Which statement is true? q1 q2 Field lines start fromq2and terminate on q1. This meansq2 is positive; q1 is negative; so, … not (a) (a)q1 and q2 have the same sign (b)q1 and q2 have the opposite signs and q1> q2 (c)q1 and q2 have the opposite signs and q1 < q2 Now, which one is bigger? Notice along a line of symmetry between the two, that the E-field still has a positiveycomponent. If they were equal, it would be zero; This indicates that q2is greater than q1**a)+q(blue)**c a b)-q(red) c)midpoint (yellow) b Electric Dipole: Lines of Force Consider imaginary spheres centered on : • All lines leavea) • All lines enterb) • Equal amounts of leaving and entering lines for c)**“S” is surface**of the box Electric Flux • Flux: Let’s quantify previous discussion about field-line “counting” Define: electric fluxFE through theclosed surface S**Flux**• How much of something is passing through some surface Ex: How many hairs passing through your scalp. • Two ways to define • Number per unit area (e.g., 10 hairs/mm2) • This is NOT what we use here. • Number passing through an area of interest • e.g., 48,788 hairs passing through my scalp. • This is what we are using here.**Electric Flux**• What does this new quantity mean? • The integral is over a CLOSED SURFACE • Since is a SCALAR product, the electric flux is a SCALAR quantity • The integration vector is normal to the surface and points OUT of the surface. is interpreted as the component of E which is NORMAL to the SURFACE • Therefore, the electric flux through a closed surface is the sum of the normal components of the electric field all over the surface. • The sign matters!! • Pay attention to the direction of the normal component as it penetrates the surface… is it “out of” or “into” the surface? • “Out of” is “+” “into” is “-”**w**z y x “S” is surface of the box surface area vector: How to think about flux • We will be interested in net flux in or out of a closed surface like this box • This is the sum of the flux through each side of the box • consider each side separately • Let E-field point in y-direction • then and are parallel and • Look at this from on top • down the z-axis**q**Flux: case 2 E-field surface area ES case 1 Case 2 is smaller! case 2 How to think about flux case 1 • Consider flux through two surfaces that “intercept different numbers of field lines” • first surface is side of box from previous slide • Second surface rotated by an angleq**right**A differential surface element, with its vector The Sign Problem left • For an opensurface we can choose the direction of S-vector two different ways • to the left or to the right • what we call flux would be different these two ways • different by a minus sign • For a closed surface we can choose the direction of S-vector two different ways • pointing “in” or “out” • define “out” to be correct • Integral of EdS over a closed surface gives net flux “out,” but can be + or -**Preflight 3:**Wire loops (1) and (2) have the same length and width, but differences in depth. 1 5) Wire loops (1) and (2) are placed in a uniform electric field as shown. Compare the flux through the two surfaces. E 2 a) Ф1 >Ф2 b) Ф1 =Ф2 c) Ф1 <Ф2**Preflight 3:**6) A cube is placed in a uniform electric field. Find the flux through the bottom surface of the cube. a) Фbottom < 0 b) Фbottom = 0 c) Фbottom > 0**(c)FEµ 6a2**(b)FEµ 2a2 (a)FE= 0 R 2B • Consider 2 spheres (of radius R and 2R)drawn around a single charge as shown. • Which of the following statements about the net electric flux through the 2 surfaces (F2Rand FR) is true? 2R (c)FR> F2R (b)FR= F2R (a)FR< F2R Lecture 3, ACT 2 2A • Imagine a cube of side a positioned in a region of constant electric field as shown • Which of the following statements about the net electric flux FE through the surface of this cube is true? a a**(c)FEµ 6a2**(b)FEµ 2a2 (a)FE= 0 • The electric flux through the surface is defined by: • is ZERO on the four sides that are parallel to the electric field. • on the bottom face is negative. (dS is out; E is in) • on the top face is positive.(dS is out; E is out) • Therefore, the total flux through the cube is: 2A • Imagine a cube of side a positioned in a region of constant electric field as shown • Which of the following statements about the net electric flux FEthrough the surface of this cube is true? a a**R**2R (c)FR> F2R (b)FR= F2R (a)FR< F2R 2B • Consider 2 spheres (of radius R and 2R) drawn around a single charge as shown. • Which of the following statements about the net electric flux through the 2 surfaces (F2RandFR) is true? • Look at the lines going out through each circle -- each circle has the same number of lines. • The electric field is different at the two surfaces, because E is proportional to1 / r2, but the surface areas are also different. The surface area of a sphere is proportional tor2. • Since flux = , ther2and 1/r2terms will cancel, and the two circles have the same flux! • There is an easier way. Gauss’ Law states the net flux is proportional to the NET enclosed charge. The NET charge is the SAME in both cases. • But, what is Gauss’ Law ??? --You’ll find out next lecture!**E(r) =**r ++++++++++++++++++++++++++ Summary • Electric Fields of continuous charge distributions • Electric Flux: • How to think about flux: number of field lines intercepting a surface, perpendicular to that surface • Next Time: Gauss’ Law**dE**y q r' r ++++++++++++++++ x dx Weuse Coulomb’s Law to find dE: Appendix What isdq in terms of dx? Infinite Line of Charge Therefore, What isr’in terms of r ?**dE**y q r' r ++++++++++++++++ x dx Infinite Line of Charge We still have x and qvariables. We are dealing with too many variables. We must write the integral in terms of only one variable (qor x). We will use q. x and q are not independent! x = rtanq dx = r sec2qdq**y**dE q Ey q r' r Ex ++++++++++++++++ x dx Infinite Line of Charge • Components: • Integrate:**y**dE q Ey q r' r Ex ++++++++++++++++ x dx Infinite Line of Charge • Now • The final result: