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5.1 2012
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5.1 2012






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5.1 2012 5.1 2012 Presentation Transcript

  •  If you want to move a charge closer to a charged sphere you have to push against the repulsive force You do work and the charge gains electric potential energy. If you let go of the charge it will move away from the sphere, losing electric potential energy, but gaining kinetic energy.
  •  When you move a charge in an electric field its potential energy changes. This is like moving a mass in a gravitational field.
  •  The electric potential V at any point in an electric field is the potential energy that each coulomb of positive charge would have if placed at that point in the field. The unit for electric potential is the joule per coulomb (J C-1), or the volt (V). Like gravitational potential it is a scalar quantity.
  •  In the next figure, a charge +q moves between points A and B through a distance x in a uniform electric field. The positive plate has a high potential and the negative plate a low potential. Positive charges of their own accord, move from a place of high electric potential to a place of low electric potential. Electrons move the other way, from low potential to high potential.
  •  In moving from point A to point B in the diagram, the positive charge +q is moving from a low electric potential to a high electric potential. The electric potential is therefore different at both points.
  •  In order to move a charge from point A to point B, a force must be applied to the charge equal to qE (F = qE). Since the force is applied through a distance x, then work has to be done to move the charge, and there is an electric potential difference between the two points. Remember that the work done is equivalent to the energy gained or lost in moving the charge through the electric field.
  •  Potential difference We often need to know the difference in potential between two points in an electric field The potential difference or p.d. is the energy transferred when one coulomb of charge passes from one point to the other point.
  •  The diagram shows some values of the electric potential at points in the electric field of a positively-charged sphere What is the p.d. between points A and B in the diagram?
  •  When one coulomb moves from A to B it gains 15 J of energy. If 2 C move from A to B then 30 J of energy are transferred. In fact:
  •  Energy transferred, This could be equal to the amount of electric potential energy gained or to the amount of kinetic energy gainedW =charge, q x p.d.., V(joules) (coulombs) (volts)
  •  One electron volt (1 eV) is defined as the energy acquired by an electron as a result of moving through a potential difference of one volt. Since W = q x V And the charge on an electron or proton is 1.6 x 10-19C Then W = 1.6 x 10-19C x 1V W = 1.6 x 10-19 J Therefore 1 eV = 1.6 x 10-19 J
  •  A copper wire consists of millions of copper atoms. Most of the electrons are held tightly to their atoms, but each copper atom has one or two electrons which are loosely held. Since the electrons are negatively charged, an atom that loses an electron is left with a positive charge and is called an ion.
  •  The diagram shows that the copper wire is made up of a lattice of positive ions, surrounded by free electrons: The ions can only vibrate about their fixed positions, but the electrons are free to move randomly from one ion to another through the lattice. All metals have a structure like this.
  •  The free electrons are repelled by the negative terminal and attracted to the positive one. They still have a random movement, but in addition they all now move slowly in the same direction through the wire with a steady drift velocity. We now have a flow of charge - we have electric current.
  •  Current is measured in amperes (A) using an ammeter. The ampere is a fundamental unit. The ammeter is placed in the circuit so that the electrons pass through it. Therefore it is placed in series. The more electrons that pass through the ammeter in one second, the higher the current reading in amps.
  •  1 amp is a flow of about 6 x 1018 electrons in each second! The electron is too small to be used as the basic unit of charge, so instead we use a much bigger unit called the coulomb (C). The charge on 1 electron is only 1.6 x 10-19 C.
  •  In fact: Or I = Δq/ Δt Current is the rate of flow of charge
  •  Which way do the electrons move? › At first, scientists thought that a current was made up of positive charges moving from positive to negative. › We now know that electrons really flow the opposite way, but unfortunately the convention has stuck. › Diagrams usually show the direction of `conventional current going from positive to negative, but you must remember that the electrons are really flowing the opposite way.
  •  Why do models change as knowledge changes (current vs electron flow)?
  •  A tungsten filament lamp has a high resistance, but connecting wires have a low resistance. What does this mean? The greater the resistance of a component, the more difficult it is for charge to flow through it.
  •  The electrons make many collisions with the tungsten ions as they move through the filament. But the electrons move more easily through the copper connecting wires because they make fewer collisions with the copper ions.
  •  Resistance is measured in ohms (Ω) and is defined in the following way: › The resistance of a conductor is the ratio of the p.d. applied across it, to the current passing through it. In fact:
  •  Resistors are components that are made to have a certain resistance. They can be made of a length of nichrome wire.
  •  Resistance depends on › Temperature › Material of conductor › Length › Cross-sectional areaTemperature The resistance of a metallic conductor increases as the temperature increases e.g. copper The resistance of a semiconductor/insulator decreases as the temperature increases e.g. thermistor.
  • Factors affecting Resistance Length Resistance of a uniform conductor is directly proportional to its length. i.e. R L Cross-sectional area Resistance of a uniform conductor is inversely proportional to its cross- sectional area. i.e. R 1 A
  •  Material › The material also affects the resistance of a conductor by a fixed amount for different materials. This is known as resistivity ( ). › R= L = constant of proportionality A Unit: ohm meter m › = R d2 (For a wire with circular cross-sectional area) 4L
  • The resistance of a conductor is directlyproportional to the length since thecurrent needs to pass through all theatoms in the length.The resistance is inversely proportional tothe cross-sectional area since there ismore room for the current to passthrough.The above observations can be combinedand the resistance, R of the conductor is: L R . A
  • Resistivity is an inherent property of amaterial, inherent in the same sense thatdensity is an inherent property.
  •  The current through a metal wire is directly proportional to the p.d. across it (providing the temperature remains constant). This is Ohms law. Materials that obey Ohms law are called ohmic conductors.
  •  When X is a metal resistance wire the graph is a straight line passing through the origin: (if the temperature is constant) This shows that: I is directly proportional to V. If you double the voltage, the current is doubled and so the value of V/I is always the same. Since resistance R =V/I, the wire has a constant resistance. The gradient is the resistance on a V against I graph, and 1/resistance in a I against V graph.
  •  Doubling the voltage produces less than double the current. This means that the value of V/I rises as the current increases. As the current increases, the metal filament gets hotter and the resistance of the lamp rises.
  •  The graphs for the wire and the lamp are symmetrical. The current-voltage characteristic looks the same, regardless of the direction of the current. TOK How can we be sure about relationships between measured quantities?
  •  P= W/t W = qV q = It or t = q/I Therefore P = qV/(q/I) P = VI
  •  Using V = IR & P = VI P = IR x I P = I 2R Or as I = V/R P = V x V/R P = V2/R P = VI = I2R = V2/R