Chapter 24. Gauss s Law

Transcription

1 Chapter 24 Gauss s Law

2 Electric Flux Electric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the field Φ E = EA

3 Defining Electric Flux EFM06AN1

4 Electric Flux, General Area The electric flux is proportional to the number of electric field lines penetrating some surface The field lines may make some angle θ with the perpendicular to the surface Then Φ E = EA cos θ

5 Electric Flux, Interpreting the Equation The flux is a maximum when the surface is perpendicular to the field The flux is zero when the surface is parallel to the field If the field varies over the surface, then Φ = EA cos θ is valid for only a small element of the area

6 Electric Flux, General In the more general case, look at a small area element E E i A i cos i E i A i In general, this becomes lim E A d E i i A 0 i E A surface

7 Electric Flux, final The surface integral means the integral must be evaluated over the surface in question In general, the value of the flux will depend both on the field pattern and on the surface The units of electric flux will be N. m 2 /C i.e. [N/C] x [m 2 ]

8 Electric Flux, Closed Surface Assume a closed surface The vectors ΔA i point in different directions At each point, they are perpendicular to the surface By convention, they point outward

9 Flux Through Closed Surface The net flux through the surface is proportional to the net number of lines leaving the surface This is the number of lines leaving the surface minus the number entering the surface If E n is the component of E perpendicular to the surface, then E da E da E n

10 Quick Quiz 24.1 Suppose the radius of a sphere, with a charge at its centre, is halved. What happens to the flux through the sphere and the magnitude of the electric field at its surface? (a) The flux and field both increase. (b) The flux and field both decrease. (c) The flux increases and the field decreases. (d) The flux decreases and the field increases. (e) The flux remains the same and the field increases. (f) The flux decreases and the field remains the same.

11 Quick Quiz 24.1 Answer: (e). The same number of field lines pass through a sphere of any size. Because points on the surface of the sphere are closer to the charge, the field is stronger.

12 Quick Quiz 24.2 In a charge-free region of space, a closed container is placed in an electric field. A requirement for the total electric flux through the surface of the container to be zero is that (a) the field must be uniform (b) the container must be symmetric (c) the container must be oriented in a certain way (d) The requirement does not exist the total electric flux is zero no matter what.

13 Quick Quiz 24.2 Answer: (d). All field lines that enter the container also leave the container so that the total flux is zero, regardless of the nature of the field or the container.

14 Gauss s Law, Introduction Gauss s law is an expression of the general relationship between the net electric flux through a closed surface and the charge enclosed by the surface The closed surface is often called a Gaussian surface Gauss s law is of fundamental importance in the study of electric fields

15 Gauss s Law General A positive point charge, q, is located at the centre of a sphere of radius r The magnitude of the electric field everywhere on the surface of the sphere is E = k e q / r 2

16 Gauss s Law General, cont. The field lines are directed radially outward and are perpendicular to the surface at every point E E da E da This will be the net flux through the gaussian surface, the sphere of radius r We know E = k e q/r 2 and A sphere = 4πr 2, E E A sphere 4 k e q q 0

17 Gauss s Law General, notes The net flux through any closed surface surrounding a point charge, q, is given by q/ε o and is independent of the shape of that surface. 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 E E da E 1 E 2 da

18 Gauss s Law Final Gauss s law states E E da q in 0 q in is the net charge inside the surface E represents the electric field at any point on the surface E is the total electric field and may have contributions from charges both inside and outside of the surface Although Gauss s law can, in theory, be solved to find E for any charge configuration, in practice it is limited to symmetric situations

19 Finding an electric field using Gauss s law EFA06AN1

20 Quick Quiz 24.3 If the net flux through a gaussian surface is zero, the following four statements could be true. Which of the statements must be true? (a) There are no charges inside the surface. (b) The net charge inside the surface is zero. (c) The electric field is zero everywhere on the surface. (d) The number of electric field lines entering the surface equals the number leaving the surface.

21 Quick Quiz 24.3 Answer: (b) and (d). Statement (a) is not necessarily true because an equal number of positive and negative charges could be present inside the surface. Statement (c) is not necessarily true - the number of field lines entering the surface will be equal to the number leaving the surface, but this does not mean the field is zero on the surface.

22 Quick Quiz 24.4 Consider the charge distribution shown in the figure. The charges contributing to the total electric flux through surface S are (a) q 1 only (b) q 4 only (c) q 2 and q 3 (d) all four charges (e) none of the charges

23 Quick Quiz 24.4 Answer: (c). The charges q 1 and q 4 are outside the surface and contribute zero net flux through S.

24 Quick Quiz 24.5 Again consider the charge distribution shown in this figure. The charges contributing to the total electric field at a chosen point on the surface S are (a) q 1 only (b) q 4 only (c) q 2 and q 3 (d) all four charges (e) none of the charges

25 Quick Quiz 24.5 Answer: (d). We don't need the surfaces to realize that any given point in space will experience an electric field due to all local source charges.

26 Conditions for a Gaussian Surface Choose a surface over which the integral can be simplified and the electric field determined. Try to choose a surface that satisfies one or more of these conditions: The value of the electric field can be argued from symmetry to be constant over the surface The dot product of E. da can be expressed as a simple algebraic product EdA because E and da are parallel The dot product is 0 because E and da are perpendicular The field can be argued to be zero over the surface

27 Field Due to a Point Charge Choose a sphere as the Gaussian surface E is parallel to da at each point on the surface E E da E da q 0 E E da E (4 r 2 ) E q 4 0 r 2 k e q r 2

28 Field Due to a Spherically Symmetric Charge Distribution Select a sphere as the Gaussian surface For r >a E E da E da Q 0 E 4 r 2 Q 0 so E Q 4 0 r 2 k e Q r 2 Q is the total charge inside the surface

29 Spherically Symmetric, cont. Select a sphere as the Gaussian surface, r < a q enclosed = (4/3πr 3 ) Charge density = Q / [4/3πa 3 ] E E da E da q enc 0 E q enc 4 0 r 2 k e q enc r 2 k e 4 3 r 3 r 2 4 k e r 3 k e Qr a 3

30 Spherically Symmetric Distribution, final Inside the sphere, E varies linearly with r E 0 as r 0 The field outside the sphere is equivalent to that of a point charge located at the centre of the sphere

31 Field Due to a Thin Spherical Shell Use spheres as the Gaussian surfaces When r > a, the charge inside the surface is Q and E = k e Q / r 2 When r < a, the charge inside the surface is 0 and E = 0

32 Field at a Distance from a Line of Charge Select a cylindrical charge distribution The cylinder has a radius of r and a length of l E is constant in magnitude and perpendicular to the surface at every point on the curved part of the surface

33 Field Due to a Line of Charge The end view confirms the field is perpendicular to the curved surface The field through the ends of the cylinder is 0 since the field is parallel to these surfaces

34 Field Due to a Line of Charge Use Gauss s law to find the field E E da E da EA q encl 0 l 0 A 2 rl, so E (2 rl) l 0 E 2 0 r 2k e r

35 Field Due to a Plane of Charge E must be perpendicular to the plane and must have the same magnitude at all points equidistant from the plane Choose a small cylinder whose axis is perpendicular to the plane for the Gaussian surface E is parallel to the curved surface, with no contribution to the surface area from this curved part of the cylinder The flux through each end of the cylinder is EA and so the total flux is 2EA

36 Field Due to a Plane of Charge The total charge in the surface is σa Applying Gauss s law E 2 EA A 0, so E 2 0 Note, this does not depend on r Therefore, the field is uniform everywhere

37 Electrostatic Equilibrium When there is no net motion of charge within a conductor, the conductor is said to be in electrostatic equilibrium

38 Properties of a Conductor in Electrostatic Equilibrium 1. The electric field is zero everywhere inside the conductor. 2. If an isolated conductor carries a charge, the charge resides on its surface. 3. The electric field just outside a charged conductor is perpendicular to the surface and has a magnitude of σ/ε o. 4. On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is the smallest.

39 Property 1: E inside = 0 Consider a conducting slab in an external field E If the field inside the conductor were not zero, free electrons in the conductor would experience an electrical force These electrons would accelerate These electrons would not be in equilibrium Therefore, there cannot be a field inside the conductor

40 Property 2: Charge Resides on the Surface Choose a Gaussian surface inside but close to the actual surface The electric field inside is zero (property 1) 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 Therefore, any charge must be on the surface.

41 Demo EA10: Gauss s law; field inside insulator, not conductor Excess charges resides on the outer surface of a conductor. A metal can is placed on top of a van der Graaff generator and filled with styrofoam packing. When the generator is turned on, the packing remains within the cup. One can show that the outside of the can has charge by drawing off a spark with the earth rod. The metal can and the generator are all at one equipotential and the charge is on the outside of the conductor. There is no field inside the can. The can is replaced by a polystyrene cup. In this case, the packing is ejected out of the cup. Because the cup is an insulator, a field can exist within it the cup makes little disturbance to the equipotential surfaces.

42 Property 3: Field s Magnitude and Direction Choose a cylinder as the Gaussian surface The field must be perpendicular to the surface, for, 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. E EA A 0, and E 0

43 Property 4: Charge density highest at sharp points The charge density is high where the radius of curvature is small And low where the radius of curvature is large The electric field is large near the convex points having small radii of curvature and reaches very high values at sharp points

44 End of Chapter

45 Quick Quiz 24.6 Your little brother likes to rub his feet on the carpet and then touch you to give you a shock. While you are trying to escape the shock treatment, you discover a hollow metal cylinder in your basement, large enough to climb inside. In which of the following cases will you not be shocked? (a) You climb inside the cylinder, making contact with the inner surface, and your charged brother touches the outer metal surface. (b) Your charged brother is inside touching the inner metal surface and you are outside, touching the outer metal surface. (c) Both of you are outside the cylinder, touching its outer metal surface but not touching each other directly.

46 Quick Quiz 24.6 Answer: (a). Charges added to the metal cylinder by your brother will reside on the outer surface of the conducting cylinder. If you are on the inside, these charges cannot transfer to you from the inner surface. For this same reason, you are safe in a metal automobile during a lightning storm.