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Continuous Charge Distributions Discrete charges Continuous charge distribution Single charge Single piece of a charge distribution +Q 3 +Q 2 +Q 1 0 0.

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Presentation on theme: "Continuous Charge Distributions Discrete charges Continuous charge distribution Single charge Single piece of a charge distribution +Q 3 +Q 2 +Q 1 0 0."— Presentation transcript:

1 Continuous Charge Distributions Discrete charges Continuous charge distribution Single charge Single piece of a charge distribution +Q 3 +Q 2 +Q

2 Finding dq Line charge Cartesian Polar Surface charge Volume charge

3 Example – Infinitely Long Line of Charge y-components cancel by symmetry

4 Example – Charged Ring (ex 22-5) perpindicular-components cancel by symmetry

5 Check a Limiting Case When: The charged ring must look like a point source. 0

6 Uniformly Charged Disk (ex. 22-7)

7 Binomial Expansion Theorem Quadratic terms and higher are small

8 Two Important Limiting Cases Large Charged Plate: Very Far From the Charged Plate:

9 Parallel Plate Capacitor -Q +Q

10 Summary Coulomb’s Law Electric Field Point Charges: These can be used to find the fields in the vicinity of continuous charge distributions: Line of Charge: Charged Plate: Dipole: Ring of Charge

11 From Chapter 23 – Coulomb’s Law We found the fields in the vicinity of continuous charge distributions by integration: Line of Charge: Charged Plate: Are you interested in learning an easier way?

12 Electric flux The field is uniform The plane is perpendicular to the field

13 Flux for a uniform electric field passing through an arbitrary plane The field is uniform The plane is not perpendicular to the field

14 Rules for drawing field lines The electric field,, is tangent to the field lines. The number of lines leaving/entering a charge is proportional to the charge. The number of lines passing through a unit area normal to the lines is proportional to the strength of the field in that region. Field lines must begin on positive charges (or from infinity) and end on negative charges (or at infinity). The test charge is positive by convention. No two field lines can cross.

15 General flux definition The field is not uniform The surface is not perpendicular to the field If the surface is made up of a mosaic of N little surfaces

16 Electric Flux, General In the more general case, look at a small area element In general, this becomes

17 Electric Flux Calculations 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 2

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

19 Closed surfaces +Q Negative Flux Zero Flux

20 Flux from a point charge through a closed sphere

21 Gauss’s Law Gauss asserts that the proceeding calculation for the flux from a point charge is true for any charge distribution!!! This is true so long as Q is the charge enclosed by the surface of integration.

22 A few questions If the electric field is zero for all points on the surface, is the electric flux through the surface zero? If the electric flux is zero, must the electric field vanish for all points on the surface? If the electric flux is zero for a closed surface, can there be charges inside the surface? What is the flux through the surface shown? Why?

23 Example P24.1 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 m wide and 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.

24 Example P24.9 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?

25 Example P23.17 A point charge Q = 5.00 μC is located at the center of a cube of edge L = m. In addition, six other identical point charges having q = –1.00 μC are positioned symmetrically around Q as shown in Figure P Determine the electric flux through one face of the cube.

26 Applying Gauss’s Law Select a surface –Try to imagine a surface where the electric field is constant everywhere. This is accomplished if the surface is equidistant from the charge. –Try to find a surface such that the electric field and the normal to the surface are either perpendicular or parallel. Determine the charge inside the surface If necessary, break the integral up into pieces and sum the results. –e.g. For a cylinder, r h

27 L r Example – a line of charge 1.Find the correct closed surface 2.Find the charge inside that closed surface

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29 Example – a charged plane +Q 1.Find the correct closed surface 2.Find the charge inside that closed surface

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31 r Example – a solid sphere of charge +Q uniformly distributed a Inside the charged sphere: 1.Find the correct closed surface 2.Find the charge inside that closed surface

32 Example – a solid sphere of charge +Q uniformly distributed a Outside the charged sphere: 1.Find the correct closed surface 2.Find the charge inside that closed surface r

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34 Example – a conducting conductor Insulators, like the previous charged sphere, trap excess charge so it cannot move. Conductors have free electrons not bound to any atom. The electrons are free to move about within the material. If excess charge is placed on a conductor, the charge winds up on the surface of the conductor. Why? The electric field inside a conductor is always zero. The electric field just outside a conductor is perpendicular to the conductors surface and has a magnitude,  

35 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

36 E inside = 0, cont. Before the external field is applied, free electrons are distributed throughout the conductor When the external field is applied, the electrons redistribute until the magnitude of the internal field equals the magnitude of the external field There is a net field of zero inside the conductor This redistribution takes about s and can be considered instantaneous

37 Charge Resides on the Surface Choose a gaussian surface inside but close to the actual surface The electric field inside is zero (prop. 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

38 Charge Resides on the Surface, cont 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

39 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

40 Field’s Magnitude and Direction, cont. The net flux through the gaussian surface is through only the flat face outside the conductor –The field here is perpendicular to the surface Applying Gauss’s law

41 Conductors in Equilibrium, example The field lines are perpendicular to both conductors There are no field lines inside the cylinder

42 r Example – a solid conducting sphere of charge surrounded by a conducting shell +2Q on inner sphere -Q on outer shell a Find the Electric Field: 1.Find the correct closed surface 2.Find the charge inside that closed surface b c

43 Example P24.31 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.

44 Example P24.35 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.

45 Example P24.43 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.

46 Example P24.47 A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge per unit length of λ, and the cylinder has a net charge per unit length of 2λ. From this information, use Gauss’s law to find (a) the charge per unit length on the inner and outer surfaces of the cylinder and (b) the electric field outside the cylinder, a distance r from the axis.


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