Polarization. Polarization is not necessarily a charge imbalance!

Transcription

1 Electrostatics

2 Polarization Polarization is the separation of charge In a conductor, free electrons can move around the surface of the material, leaving one side positive and the other side negative. In an insulator, the electrons realign themselves within the atom (or molecule), leaving one side of the atom positive and the other side of the atom negative. Polarization is not necessarily a charge imbalance!

3 Charging by Conduction When a charged conductor makes contact with a neutral conductor there is a transfer of charge. CHARGING NEGATIVELY CHARGING POSITIVELY Electrons are transferred from the rod to the ball, leaving them both negatively charged. Electrons are transferred from the ball to the rod, leaving them both positively charged. Remember, only electrons are free to move in solids. Notice that the original charged object loses some charge. click for animation

4 Charging by Induction Induction uses the influence of one charged object to coerce charge flow. Step 1. A charged rod is brought near an isolated conductor. The influence of the charge object polarizes the conductor but does not yet charge it. Step 2. The conductor is grounded to the Earth, allowing charge to flow out between it and the Earth.

5 Charging by Induction (cont.) Step 3. The ground is removed while the charge rod is still nearby the conductor. Step 4. The rod is removed and the conductor is now charge (opposite of rod). An object charged by induction has the opposite sign of the influencing body. Notice that the original charged object does not lose charge.

6 The Electrostatic Force Charles Coulomb s Torsion Balance A torsion balance measures the force between small charges. The electrostatic force depends directly on the magnitude of the charges. The force depends inversely on the square of distance between charges (another inverse square law )! TORSION BALANCE COULOMB S LAW OF ELECTROSTATIC FORCE electrostatic force constant F e = kq 1q 2 r 2 charges distance The constant of proportionality, k, is equal to 9.0 x 10 9 Nm 2 /C 2. A negative force is attractive, and a positive force is repulsive. The sign (+ or ) is different from a vector direction (left or right)

7 The Electrostatic Force EXAMPLE 1 - Find the force between these two charges ( ) C F e = ( 0.04 m) 2 ( )( C) F e = 225 N The negative signs means force of attraction, but does not indicate left or right direction EXAMPLE 2 - Find the net force on the left charge ( ) C F e = ( m) 2 ( )( C) F e = 360 N (force of repulsion) F net = F left F right F net = 360 N 225 N = 135 N, to the left

8 Electric Field Strength Field Theory Visualizes Force At A Distance DEFINITION OF ELECTRIC FIELD E field = force charge E = F e q 0 q 0 is a small, positive test charge Electric field is a vector quantity E field points toward negative charges E field points away from positive charges SI unit of electric field newton coulomb = N C

9 Electric Field Lines Single Point Charges POSITIVE CHARGE NEGATIVE CHARGE Density of field lines indicates electric field strength Inverse square law obeyed click for applet Definition of E Field for single point charge E = F e q 0 = kq 0 q / r 2 q 0 electric field constant E = kq r 2 charge distance

10 Electric Field Lines Electric fields for multiple point charges POSITIVE AND NEGATIVE POINT CHARGES TWO POSITIVE POINT CHARGES OPPOSITE MAGNETIC POLES ALIKE MAGNETIC POLES

11 Electric Fields EXAMPLE 1 E Find the electric field strength at 2 meters from the 5 millicoulomb charge. E = kq r 2 ( )( C) E = Nm 2 /C 2 ( 2 m) 2 E= N/C, to the right EXAMPLE 2 Find the force on an proton placed 2 meters from the 5 millicoulomb charge in the problem above. E = F e q F e = Nm 2 /C 2 F e = qe = ( C) ( N/C) = N, to the right OR ( )( C) ( C) = N, to the right ( 2 m) 2

12 PE for Constant Electric Field PE = Work done to move a charge through a distance = Force * distance CONSTANT ELECTRIC FIELD PE = qed electric potential energy charge E field distance Example How much potential energy is converted when an electron is accelerated through 0.25 m in a cathode ray tube (TV set) with an electric field strength of 2 x 10 5 N/C? PE = qed = ( ) ( )(0.25) = J

13 PE for Two Point Charges Potential Energy is force times distance constant charges PE = F e d = kq 1 q 2 r 2 r electric potential energy Potential energy is positive for like charges Potential energy is negative for opposite charges Potential energy is zero at infinite distance PE = kq 1 q 2 r distance Example How much electrostatic potential energy in a hydrogen atom, which consists of one electron at a distance of 5.3 x meters from the nucleus (proton). PE = kq 1 q 2 r = (9 109 )( )( ) = J

14 Potential Difference (Voltage) Electric is average energy per charge. Energy is a rela@ve quan@ty (absolute energy doesn t exist), so the change in electric poten@al, called poten@al difference, is meaningful. A good analogy: poten&al is to temperature, as poten&al energy is to heat. Poten@al difference is ocen called voltage. Voltage is only dangerous when a lot of energy is transferred. SI Units Voltage, like energy, is a scalar. A volt (v) is the unit for voltage named in honor of Alessandro Volta, inventor of the first ba9ery. 1 volt = 1 joule V = J 1 coulomb C source Potential = Energy Charge ΔV = ΔPE q voltage (V) common dry cell 1.5 car battery 12 household (US) 120 comb through hair 500 utility pole 4,400 transmission line 120,000 Van de Graaff 400,000 lightning 1,000,000,000

15 Potential Difference for Constant Electric Field Potential energy is often stored in a capacitor. Capacitors are made by putting an insulator in between two conductors. Most capacitors have constant electric fields. ΔV = ΔPE q Example = qed q ΔV = Ed voltage E field distance Calculate the magnitude of the electric field set up in a 2- millimeter wide capacitor connected to a 9-volt battery. V = Ed 9 = E(0.002) E = 4500 N/C

16 Potential Difference for Point Charge (Honors only) Consider a test charge to measure potential V = PE q 0 = kqq 0 / r q 0 potential difference constant charge ΔV = kq r distance Example 0.3 m -4 nc 0.4 m 10 nc 6 nc find V here ΔV 1 = kq 1 r = (9 109 )( ) 0.3 ΔV 2 = kq 2 r = (9 109 )( ) 0.4 = (9 109 )( ) 0.5 = 180 V = 90 V ΔV 3 = kq 3 = 180 V r ΔV = ΔV 1 + ΔV 2 + ΔV 3 = = 270 V

17 Summary of Electrostatic Equations Electrostatic Force F e = kq 1 q 2 r 2 force between two charges Electric Field E = F e q 0 definition E = kq r 2 for point charge Potential Energy PE = qed for constant E field PE = kq 1 q 2 r for two charges Potential Difference ΔV = ΔPE q definition ΔV = Ed for constant E field ΔV = kq r for point charge