Charge and Polarization Monday, January 14, 2008 Charge and Polarization

Demonstration #1 Demonstrate how you can pick up the tissue without touching it in any way with your body. What is occurring on the atomic level that lets you do this?

The atom The atom has positive charge in the nucleus, located in the protons. The positive charge cannot move from the atom unless there is a nuclear reaction. The atom has negative charge in the electron cloud on the outside of the atom. Electrons can move from atom to atom without all that much difficulty.

Question You charge the balloon by rubbing it on hair or on a sweater, and the balloon becomes negative. How can it pick up a neutral tissue?

This is an electroscope Pole The electroscope is made from a metal or other conductor, and may be contained within a flask. The vanes are free to move. Vanes

Demonstration #2 Rub the black rod with the fur. Bring the rod toward the pole of the electroscope. What happens to the vanes? Come up with an atomic-level explanation for your observations.

Demonstration #3 Rub the glass rod with the silk. Bring the rod toward the pole of the electroscope. What happens to the vanes? Come up with an atomic-level explanation for your observations.

Demonstration #4 What happens when your touch the electroscope with the glass rod?

Charge Charge comes in two forms, which Ben Franklin designated as positive (+) and negative(–). Charge is quantized. The smallest possible stable charge, which we designate as e, is the magnitude of the charge on 1 electron or 1 proton. We say a proton has charge of e, and an electron has a charge of –e. e is referred to as the “elementary” charge. e = 1.602  10-19 Coulombs. The coulomb is the SI unit of charge.

Sample Problem A certain static discharge delivers -0.5 Coulombs of electrical charge. How many electrons are in this discharge?

Sample Problem How much positive charge resides in two moles of hydrogen gas (H2)? How much negative charge? How much net charge?

Sample Problem The total charge of a system composed of 1800 particles, all of which are protons or electrons, is 31x10-18 C. How many protons are in the system? How many electrons are in the system?

Coulomb’s Law and Electrical Force Tuesday, January 15, 2008 Coulomb’s Law and Electrical Force

Electric Force Charges exert forces on each other. Like charges (two positives, or two negatives) repel each other, resulting in a repulsive force. Opposite charges (a positive and a negative) attract each other, resulting in an attractive force.

Coulomb’s Law – form 1 Coulomb’s law tells us how the magnitude of the force between two particles varies with their charge and with the distance between them. k = 8.99  109 N m2 / C2 q1, q2 are charges (C) r is distance between the charges (m) F is force (N) Coulomb’s law applies directly only to spherically symmetric charges.

Coulomb’s Law – form 2 Sometimes you see Coulomb’s Law written in a slightly different form eo = 8.85  10-12 C2/ N m2 q1, q2 are charges (C) r is distance between the charges (m) F is force (N) This version is theoretically derived and less practical that form 1

Spherically Symmetric Forces Newton’s Law of Gravity Coulomb’s Law

Sample Problem A point charge of positive 12.0 μC experiences an attractive force of 51 mN when it is placed 15 cm from another point charge. What is the other charge?

Sample Problem qA = 1.50 nC A B qB = -0.50 nC 1.3 m Calculate the mass of ball B, which is suspended in midair. A qA = 1.50 nC 1.3 m B qB = -0.50 nC

Superposition Electrical force, like all forces, is a vector quantity. If a charge is subjected to forces from more than one other charge, vector addition must be performed. Vector addition to find the resultant vector is sometimes called superposition.

y (m) 2.0 1.0 -3 mC 2 mC 4 mC 1.0 2.0 x (m) Sample Problem What is the force on the 4 mC charge? y (m) 2.0 1.0 -3 mC 2 mC 4 mC 1.0 2.0 x (m)

y (m) 2.0 -3 mC 1.0 2 mC 4 mC 1.0 2.0 x (m) Sample Problem What is the force on the 4 mC charge? y (m) 2.0 -3 mC 1.0 2 mC 4 mC 1.0 2.0 x (m)

Wednesday, January 16, 2008 The Electric Field

The Electric Field The presence of + or – charge modifies empty space. This enables the electrical force to act on charged particles without actually touching them. We say that an “electric field” is created in the space around a charged particle or a configuration of charges. If a charged particle is placed in an electric field created by other charges, it will experience a force as a result of the field. Sometimes we know about the electric field without knowing much about the charge configuration that created it. We can easily calculate the electric force from the electric field.

Why use fields? Forces exist only when two or more particles are present. Fields exist even if no force is present. The field of one particle only can be calculated.

Field around + charge The arrows in a field are not vectors, they are “lines of force”. The lines of force indicate the direction of the force on a positive charge placed in the field. Negative charges experience a force in the opposite direction.

Field around - charge

Field between charged plates + + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Field Vectors from Field Lines The electric field at a given point is not the field line itself, but can be determined from the field line. The electric field vectors is always tangent to the line of force at that point. Vectors of any kind are never curvy!

Field Vectors from Field Lines

Force from Electric Field The force on a charged particle placed in an electric field is easily calculated. F = E q F: Force (N) E: Electric Field (N/C) q: Charge (C)

Sample Problem The electric field in a given region is 4000 N/C pointed toward the north. What is the force exerted on a 400 μg styrafoam bead bearing 600 excess electrons when placed in the field?

Sample Problem A 400 μg styrofoam bead has 600 excess electrons on its surface. What is the magnitude and direction of the electric field that will suspend the bead in midair?

Thursday, January 17, 2008 Superposition

Sample Problem A proton traveling at 440 m/s in the +x direction enters an an electric field of magnitude 5400 N/C directed in the +y direction. Find the acceleration.

For Spherical Electric Fields The Electric Field surrounding a point charge or a spherical charge can be calculated by: E = k q / r2 E: Electric Field (N/C) k: 8.99 x 109 N m2/C2 q: Charge (C) r: distance from center of charge q (m) Remember that k = 1/4peo

Principle of Superposition When more than one charge contributes to the electric field, the resultant electric field is the vector sum of the electric fields produced by the various charges. Again, as with force vectors, this is referred to as superposition.

Remember… Electric field lines are NOT VECTORS, but may be used to derive the direction of electric field vectors at given points. The resulting vector gives the direction of the electric force on a positive charge placed in the field.

Sample Problem A particle bearing -5.0 μC is placed at -2.0 cm, and a particle bearing 5.0 μC is placed at 2.0 cm. What is the field at the origin?

Sample Problem A particle bearing 10.0 mC is placed at the origin, and a particle bearing 5.0 mC is placed at 1.0 m. Where is the field zero?

Sample Problem What is the charge on the bead? It’s mass is 32 mg. E = 5000 N/C 40o

Electric Potential and Potential Energy Friday, January 18, 2008 Electric Potential and Potential Energy

Electric Potential Energy Electrical potential energy is the energy contained in a configuration of charges. Like all potential energies, when it goes up the configuration is less stable; when it goes down, the configuration is more stable. The unit is the Joule.

Electric Potential Energy Electrical potential energy increases when charges are brought into less favorable configurations ΔU > 0 + + - +

Electric Potential Energy Electrical potential energy decreases when charges are brought into more favorable configurations. ΔU < 0 + - + +

Electric Potential Energy + – ΔU = ____ ΔU = ____ + + + - Work must be done on the charge to increase the electric potential energy.

Work and Charge For a positive test charge to be moved upward a distance d, the electric force does negative work. The electric potential energy has increased and ΔU is positive (U2 > U1) + d + E F

Work and Charge If a negative charge is moved upward a distance d, the electric force does positive work. The change in the electric potential energy ΔU is negative (U2 < U1) - d F - E

Electric Potential Electric potential is hard to understand, but easy to measure. We commonly call it “voltage”, and its unit is the Volt. 1 V = 1 J/C Electric potential is easily related to both the electric potential energy, and to the electric field.

Electrical Potential and Potential Energy The change in potential energy is directly related to the change in voltage. U = qV U: change in electrical potential energy (J) q: charge moved (C) V: potential difference (V) All charges will spontaneously go to lower potential energies if they are allowed to move.

Electrical Potential and Potential Energy Since all charges try to decrease UE, and DUE = qDV, this means that spontaneous movement of charges result in negative DU. ΔV = ΔU / q Positive charges like to DECREASE their potential (DV < 0) Negative charges like to INCREASE their potential. (DV > 0)

Sample Problem A 3.0 μC charge is moved through a potential difference of 640 V. What is its potential energy change?

Electrical Potential in Uniform Electric Fields The electric potential is related in a simple way to a uniform electric field. V = -Ed V: change in electrical potential (V) E: Constant electric field strength (N/C or V/m) d: distance moved (m) d E DV

Sample Problem An electric field is parallel to the x-axis. What is its magnitude and direction of the electric field if the potential difference between x =1.0 m and x = 2.5 m is found to be +900 V?

Sample Problem y(m) C 1.0 A B 1.0 2.0 x(m) What is the voltmeter reading between A and B? Between A and C? Assume that the electric field has a magnitude of 400 N/C. y(m) C 1.0 A B 1.0 2.0 x(m)

Sample Problem y(m) C 1.0 A B 1.0 2.0 x(m) How much work would be done BY THE ELECTRIC FIELD in moving a 2 mC charge from A to C? From A to B? from B to C?. How much work would be done by an external force in each case? y(m) C 1.0 A B 1.0 2.0 x(m)

Electric Field Lines and Shielding Tuesday, January 22, 2008 Electric Field Lines and Shielding

More on Electric Field Maps

More on Electric Field Maps

Review: Electric Fields and Equipotential Lines Java Simulation http://phet.colorado.edu/new/simulations/sims.php?sim=Charges_and_Fields

Excess Charges on Conductors Excess charges reside on the surface of a charged conductor. If excess charges were found inside a conductor, they would repel one another until the charges were as far from each other as possible… the surface! +

Electric Field and Lightning Rods Electric field lines are more dense near a sharp point, indicating the electric field is more intense in such regions. All lightning rods take advantage of this by having a sharply pointed tip. During an electrical storm, the electric field at the tip becomes so intense that charge is given off into the atmosphere, discharging the area near a house at a steady rate and preventing a sudden blast of lightning.

Electric Field inside a Conductor The electric field inside a conductor must be zero. + + + + + + E = 0 + + + + + +

Conductor in an electric field If a conductor is placed in an electric field, then the charges polarize to nullify the external field. - + - + - + - E = 0 + + - + - + - - +

Energy Conservation in Electric Fields Wednesday, January 23, 2008 Energy Conservation in Electric Fields

Conservation of Energy Review In a conservative system, energy changes from one form of mechanical energy to another. When only the conservative electrostatic force is involved, a charged particle released from rest in an electric field will move so as to lose potential energy and gain an equivalent amount of kinetic energy. The change in electrical potential energy can be calculated by DUE = qDV.

Sample Problem If a proton is accelerated through a potential difference of -2,000 V, what is its change in potential energy? How fast will this proton be moving if it started at rest?

Sample Problem A proton at rest is released in a uniform electric field. How fast is it moving after it travels through a potential difference of -1200 V? How far has it moved?

Electric Potential Energy for Spherical Charges Electric potential energy is a scalar, like all forms of energy. U = kq1q2/r U: electrical potential energy (J) k: 8.99  109 N m2 / C2 q1, q2 : charges (C) r: distance between centers (m) This formula only works for spherical charges or point charges.

Drawing Parallels Gravitation Electrostatics

Sample Problem y (m) 2.0 1.0 2 mC 4 mC x (m) 1.0 2.0 What is the potential energy of the configuration shown below? y (m) 2.0 1.0 2 mC 4 mC x (m) 1.0 2.0

Sample Problem y (m) 2.0 -3 mC 1.0 2 mC 4 mC x (m) 1.0 2.0 How much work was done in assembling the charge configuration shown below? y (m) 2.0 -3 mC 1.0 2 mC 4 mC x (m) 1.0 2.0

Potential and Potential Energy of Configurations of Point Charges Wednesday, January 23, 2008 Potential and Potential Energy of Configurations of Point Charges

Absolute Electric Potential (spherical) For a spherical or point charge, the electric potential can be calculated by the following formula V = kq/r V: potential (V) k: 8.99 x 109 N m2/C2 q: charge (C) r: distance from the charge (m) Remember, k = 1/(4peo)

Sample Problem y (m) 2.0 -3 mC 1.0 2 mC 4 mC x (m) 1.0 2.0 What is the electric potential at (2,2)? y (m) 2.0 -3 mC 1.0 2 mC 4 mC x (m) 1.0 2.0

Equipotential surfaces high highest medium low lowest positive negative

Equipotential surfaces High potential Low potential

Question What can you say about the intersection between field lines and equipotential surfaces?

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Sample Problem Draw field lines for the charge configuration below. The field is 600 V/m, and the plates are 2 m apart. Label each plate with its proper potential, and draw and label 3 equipotential surfaces between the plates. You may ignore edge effects. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

Sample Problem Draw a negative point charge of -Q and its associated electric field. Draw 4 equipotential surfaces such that DV is the same between the surfaces, and draw them at the correct relative locations. What do you observe about the spacing between the equipotential surfaces?

Fill in the following table for spherical charges Force Potential Energy Field Potential

What is magnitude and direction of electric field What is magnitude and direction of electric field? b) What is shortest distance one can go to undergo a change of 5.00 V?

Thursday, January 24, 2008 Review