# Date Review of Coulomb’s Law. Fundamentals of Electrical Charge If a particle bears an electrical charge, we can generally assume it has too many electrons.

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date Review of Coulomb’s Law

Fundamentals of Electrical Charge If a particle bears an electrical charge, we can generally assume it has too many electrons (negative charge) or too few electrons (positive charge). This is because electrons move around and protons stay put. Charge on electron: -e Charge on proton: +e e = 1.602  10 -19 Coulombs, C Any charge q is a multiple of the number of extra electrons or missing electrons such that q = ne

Coulomb’s Law k = 8.99  10 9 Nm 2 /C 2 q 1 and q 2 are spherical or point charges in Coulombs. r is the separation between the centers of the charges  o = 8.854  10 -12 C 2 /(N m 2 )

For Two Charges The electrostatic force is repulsive if the charges are of the same sign and attractive if the charges are of unlike sign. The force exerted on charge A by charge B is always equal and opposite of the force exerted by charge B on charge A (Newton’s 3 rd Law)

For More Than Two Charges Determine the force on a charge due to each of the other charges in the vicinity. Do a vector addition. q “It’s all about me! Who is nearby who can affect me?” When you are determining electrostatic force, you focus on one charge.

Static Electricity Experiment Cut 2 20-cm strips of transparent tape (mass of each 65 mg). Fold about 1 cm tape over at one end of each strip to create a handle. Press both pieces of tape side-by-side on your lab table and rub your finger back and forth across the strips. Quickly pull the strips off the lab table. Hold the handles together and the strips will repel each other, forming an inverted “V”. Estimate the charge on each strip. Assume the charges act as though they are at the center of mass of the strip. Hint: Begin by drawing a Free Body Diagram!

date Electric Field Review

Two identical balls of mass m and charge q are hanging from strings of length L. Derive an expression for q in terms of m, , L, and fundamental constants. L  L m,q

The Electric Field A charge distribution changes the empty space around it such that this empty space will affect other charges brought nearby. We say that a charge distribution creates an electric field in the empty space surrounding it.

Electric Field This equation can be used to calculate the electric field a distance r away from a the center of a spherically symmetric charge distribution of q o Coulombs. Another charge q entering the electric field created by q o will experience a force F, which can be calculated by the equation F = qE.

Electric Field Direction The direction of the electric field at a point in space is the direction that a small positive charge at rest (a “test charge”) wishes to move if it is placed at that location. Thus, the electric field points away from positive charges and toward negative charges.

Electric Field Direction The electric field around point charges is spherically symmetric. When multiple point charges are present, the fields due to the individual charges “superimpose” to create a more complex picture. http://www.edumedia-sciences.com/m195_l2- electric-field.htmlhttp://www.edumedia-sciences.com/m195_l2- electric-field.html http://www.falstad.com/emstatic/ http://www.cco.caltech.edu/~phys1/java/phys1/EF ield/EField.htmlhttp://www.cco.caltech.edu/~phys1/java/phys1/EF ield/EField.html

Caution… Electric field lines are NOT VECTORS, but may be used to derive the direction of electric field vectors at given points. Electric field vectors are always tangent to the electric field line at any given point in space.

-q 1 q2q2 P Superposition problems involve vector addition. E2E2 E1E1 E

Static Electricity Experiment Blow up the balloon and charge it by rubbing it against your hair. Can you deflect a flowing stream of water with the charged balloon? Is the water charged? If not, why is it being deflected? See if your group can come up with a plausible answer.

date Charge Distributions

Limitations of Coulomb’s Law Coulomb’s Law equations for Force and Field can only be used directly for point charges or spherically symmetric charges. For more complicated “continuous charge distributions” we need to break up the charge distribution into little bitty pieces and use Coulomb’s Law and superposition together to determine the electric field at a given location in space near the charge distribution.

Linear Charge Distribution When charge resides on a long thin object such as a wire or a ring, we call that a linear charge distribution. It is sometimes convenient for us to define a linear charge density,, which is charge per unit length. = Q/L = dQ/dL +++++ + + + + + + + + + + + + + + + + +

Surface Charge Distribution +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ + + + + + + ++++ + When charge resides on larger surface, we call it a surface charge distribution. It is sometimes convenient for us to define a surface charge density, σ, which is charge per unit area. σ = Q/A = dQ/dA

General Procedure Each little infinitesimally small charge dq in a charge distribution containing Q total Coulombs creates its own tiny electric field dE at a point P in space a distance r from dq. I can add all these little infinitesimal fields dE together to get the field at point P. What does this sound like to you?

General Procedure - continued You need to integrate over a spatial variable (not charge!). Appropriate choices are linear distance, arc length, or angle (x,y,s, ,  ) Find a common variable that r and/or dq both depend on. See if symmetry (and trig) can be used to simplify the problem by elimination of off-axis components of E. Find the appropriate limits to the integral. Don’t skip set-up steps. The physics is in the setup!

Sample Problem Determine the electric field magnitude and direction a distance y away from an extremely long, straight wire of charge density. y + +++++++++++++++++++++++++++++++++++++++++++

date Charge Distributions- continued

Sample Problem Determine the electric field magnitude and direction a distance x away from a ring of radius R bearing charge Q. x R

date Electric Field Visualization

date Motion of Particles in Electric Field

Motion of Charged Particles in Electric Fields If the electric field is constant, acceleration will be constant, and kinematic equations can be employed. The motion is not unlike projectile motion.

Sample problem What is the speed and position of an electron released from rest in this electric field after 3.0 ns? e- E = 320 N/C

Sample problem What is the velocity and position of this electron 3.0 ns after it enters the field? e- E = 320 N/C v = 20,000 m/s x y

Non-constant Fields http://www.edumedia- sciences.com/m195_l2-electric-field.htmlhttp://www.edumedia- sciences.com/m195_l2-electric-field.html http://www.falstad.com/mathphysics.html

02/03/2010 Electric Flux

Flux Flux means “flow”. Consider three rectangular wire loops in a vector field. Which one has maximum flux (or flow) of the field lines through it? Consider a Vector Field, v max flux no flux intermediate flux

To Increase Flux Increase the field Increase the area of the loop Make sure the hoop is appropriately angled Consider a Vector Field, v max flux no flux intermediate flux

The Area Vector Consider a Vector Field, v A A A The “area vector” is defined as a vector perpendicular to a surface with magnitude equal to the scalar area of the surface. Consider the angle between v and A. For what angle is the flux maximum?

Flux Equation Consider a Vector Field, v A A A The flux is proportional to field vector magnitude, area vector magnitude, and the cosine of the angle between them. What vector operation does this sound like?

Flux Equation Consider a Vector Field, v A A A What vector operation does this sound like?

Flux Equation for Electric Field E A A A Units: Nm 2 /C

Area Vectors For a Closed Shape This rectangular prism has six surfaces. Each surface has an area vector that points outward from center of the prism, and is normal to the surface. A A A

Another Example This cylinder is a bit more complicated. The top and bottom have areas that can easily be calculated, and the corresponding vectors point outward. On the sides, we must define and infinite number of infinitesimally small areas, each of which defines a little vector (dA) that points outward. A dAdA dAdA

The Calculation of Flux Over a Closed Surface in a Vector Field At each point on the closed surface, we must take the dot product with the vector field to get the flux for that small area. Then we add all these dot products up together to get the flux for the entire surface. This leads to some interesting observations. If there is a “source” of the vector field in the closed shape, the flux over its surface is positive. If there is a “sink” of the vector field in the closed shape, the flux over its surface is negative. If there is neither a source or sink of the vector field in the closed shape, the flux over its surface is zero.

What do we mean by “source” and “sink” of an electric field? The source is where the field starts, and the sink is where the field terminates. In an electric field, the source is the positive charge, and the sink is the negative charge. Therefore, –If a closed shape encloses a positive charge, the flux is positive. –If a closed shape encloses a negative charge, the flux is negative. –If the closed shape encloses no net charge, the flux is zero.

Mathematical Representation For a general vector field, v: For an electric field, E: In an electric field, the closed shape we integrate the flux over is referred to as a “Gaussian surface”.

Sample Problem Calculate the electric flux through a spherical surface of radius 2.0 m containing a point charge of 3mC at its center.

Sample Problems 1.Draw an electric dipole, and sketch three Gaussian surfaces for which one has positive electric flux, one has negative electric flux, and one has zero electric flux.

Sample Problem 2 2. A cube of side L is placed in a region of uniform electric field E. Find the electric flux through each face of the cube and the total flux when (see diagram) –it is oriented with two of its faces perpendicular to E

HW for tomorrow (yes, tomorrow) Ch. 23 Problems 1-5.

date Gauss’s Law of Electricity

Gauss’ Law of Electricity q: net charge (C) enclosed inside a given Gaussian surface. This is a sum of all + and - charges  o : electrical permittivity of free space –8.85  10 -12 F/m –8.85  10 -12 C 2 /Nm 2 –  E : electrical flux over the Gaussian surface +

Other forms of Gauss’s Law Integral forms give indication of enclosed charge. Differential form gives indication charge density. For more information see http://hyperphysics.phy- astr.gsu.edu/hbase/electric/ maxeq2.html http://hyperphysics.phy- astr.gsu.edu/hbase/electric/ maxeq2.html

Gaussian Surface A Gaussian surface is simply any closed shape in space, which can be of any arbitrary shape. All Gaussian surfaces give the same answer in Gauss’s Law if they enclose the same net charge! So, to make the math easier, Gaussian surfaces are typically chosen for convenience and high symmetry with regard the electric field.

What’s Gauss’s Law Good For? Gauss’s Law can be used to determine how much charge is enclosed in a surface. More commonly, Gauss’s Law is used to determine the electric field at a point in space.

Sample Problem A point charge q is located a distance d from an infinite plane. Determine the electric flux through the plane due to the point charge.

date Application of Gauss’s Law (I)

Sample Problem Consider two Gaussian surfaces, a sphere of radius R and a cube of side 4R. In each is a positive point charge of +q. How does the electric flux compare for the two surfaces?

Sample Problem Derive the electric field outside a charged spherical non-conductor with an even charge distribution.

Sample Problem Derive the electric field INSIDE a charged spherical non-conductor with an even charge distribution.

date Application of Gauss’s Law (II)

date Review for exam

date Exam

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