PES 1120 Spring 2014, Spendier Lecture 7/Page 1 Lecture today: 1) Electric field due to a uniformly charged disk (Chapter 22) 2) Electric Field in an ideal Parallel Plate Capacitor 3) Electric Flux (Chapter 23) Electric field due to a uniformly charged disk A uniformly charged disk of radius R with a total charge Q lies in the xy-plane. We want to find the electric field at a point P, along the positive z-axis that passes through the center of the disk perpendicular to its plane. We will also discuss the limit where R >> z. 1) By symmetry arguments, the electric field at P points in the z+-direction. 2) We treat the disk as a set of concentric uniformly charged rings of radius r′ and thickness dr′, as shown in Figure above. Each of theses rings has a charge distribution dq. 3) Write an expression for the electric charge dq in that charged ring of thickness dr′. dq = σ dA = σ (2π r′ dr′) 4) Write expressions for the non-zero components of the electric field at P caused by the designated charge dq shown above.
Transcript
PES 1120 Spring 2014, Spendier Lecture 7/Page 1 Lecture today:
1) Electric field due to a uniformly charged disk (Chapter 22) 2) Electric Field in an ideal Parallel Plate Capacitor 3) Electric Flux (Chapter 23)
Electric field due to a uniformly charged disk A uniformly charged disk of radius R with a total charge Q lies in the xy-plane. We want to find the electric field at a point P, along the positive z-axis that passes through the center of the disk perpendicular to its plane. We will also discuss the limit where R >> z.
1) By symmetry arguments, the electric field at P points in the z+-direction. 2) We treat the disk as a set of concentric uniformly charged rings of radius r′ and thickness dr′, as shown in Figure above. Each of theses rings has a charge distribution dq. 3) Write an expression for the electric charge dq in that charged ring of thickness dr′. dq = σ dA = σ (2π r′dr′) 4) Write expressions for the non-zero components of the electric field at P caused by the designated charge dq shown above.
PES 1120 Spring 2014, Spendier Lecture 7/Page 2 5) Write the integral with the appropriate limits and complete the integral. Integrating from r′ = 0 to r′ = R, the total electric field at P becomes:
6) Consider the limit of R >> z. Physically this means that the charged plane is very large (infinite sheet of charge), or the point P is extremely close to the surface of the plane. The electric field in this limit becomes
Hence:
02E s
e
This is the electric field produced by an infinite sheet of uniform charge located on one side of a nonconductor such as plastic.
PES 1120 Spring 2014, Spendier Lecture 7/Page 3 Electric Field in an ideal Parallel Plate Capacitor: Parallel-Plate Capacitor: Two oppositely-charge conducting sheets separated by a small distance used to store energy electrostatically in an electric field. The Electric Field due to a charged infinite sheet is very simple in that the E-field is constant at any distance from the sheet.
When two plate of different charge are placed near each other, the two E-fields between the plates add while the E-field outside the plate cancels.
Capacitor: When the plates are close to each other to form a capacitor, the E-field between the plates is constant through out the interior of the capacitor as long as one is not near the edges of the plates.
PES 1120 Spring 2014, Spendier Lecture 7/Page 4 Starting Chapter 23 Now we are going to talk about another way of relating electric fields and charges. This relation is called "Gauss' Law." But before we can tackle Gauss' Law, we need to understand something called "flux". To understand flux, let's talk first about a general example before talking about the flux of an electric field. Flux (Latin - fluxus - for flow) To understand flux, we need to return to the idea of a vector fields, and, in particular, I will use the idea of fluid (water) flow. Open Surface: Let’s first consider a case of steady water flow (say, a deep river moving slowly with constant velocity throughout). (Show sheet of paper – as loop) Imagine this is a rectangular surface bounded by a wire. If this wire loop is in the river, I can ask : How do I need to orient the loop to have the most water flowing through?
Answer: A perpendicular to flow direction = most water will flow through We can actually calculate the rate of water that flows through this loop as a given time. Let’s call this quantity velocity flux For A:
velThrough surface
dV dxvolume flow rate A vA velocity fluxdt dt
PES 1120 Spring 2014, Spendier Lecture 7/Page 5
A
A area A
For B: We need to compute the area perpendicular to the flow direction
cosvel vA vA q Let’s make the notion of flux more precise. Since the water is flowing – has direction – we can assign a velocity vector to each point flowing through A (components of all these vectors is a vector field). Since the flux, they way we defined it is a scalar, the area element also needs to have a vector – call this area vector Area vector: Definition: area vector ˆA A n
, associated with a flat surface of area A.
Magnitude of vector A
= area A of surface.
Direction of vector A
= direction perpendicular (normal) to surface = direction of unit
normal n̂ .
Using this notion of area vector, we can rewrite the velocity flux as a dot product between the velocity vector and area vector:
cosvel vA v A q
Flux of a vector field through a loop of area A. This expression has a maximum when
||A v and zero when A v
PES 1120 Spring 2014, Spendier Lecture 7/Page 6 Notice that there is an ambiguity in the direction n̂ . Every flat surface has two perpendicular directions. We will get rid of this ambiguity later! The above expression assumes that the area is a flat surface. What happens if the area id curved and keeping v uniform through space? Flux with Calculus • So far we have worked with constant velocity fields. • How do we handle fields that vary over a surface (either in strength, or in direction)? • We have to break up the surface into small parts, and then add v A
up for each part.
To understand a surface integral, do this: in your imagination, break the total surface up into many little segments, labeled with an index i. The surface vector of segment i is idA
.
If the segment is very, very tiny, it is effectively flat and the velocity field is constant over that tiny surface, so we can use our special case formula vel v A
.
The flux through segment i is therefore vel ,i i iv dA .
The total flux is the sum:
vel i ii
approximate : v dA
velexact : v dA
If we now assume that the velocity field varies in space, then we have the most general expression:
vel v r dA
PES 1120 Spring 2014, Spendier Lecture 7/Page 7 Electric Flux So far I have used the analogy of water flow to describe flux in more intuitive terms.
- For a Electric field nothing is “flowing” - But it also can be described by a vector field E-field mathematics is the
same - E-field lines can be though of flow lines
The net electric flux through a closed surface is proportional to the net number of field lines entering/exiting the surface. flux the number of electric field lines crossing the surface.
E r dA
units [N m2/C]
The loop on the integral sign indicates that the integration is to be taken over the entire (closed) surface = called Gaussian surface Example 1: Calculate the net electric field flux through a cubic surface in a uniform electric field. All faces of the cube have area A.
PES 1120 Spring 2014, Spendier Lecture 7/Page 8 Example 2: Calculate the electric field flux through flat area A at 25 degrees from a normal in a uniform electric field.
