chapter 24 gauss law
Chapter 24 electric flux 24.1 Electric Flux 24.2 Gauss’s Law 24.3 Application of Gauss’s Law to Various Charge Distributions A Spherically Symmetric Charge Distribution (An insulating solid sphere ) The Electric Field Due to a Point Charge The Electric Field Due to a Thin Spherical Shell The Electric Field Due to A Cylindrically Symmetric Charge Distribution the electric field due to an infinite plane 24.4 Conductors in Electrostatic Equilibrium Norah Ali Al moneef 1 We will use a “charge density” to describe the distribution of charge. This charge density will be different depending on the geometry Symbol Name Charge per length Charge per area Charge per volume Norah Ali Almoneef 2Norah Ali Al moneef 2 Unit C/m C/m 2 C/m 3 Definitions Symmetry—The balanced structure of an object, the halves of which are alike Open and Closed Surfaces A rectangle is an open surface — it does NOT contain a volume A sphere is a closed surface — it DOES contain a volume Closed surface—A surface that divides space into an inside and outside region, so one can’t move from one region to another without crossing the surface Gaussian surface—A hypothetical closed surface that has the same symmetry as the problem we are working on— note this is not a real surface it is just an mathematical one Norah Ali Al moneef 3 24-1 Electric Flux • Like the flow of water, or light energy, we can think of the electric field as flowing through a surface (although in this case nothing is actually moving). • We represent the flux of electric field as F (greek letter phi), The units for electric flux are Nm2/C. Norah Ali Al moneef 4 24.1Electric Flux The number of lines per unit area is proportional to the magnitude of the electric field. This means that the total number of lines that penetrate a given area is proportional to the magnitude of the electric field times the area which is being penetrated. E x A The product of the electric field (E) and surface area (A) which is perpendicular to the field is called the electric flux (ΦE). Consider a uniform electric field, and a surface area through which the electric field is passing. We define the angle of the given area, relative to the direction of the electric field, by a normal vector which is perpendicular to the surface. Norah Ali Al moneef 5 The electric flux passing through a surface is the number of electric field lines that pass through it. Because electric field lines are drawn arbitrarily, we quantify electric flux like this: FE=EA, except that… We define A to be a vector having a magnitude equal to the area of the surface, in a direction normal to the surface. The “amount of surface” perpendicular to the electric field is A cos. Because A is perpendicular to the surface, the amount of A parallel to the electric field is A cos . A = A cos so FE = EA = EA cos . Remember the dot product? FE E A Norah Ali Al moneef 6 A E Norah Ali Al moneef 7 • If the electric field is not uniform, or the surface is not flat… • divide the surface into infinitesimal surface elements and add the flux through each… • so the flux of the electric field through an element of area DA is DF E DA E DA cos • When < 90˚, the flux is positive (out of the surface), and when > 90˚, the flux is negative. • When we have a complicated surface, we can divide it up into tiny elemental areas: dF E dA E dA cos F E lim DA i 0 Norah Ali Al moneef E i F DA i i 8 E E .dA Electric Flux Norah Ali Al moneef 9 • We are going to be most interested in closed surfaces, in which case the outward direction becomes self-evident. • We can ask, what is the electric flux out of such a closed surface? Just integrate over the closed surface: dF E dA Flux positive => out F E Flux negative => in • The symbol has a little circle to indicate that the integral is over a closed surface. • The closed surface is called a gaussian surface, because such surfaces are used by Gauss’ Law, which states that: Norah Ali Al moneef 10 Spherical Gaussian surfaces around (a) positive and (b) negative point charge. Norah Ali Al moneef 11 The flux is a maximum when the surface is perpendicular to the field Φ = EA The flux is zero when the surface is parallel to the field If the field varies over the surface, Φ = EA cos θ is valid for only a small element of the area In general, the value of the flux will depend both on the field pattern and on the surface For a uniform electric field perpendicular to a rectangular surface, ΦE the electric flux is ΦE = EA (Nm^2/C) Norah Ali Al moneef 12 Flux of Electric Field 1. Which of the following figures correctly shows a positive electric flux out of a surface element? A. I. DA B. II. I. C. III. E D. IV. E. I and III. III. Norah Ali Al moneef DA 13 E II. E DA IV. DA E Example 1: flux through a cube of a uniform electric field The field lines pass through two surfaces perpendicularly and are parallel to the other four surfaces For side 1, ΦE = -El 2 For side 2, ΦE = El 2 For the other sides, ΦE = 0 Therefore, Φtotal = 0 Norah Ali Al moneef 14 24-2Gauss’s Law Gauss’s Law relates the electric flux through a closed surface with the charge Qin inside that surface. This is a useful tool for simply determining the electric field, but only for certain situations where the charge distribution is either rather simple or possesses a high degree of symmetry. More general and elegant form of Coulomb’s law. The electric field by the distribution of charges can be obtained using Coulomb’s law by summing (or integrating) over the charge distributions. Gauss’ law, however, gives an additional insight into the nature of electrostatic field and a more general relationship between the charge and the field Norah Ali Al moneef 15 Electric lines of flux and Derivation of Gauss’ Law using Coulombs law Consider a sphere drawn around a positive point charge. Evaluate the net flux through the closed surface. Net Flux = F For a Point charge F EdA E dA E cosdA EdA E=kq/r2 2 kq/r dA F kq/r 2 dA kq/r 2 (4r 2 ) F 4kq 4k 1/0 where 0 8.85x10 12 Fnet qenc 0 Norah Ali Al moneef C2 Nm 2 Gauss’ Law 16 Gauss’ Law The flux of electric field through a closed surface is proportional to the charge enclosed. 16 The Gaussian Surface and Gauss’s Law Closed surfaces of various shapes can surround the charge Only S1 is spherical The flux through all other surfaces (S2 and S3) are the same. These surfaces are all called the Gaussian Surface. Gauss’s Law The net flux through any closed surface surrounding a charge q is given by q/εo and is independent of the shape of that surface q F E E dA 0 The net electric flux through a closed surface that surrounds no charge is zero Since the electric field due to many charges is the vector sum of the electric fields produced by the individual charges, the flux through any closed surface can be expressed as q1 q2 ... F E E dA (E1 E 2 ...) dA • 0 Gauss’s Law connects electric field with its source charge Norah Ali Al moneef 17 in q out Flux lines & Flux Just what we would expect because the number of field lines passing through each sphere is the same N F and number of lines passing through each sphere is the same F1 out Norah Ali Al moneef F S F 2 F1 In fact the number of flux lines passing through any surface surrounding this charge is the same F2 Fs 18 in out F N Q 0 even when a line passes in and out of the surface it crosses out once more than in Flux through a sphere from a point charge The electric field around a point charge 1 Q | E | 2 40 | r1 | E r1 Thus the flux on a sphere is E × Area Cancelling we get Norah Ali Al moneef 19 1 Area Q 2 F 4 | r | 1 2 40 | r1 | F Q 0 Flux through a sphere from a point charge Now we change the radius of sphere The electric field around a point charge r1 | E | 1 Q 4 0 | r1 |2 E Area Thus the 1 Q flux on a F 4 | r1 |2 2 4 | r | 0 1 sphere is E × Area Cancelling we get F Q 0 1 Q | E | 2 40 | r2 | r2 1 Q 2 F2 4 | r | 2 40 | r2 |2 The flux is Q Q the same as F2 F 2 F1 before 0 Norah Ali Al moneef 20 0 Principle of superposition: What is the flux from two charges? Since the flux is related to the number of field lines passing through a surface the total flux is the total from each charge FS Q1 0 Q2 0 In general FS Q1 Q2 Qi Fs Gauss’s Law Norah Ali Al moneef 21 For any 0 surface The total flux through the below spherical surface is 1. 2. 3. 4. positive (net outward flux) negative (net inward flux) zero. I don’t know +q No enclosed charge no net flux. Flux in on left cancelled by flux out on right Norah Ali Al moneef 22 22 Example The net outward flux through the surface of a black box is 8 ×103 Nm2C–1. (a) What is the net charge inside the box? (b) If the net outward flux were zero, could you say that there were no charges inside the box? (a) Ф = 8 ×103 Nm2C–1 If the net charge inside is q, then, F q in q in 0 F 0 8 103 8.85 1012 7 108 C (b) Not necessarily. We can only say that the net charge inside the box is zero. Norah Ali Al moneef 23 Example: A surface is so constructed that, at all points on the surface, the E vector points inward. Therefore, it can be said that A.the surface encloses a net positive charge. B.the surface encloses a net negative charge. C.the surface encloses no net charge. Norah Ali Al moneef 24 Example: The electric field E in Gauss’s Law is A. only that part of the electric field due to the charges inside the surface. B. only that part of the electric field due to the charges outside the surface. C. the total electric field due to all the charges both inside and outside the surface. Norah Ali Al moneef 25 Example: The figure shows a surface enclosing the charges q and –q. The net flux through the surface surrounding the two charges is A. q/0 B. 2q/0 C. –q/0 D. zero E. –2q/0 Norah Ali Al moneef 26 Example: The figure shows a surface enclosing the charges 2q and – q. The net flux through the surface surrounding the two charges is q A. 0 B. 2q ε0 C. q D. zero 3q E. 0 0 Norah Ali Al moneef 27 Example: The figure shows a surface, S, with two charges q and –2q. The net flux through the surface is q A. 0 B. C. D. E. 2q ε0 s q 0 +q zero 3q 0 Norah Ali Al moneef 28 2q Example: These are two-dimensional cross sections through threedimensional closed spheres and a cube. Rank order, from largest to smallest, the electric fluxes Фa to Фe through surfaces a to e. 1. 2. 3. 4. 5. Φa > Φc > Φb > Φd > Φe Φb = Φe > Φa = Φc = Φd Φe > Φd > Φb > Φc > Φa Φb > Φa > Φc > Φe > Φd Φd = Φe > Φc > Φa = Φb Norah Ali Al moneef 29 Sample Problem Four closed surfaces, S1 through S4, together with the charges -2Q, +Q, and –Q are sketched in the figure. Find the electric flux through each surface. S1 -2Q +Q -Q S3 S4 Norah Ali Al moneef 30 S2 Example: a point charge q1 = 4.00 nC is located on the x-axis at x = 2.00 m, and a second point charge q2 = -6.00 nC is on the y-axis at y = 1.00 m. What is the total electric flux due to these two point charges through a spherical surface centered at the origin if the radius is (a) 0.500 m? (b) 1.50 m? (c) 2.50 m? y q2 1m x 2m (a) FE 0 Norah Ali Al moneef q1 6nC (b) F E 0 31 2nC (c ) F E 0 Example: At what angle will the flux be zero and what angle will it be a maximum. 1. π/4 ; π/2 2. π/2 ; 0 3. 0 ; π/2 Norah Ali Al moneef 32 Example An electric field with a magnitude of 3.50 kN/C is applied along the x axis. Calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m long assuming that (a) the plane is parallel to the yz plane; (b) the plane is parallel to the xy plane; (c) the plane contains the y axis, and its normal makes an angle of 40.0° with the x axis. FE EA cos 3.50 103 0.350 0.700 cos0 858 N m 2 C 90.0 FE 0 FE 3.50 103 0.350 0.700 cos40.0 657 N m Norah Ali Al moneef 33 2 C example What is the net electric flux through the surface if Q1=q4=+3.1nC, q2=q5=-5.9nC, and q3=-3.1nC? F Norah Ali Al moneef qenc 0 q1 q2 q3 0 34 670 N m / C 2 Problem What is the flux Φ of The electric field through This closed surface? Step one: F EA E dA E dA E dA a Step two: b c 0 E d A E cos 180 dA E dA EA a E dA c E dA E cos 0 dA EA E cos 90 0 dA 0 b Step three: Norah Ali Al moneef F EA 0 EA 0 35 Remember – electric field lines must start and must end on charges! If no charge is enclosed within Gaussian surface – flux is zero! Electric flux is proportional to the algebraic number of lines leaving the surface, outgoing lines have positive sign, incoming - negative Norah Ali Al moneef 36 Example The following charges are located inside a submarine: 5.00 μC, –9.00 μC, 27.0 μC, and –84.0 μC. (a) Calculate the net electric flux through the hull of the submarine. (b) Is the number of electric field lines leaving the submarine greater than, equal to, or less than the number entering it? qin 5.00 C 9.00 C 27.0 C 84.0 C 6 2 2 FE 6. 89 10 N m C 0 8.85 1012 C 2 N m 2 Norah Ali Al moneef 37 Example: Consider a cube with each edge = 55cm. There is a 1.8 C charge In the center of the cube. Calculate the total flux exiting the cube. 6 1.8 10 5 2 F 2 . 03 10 Nm /C 12 0 8.85 10 q NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE! Norah Ali Al moneef 38 Example A point charge Q = 5.00 μC is located at the center of a cube of edge L = 0.100 m. In addition, six other identical point charges having q = –1.00 μC are positioned symmetrically around Q as shown in Figure P24.17. Determine the electric flux through one face of the cube. Norah Ali Al moneef 39 Problem solving guide for Gauss’ law Use the symmetry of the charge distribution to determine the pattern of the field lines. Choose a Gaussian surface for which E is either parallel to or perpendicular to dA. If E is parallel to dA, then the magnitude of E should be constant over this part of the surface. The integral then reduces to a sum over area elements. Norah Ali Al moneef 40 24-3 Application of gauss s law to various charge distributions Gauss’s law is useful in determining electric fields when the charge distribution is characterized by a high degree of symmetry. The following examples demonstrate ways of choosing the Gaussianqsurface over which the surface integral given by Equation E d A can be simplified and the electric field determined. In choosing the surface, we should always take advantage of the symmetry of the charge distribution so that we can remove E from the integral and solve for it. The goal in this type of calculation is to determine a surface that satisfies one or more of the following conditions: 1. The value of the electric field can be argued by symmetry to be constant over the surface. 2. The dot product in the Equation can be expressed as a simple algebraic product E dA because E and dA are parallel. 3. The dot product in the Equation is zero because E and d A are perpendicular. 4. The field can be argued to be zero over the surface. All four of these conditions are used in examples throughout the remainder of this chapter. in E 0 Norah Ali Al moneef 41 Gauss’ Law and Coulomb law Example 24.4 The Electric Field Due to a Point Charge Starting with Gauss’s law, calculate the electric field due to an isolated point charge q. The field lines are directed radially outwards and are perpendicular to the surface at every point, so The angle between E and dA is zero at any point on the surface, we can re-write Gauss’ Law as F E E dA E n dA EdA E dA E 4r 2 FE E .dA Qin 0 E has the has same value at all points on the surface E is can be moved out A spherical Gaussian surface centered on a point charge q Norah Ali Al moneef Integral is the sum of surface area Coulomb’s Law 42 Example 24.5 A Spherically Symmetric Charge Distribution An insulating solid sphere of radius a has a uniform volume charge density ρ and carries total charge Q. (A) Find the magnitude of the E-field at a point outside the sphere E (B) Find the magnitude of the E-field at a point inside the sphere Since the charge distribution is spherically symmetric we select a spherical Gaussian surface of radius r > a a centered on the charged sphere. Since the charged sphere has a positive charge, the field will be directed radially outward. On the Gaussian sphere E is always parallel to dA, and is constant. Qin E . d A r dA Q Right side: Qin 0 Q 0 2 E d A E d A E ( 4 ) r E 4 r 2 Norah Ali Al moneef 0 43 Q 0 or E 1 Q Q k e 4 0 r 2 r2 Example 3 Find the electric field at a point inside the sphere. Now we select a spherical Gaussian surface with radius r < a. Again the symmetry of the charge distribution allows us to simply evaluate the left side of Gauss’s law just as before. r a The charge inside the Gaussian sphere is no longer Q. If we call the Gaussian sphere volume V’ then“V Q 2 E d A E d A E ( 4 r ) Volume charge density“: ρ = charge / unit volume is used to characterize the charge distribution 4 3 Qin 4 r 3 2 Right side: Qin V r E 4 r 0 3 0 3 4 r 3 Q 1 Q Q E r but so E r ke 3 r 3 2 4 3 0 4 0 a a 3 0 4 r a3 3 Norah Ali Al moneef 44 The electric field of a uniformly charged INSULATING sphere. E (r ) = outside inside Norah Ali Al moneef Q 4o a 3 rrˆ Q We found for r > a , E ke 2 r ke Q and for r < a , E 3 r a 45 same as a point charge! Example 24.6 The Electric Field Due to a Thin Spherical Shell A thin spherical shell of radius a has a total charge Q distributed uniformly over its surface Find the electric field at points (A) outside and (B) inside the shell. Norah Ali Al moneef 46 Let’s start with the Gaussian surface outside the sphere of charge, r > a We know from symmetry arguments that the electric field will be radial outside the charged sphere If we rotate the sphere, the electric field cannot change Spherical symmetry Thus we can apply Gauss’ Law and get Flux E dA E 4π r 2 q / 0 (Gauss) … so the electric field is E Norah Ali Al moneef 1 4 0 47 q r2 Let’s let’s take the Gaussian surface inside the sphere of charge, r < a We know that the enclosed charge is zero so Flux F E EA 0 We find that the electric field is zero everywhere inside spherical shell of charge E 0 Thus we obtain two results The electric field outside a spherical shell of charge is the same as that of a point charge. The electric field inside a spherical shell of charge is zero. Norah Ali Al moneef 48 Let’s let’s take the Gaussian surface inside the sphere of charge, r < a We know that the enclosed charge is zero so Flux F E EA 0 We find that the electric field is zero everywhere inside spherical shell of charge E 0 Thus we obtain two results The electric field outside a spherical shell of charge is the same as that of a point charge. The electric field inside a spherical shell of charge is zero. Norah Ali Al moneef 49 24-6 The Electric Field Due to A Cylindrically Symmetric Charge Distribution Find the E-field a distance r from a line of positive charge of infinite length and constant charge per unit length λ. Symmetry E field must be ^ to line and can only depend on distance from line Construct a “Gaussian Surface” that reflects the symmetry of the charge - cylindrical in this case, then evaluate Norah Ali Al moneef 50 q L E .dA Qin Gauss s law L E d A E d A 0 E ( 2 r L ) 0 0 E 2 0 r 2k r E 20r 2k E r̂ r Norah Ali Al moneef 51 Line of Charge Example 24.8 A Plane of Charge •Find the electric field due to an infinite plane of positive charge with uniform surface charge density σ. Assume that we have a thin, infinite non-conducting sheet of positive charge The charge density in this case is the charge per unit area, From symmetry, we can see that the electric field will be perpendicular to the surface of the sheet Norah Ali Al moneef 52 Planar Symmetry (2) To calculate the electric field using Gauss’ Law, we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height 2r, chosen to cut through the plane perpendicularly. Because the electric field is perpendicular to the plane everywhere, the electric field will be parallel to the walls of the cylinder and perpendicular to the ends of the cylinder. Using Gauss’ Law we get Flux F E E dA EA EA q / 0 A / 0 (Gauss) … so the electric field from an infinite non-conducting sheet with charge density Norah Ali Al moneef 53 Example A conducting spherical shell of inner radius a and outer radius b with a net charge -Q is centered on point charge +2Q. Use Gauss’s law to find the electric field everywhere, and to determine the charge distribution on the spherical shell. First find the field for 0 < r < a This is the same as Ex. 2 and is the field due to a point charge with charge +2Q. -Q a +2Q b 2Q E ke 2 r Now find the field for a < r < b The field must be zero inside a conductor in equilibrium. Thus from Gauss’s law Qin is zero. There is a + 2Q from the point charge so we must have Qa = -2Q on the inner surface of the spherical shell. Since the net charge on the shell is -Q we can get the charge on the outer surface from Qnet = Qa + Qb. Qb= Qnet - Qa = -Q - (-2Q) = + Q. Norah Ali Al moneef 54 Example Find the field for r > b From the symmetry of the problem, the field in this region is radial and everywhere perpendicular to the spherical Gaussian surface. Furthermore, the field has the same value at every point on the Gaussian surface so the solution then proceeds exactly as in Ex. 2, but Qin=2Q-Q. 2 E d A E d A E ( 4 r ) Gauss’s law now gives: E 4 r Norah Ali Al moneef 2 Qin 0 2Q Q 0 Q 0 55 1 Q Q or E ke 2 2 4 0 r r Example A spherical shell of radius 10 cm has a charge 2×10–6 C distributed uniformly over its surface. Find the electric field (a) Inside the shell (b) Just outside the shell (c) At a point 15 cm away from the centre q = 2 ×10–6 C, R = 0.1 m, r = 0.15 m (a) Inside the shell, electric field is zero 1 q 2×10–6 9 (b) E = = 9×10 × 2 4ε0 R 0.12 = 1.8×106 NC–1 1 q 2×10–6 9 (c) E' = = 9×10 × 4ε0 r2 0.152 =8×105 NC–1 Norah Ali Al moneef 56 Example An infinite line charge produces a field of 9×104 N C–1 at a distance of 2 cm. Calculate the linear charge density. E = 9×104 N C–1, E= r = 2×10–2 m 1 λ 2 ε 0 r λ = 4 ε 0 × Er 2 1 9×104 ×2×10–2 –7 –1 = × =10 Cm 9×109 2 Norah Ali Al moneef 57 Conductor + + +- + ++++++ Norah Ali Al moneef - - + + + + Conductors.. In these materials, the charges ARE FREE TO MOVE. 58 24-4Conductors in Electrostatic Equilibrium By electrostatic equilibrium we mean a situation where there is no net motion of charge within the conductor The electric field is zero everywhere inside the conductor Any net charge resides on the conductor’s surface The electric field just outside a charged conductor is perpendicular to the conductor’s surface •On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature of the surface is smallest. Norah Ali Al moneef 59 24-4Conductors in Electrostatic Equilibrium The electric field is zero everywhere inside the conductor Why is this so? Place a conducting slab in an external field, E. If there was a field in the conductor the charges would accelerate under the action of the field. ++++++++++++ --------------------- The charges in the conductor move creating an internal electric field that cancels the applied field on the inside of the conductor Ein E Norah Ali Al moneef E 60 When electric charges are at rest, the electric field within a conductor is zero. On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature of the surface is smallest. The electric field is stronger where the surface is more sharply curved. Norah Ali Al moneef 61 •Charge Resides on the Surface Choose a Gaussian surface inside but close to the actual surface The electric field inside is zero There is no net flux through the gaussian surface Because the gaussian surface can be as close to the actual surface as desired, there can be no charge inside the surface Since no net charge can be inside the surface, any net charge must reside on the surface Gauss’s law does not indicate the distribution of these charges, only that it must be on the surface of the conductor Norah Ali Al moneef 62 Charges on Conductors Field within conductor E=0 Charge on solid conductor resides on surface. Charge in cavity makes a equal but opposite charge reside on inner surface of conductor. Norah Ali Al moneef 63 E-Field’s Magnitude and Direction Choose a cylinder as the Gaussian surface The field must be perpendicular to the surface If there were a parallel component to E, charges would experience a force and accelerate along the surface and it would not be in equilibrium The net flux through the surface is through only the flat face outside the conductor The field here is perpendicular to the surface Applying Gauss’s law A F E EA 0 Norah Ali Al moneef FE 0 64 Under electrostatic conditions the electric field inside a solid conducting sphere is zero. Outside the sphere the electric field drops off as 1 / r2, as though all the excess charge on the sphere were concentrated at its center. Electric field = zero (electrostatic) inside a solid conducting sphere Norah Ali Al moneef 65 Example • A spherical conducting shell has an excess charge of +10 C. • A point charge of 15 C is located at center of the sphere. • Use Gauss’ Law to calculate the charge on inner and outer surface of sphere R1 R2 -15 C • Since E = 0 inside the metal, flux through this surface = 0 • Gauss’ Law says total charge enclosed = 0 -5 C • Charge on inner surface = +15 C Since TOTAL charge on shell is +10 C, Charge on outer surface = +10 C 15 C = 5 C! -15C a) Inner: +15 C; outer: 0 (b) Inner: 0; outer: +10 C (c) Inner: +15 C; outer: -5 C Norah Ali Al moneef +15C 66 A line charge (C/m) is placed along the axis of an uncharged conducting cylinder of inner radius ri = a, and outer radius ro = b as shown. What is the value of the charge density o (C/m2) on the outer surface of the cylinder? (a) Norah Ali Al moneef b a (c) (b) 67 Example The uniform surface charge density on a spherical copper shell is . What is the electric field strength on the surface of the shell? The electric field on the surface of a uniformly charged spherical conductor is given by, E= 1 q 2 4ε0 R o Norah Ali Al moneef 68 Example A spherical charged conductor has a uniform surface charge density s . The electric field on its surface is E. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ? The electric field on the surface of a uniformly charged spherical conductor is given by, 1 q 4R2 E= 4ε0 R2 40R2 o Norah Ali Al moneef 69 Example Consider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 μC distributed uniformly on its surface. Find the electric field (a) 10.0 cm and (b) 20.0 cm from the center of the charge distribution. E Norah Ali Al moneef keQ 2 r 8.99 10 32.0 10 7.19 M N 6 9 0.200 70 2 C Example A uniformly charged, straight filament 7.00 m in length has a total positive charge of 2.00 μC. An uncharged cardboard cylinder 2.00 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find (a) the electric field at the surface of the cylinder and (b) the total electric flux through the cylinder. 9 2ke 2 8.99 10 N m E r 2 C 2 2.00 106 C 7.00 m 0.100 m E 51.4 kN C ,radially outw ard FE EA cos E 2 r cos0 FE 5.14 104 N C 2 0.100 m 0.020 0 m 1.00 646 N m 2 C Norah Ali Al moneef 71 Example A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find (a) the charge density of each face of the plate and (b) the total charge on each face. 8.00 104 8.85 1012 7.08 107 C m 2 Q A 7.08 107 0.500 C 2 Q 1.77 107 C 177 nC Norah Ali Al moneef 72 Two infinite parallel sheets of charge surface E ^ of the sheet Due to uniform surface charge density, Region II E = E1 – E2 Region I E = – E1 + E2 =– = (σ A + σ B ) 2ε0 A Norah Ali Al moneef (σ A – σB ) 2ε0 73 B Two infinite parallel sheets of charge A B Region III E = E1 + E2 = Norah Ali Al moneef 74 (σ A + σ B ) 2ε0 Two Parallel Nonconducting Sheets The situation is different if you bring two nonconducting sheets of charge close to each other. In this case, the charges cannot move, so there is no shielding, but now we can use the principle of superposition. In this case, the electric field on the left due to the positively charged sheet is canceled by the electric field on the left of the negatively charged sheet, so the field there is zero. Likewise, the electric field on the right due to the negatively charged sheet is canceled by the electric field on the right of the positively charged sheet. Norah Ali Al moneef 75 The result is much the same as before, with the electric field in between being twice what it was previously. Special Case The sheets have equal and opposite charge density If A and B – I In regions I and III, E = 0 In region II, Norah Ali Al moneef E E0 then E II as 0 A B ( –) 0 – (– ) 20 0 76 III – E0 Example Two large thin metal plates with surface charge densities of opposite signs but equal magnitude of 44.27 ×10–20 Cm–2 are placed parallel and close to each other. What is the field (i) To the left of the plates? (ii) To the right of the plates? (iii) Between the plates? Electric field exists only in the region between the plates. Therefore, (i), (ii) E = 0 (iii) σ = 44.27×10–20 Cm–2 , σ 44.27×10–20 E= = =5×108 NC–1 –12 ε0 8.854×10 Norah Ali Al moneef 77 Example: A uniform electric field of magnitude 1 N/C is pointing in the positive y direction. If the cube has sides of 1 meter, what is the flux through sides A, B, C? A) 1, 0, 1 B) 0, 0, 1 B 1, 0, 0 D) 0, 0, 0 E) 1, 1, 1 C) Norah Ali Al moneef A 78 C Example: Consider three Gaussian surfaces. Surface 1 encloses a charge of +q, surface 2 encloses a charge of –q and surface 3 encloses both charges. Rank the 3 surfaces according to the flux, greatest first. A) 1, 2, 3 B) 1, 3, 2 +q C) 2, 1, 3 -q 1 2 D) 2, 3, 1 E) 3, 2, 1 Norah Ali Al moneef 3 79 Example: Rank the following Gaussian surfaces by the amount of flux that passes through them, greatest first (q is at the center of each). A) 1, 2, 3 B) 1, 3, 2 C) 2, 1, 3 D) 3, 2, 1 E) All tie q q q 1 2 3 Norah Ali Al moneef 80 Rank the following Gaussian surfaces by the strength of the field at the surface at the point direction below q (where the numbers are). A) 1, 2, 3 B) 1, 3, 2 C) 2, 1, 3 D) 3, 2, 1 q q 1 2 E) All tie Norah Ali Al moneef q 3 81 Problem What is the electric flux through the right the face, the left face,and the top face? E 3.0iˆ 4.0 ˆj Right face: dA dAiˆ F r E dA 3.0 iˆ 4.0 ˆj 3.0 xdA 3.0 3.0dA 9.0 dA 36 N m 2 / C Left face: Top face: Norah Ali Al moneef F l 12 N m 2 / C Ft 3.0iˆ 4.0 ˆj dAˆj 16 N m 2 / C 82 Example: A uniform electric field of magnitude 6000 N/C points upward in a charge-free region. What is the electric flux through the shaded side of the square pyramid shown if its base is 80 m on a side and it is 50 m high? . All flux lines entering the base must leave the top, so the flux through the four sides is the flux through the base: ФE = EA= (6000 N/C)(80 m)2= 38.4 MNm2/C. By symmetry, the flux through each side is the same: ФE= 9.6 MNm2/C Norah Ali Al moneef 83 Example: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.80 x104 N/C as shown in Figure. Calculate the electric flux through (a) the vertical rectangular surface, (b) (b) the slanted surface, (c) (c) the entire surface of the box Norah Ali Al moneef 84 Norah Ali Al moneef 85 Example: A uniform electric field aiˆ + bjˆ intersects a surface of area A. What is the flux through this area if the surface lies (a) in the yz plane? (b) in the xz plane? (c) in the xy plane? Norah Ali Al moneef 86 Example: Consider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 μC distributed uniformly on its surface. Find the electric field (a) 10.0 cm (b) 20.0 cm from the center of the charge distribution Norah Ali Al moneef 87 Example - Charge in a Cube Q=3.76 nC is at the Q center of a cube. What is the electric flux through one of the sides? Gauss’ Law: F Q / 0 Since a cube has 6 identical sides and the point charge is at the center F one face F 1Q Nm 2 (substitute the numerical values) 70.8 6 6 0 C Norah Ali Al moneef 88 Example: Shown is an arrangement of five charged pieces of plastic (q1=q4=3nC, q2=q5=-5.9nC and q3=-3.1nC). A Gaussian surface S is indicated. What is the net electric flux through the surface? A: F=-6 x 10-9C/0= -678 Nm2/C B: F= x 10-9C/0= -1356 Nm2/C C: F=0 D: F= .9 x 10-9C/0= 328 Nm2/C enclosed charge Norah Ali Al moneef 89 Example Two large thin metal plates with surface charge densities of opposite signs but equal magnitude of 44.27 ×10–20 Cm–2 are placed parallel and close to each other. What is the field (i) To the left of the plates? (ii) To the right of the plates? (iii) Between the plates? Electric field exists only in the region between the plates. (i), (ii) E = 0 (iii) σ = 44.27×10–20 Cm–2 , σ 44.27×10–20 8 –1 E= = =5×10 NC ε0 8.854×10–12 Norah Ali Al moneef 90 Example A spherical shell of radius 10 cm has a charge 2×10–6 C distributed uniformly over its surface. Find the electric field (a) Inside the shell (b) Just outside the shell (c) At a point 15 cm away from the centre q = 2 ×10–6 C, R = 0.1 m, r = 0.15 m (a) Inside the shell, electric field is zero 1 q 2×10–6 9 (b) E = = 9×10 × 2 4ε 0 R 0.12 = 1.8×106 NC–1 1 q 2×10–6 9 (c) E' = = 9×10 × 4ε0 r2 0.152 =8×105 NC–1 Norah Ali Al moneef 91 Example Ex) What is the electric flux through this disk of radius r = 0.10 m if the uniform electric field has a magnitude E = 2.0x103 N/C? F E EAcos (2.0 10 ) ( r ) cos30 3 2 2 0 (2.0 10 ) (3.14 10 ) 0.866 54 N m / C 3 Norah Ali Al moneef 2 92 Example A sphere of radius 8.0 cm carries a uniform volume charge density = 500 nC/m3. What is the electric field magnitude at r = 8.1 cm? A. 0.12 kN/C B. 1.5 kN/C C. 0.74 kN/C D.2.3 kN/C E. 12 kN/C Norah Ali Al moneef 93 Example A spherical shell of radius 9.0 cm carries a uniform surface charge density = 9.0 nC/m2. The electric field magnitude at r = 4.0 cm is approximately A. 0.13 kN/C B. 1.0 kN/C C. 0.32 kN/C D.0.75 kN/C E. zero Norah Ali Al moneef 94 Example A spherical shell of radius 9.0 cm carries a uniform surface charge density = 9.0 nC/m2. The electric field magnitude at r = 9.1 cm is approximately A. zero B. 1.0 kN/C C. 0.65 kN/C D.0.32 kN/C E. 0.13 kN/C Norah Ali Al moneef 95 Example An infinite plane of surface charge density = +8.00 nC/m2 lies in the yz plane at the origin, and a second infinite plane of surface charge density = –8.00 nC/m2 lies in a plane parallel to the yz plane at x = 4.00 m. The electric field magnitude at x = 3.50 m is approximately A. 226 N/C B. 339 N/C C. 904 N/C D.452 N/C E. zero Norah Ali Al moneef 96 Example An infinite plane of surface charge density = +8.00 nC/m2 lies in the yz plane at the origin, and a second infinite plane of surface charge density = –8.00 nC/m2 lies in a plane parallel to the yz plane at x =4.00 m. The electric field magnitude at x = 5.00 m is approximately A. 226 N/C B. 339 N/C C. 904 N/C D.452 N/C E. zero Norah Ali Al moneef 97 Example A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for ra < r < rb1 is A. B. C. D. E. kQ rˆ 2 r kQ rˆ r2 2kQ rˆ r 2kQ rˆ r zero Norah Ali Al moneef ra Q rb1 rb2 98 Example A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for rb1 < r < rb2 is kQ A. B. C. D. E. Norah Ali Al moneef 2 rˆ r kQ rˆ 2 r 2kQ rˆ r 2kQ rˆ r zero ra Q rb1 rb2 99 Example A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for r > rb1 is A. B. C. D. E. Norah Ali Al moneef kQ 2 rˆ r kQ rˆ 2 r 2kQ rˆ r 2kQ rˆ r zero ra Q rb1 rb2 100 Example Consider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 μC distributed uniformly on its surface. Find the electric field (a) 10.0 cm and (b) 20.0 cm from the center of the charge distribution. E keQ 2 r 8.99 10 32.0 10 7.19 M N 6 9 0.200 2 C Summary: Gauss’ Law · Gauss’ Law depends on the enclosed charge only qenc F E dA o 1. 2. 3. If there is a positive net flux there is a net positive charge enclosed If there is a negative net flux there is a net negative charge enclosed If there is a zero net flux there is no net charge enclosed Gauss’ Law works in cases of symmetry Norah Ali Al moneef 102 GAUSS LAW – SPECIAL SYMMETRIES SPHERICAL CYLINDRICAL PLANAR (point or sphere) (line or cylinder) (plane or sheet) Depends only on radial distance from central point Depends only on perpendicular distance from line Depends only on perpendicular distance from plane GAUSSIAN SURFACE Sphere centered at point of symmetry Cylinder centered at axis of symmetry Pillbox or cylinder with axis perpendicular to plane ELECTRIC FIELD E E constant at surface E ║A - cos = 1 E constant at curved surface and E║A E ┴ A at end surface cos = 0 E constant at end surfaces and E ║ A E ┴ A at curved surface cos = 0 CHARGE DENSITY Norah Ali Al moneef 103 Geometry Charge Density Gaussian surface Linear = q/L Cylindrical, with axis along line of charge Electric field E Sheet or Plane = q/A Cylindrical, with axis along E. E Spherical = q/V Spherical, with center on center of sphere E Norah Ali Al moneef 104 0 20 r Conducting Line of Charge E Nonconducting 2 0 q q r E r R 3 2 4 R 40 r 0 r<R Norah Ali Al moneef 105
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