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-. +. Chapter 20 Static Electricity. Charge by Conduction. + + +. -. +. + + +. + + +. Chapter 20 Static Electricity. Charge by Induction. Chapter 21 Electric Fields. Electric Field A property of space around a charged object that causes forces on other charged objects.
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- + Chapter 20Static Electricity Charge by Conduction
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Chapter 21Electric Fields • Electric Field A property of space around a charged object that causes forces on other charged objects. • Vector quantity It has both direction and magnitude • The direction of the force is away from the positive and towards the negative. • The electric field is the strongest when the lines are close together • Field lines do not exist-only a pictorial guide.
Chapter 21Electric Fields Van de Graaf Generator In the Van de Graaf generator, charge is transferred onto a moving belt A, and then onto the metal dome B, An electric motor does the work needed to increase the electric potential energy.
Chapter 21Electric Fields Van de Graaf Generator
Chapter 21Electric Fields Electric Field Intensity E Electric field intensity (N/C) F Force (Newtons) q’ Test Charge (Coulombs) Similar to Gravitational Field Intensity g Gravitational field intensity (N/kg) F Force (Newtons) m mass (kg)
Increase in gravitational potential energy Chapter 21Electric Fields + Increase in electric potential energy E g + -
Chapter 21Electric Fields The electric potential difference (V) is the work done in moving a test charge in an electric field divided by the magnitude of the test charge. Electric potential difference is measured in joules per coulomb. One joule per coulomb is a volt.
+ Chapter 21Electric Fields _ HIGH V _ + LOW V
+ + Chapter 21Electric Fields HIGH V + + LOW V
Chapter 21Electric Fields Electric potential difference in a uniform field And E = F/q so
Chapter 21Electric Fields A force of .032 N is required to move a charge of 4.2 x 10-5 C in an electric field between two points that are 25 cm apart. What potential difference exists between the points? V = 190 volts
Chapter 21Electric Fields An electron is accelerated by a machine that subjects it to a potential difference of 50 Megavolts. What energy has the electron acquired? W = (50 x 106V)(1.6 x 10-19C) = 8 x 10-12J
_ _ _ _ _ _ _ _ _ _ _ _ _ pump Chapter 21Electric Fields • Electric current The flow of electrons • Electric current can be maintained only if the electrons are • returned to areas of high electron concentration ++ ++ ++ _ _ _ _ _ _
+ + + + + + + F1 - F2 Chapter 21Electric Fields • Millikan’s oil drop experiment • Early 1900s – Determined electric charge • When the forces are balanced, F1 = F2 • Eq = mg so q = mg/E • Found that the charge is quantified • Multiples of 1.6 x 10-19C
+ + + + + + + F1 - F2 Chapter 21Electric Fields An oil drop has a mass of 1.9 x 10-16 kg and is suspended in an electric field with intensity of 6000 N/C. Find the charge on the drop and the number of excess electrons. F1= F2 so Eq = mg = 3.1 x 10-19C one extra electron
A B Chapter 21Electric Fields All systems are in equilibrium when the energy of the system is a minimum. The ball comes to rest when the potential energy is the least. It is the greatest at A and the least at B.
Chapter 21Electric Fields A charged sphere B neutral sphere A is the charged sphere with high potential energy. B is neutral with zero potential energy • - - • - - The potential of A decreases and the potential of B increases and both are at the same potential • - - • - -
Chapter 21Electric Fields What happens with a large sphere and a small sphere? • - • - • - • - • - • - • - • - • - High q Low q same V Low V High V same q
Chapter 21Electric Fields • Capacitor A device that stores a charge • As charge is added, the potential of the body increases. • For a given charge, the ratio of the charge to the potential • q/V is a constant. • Capacitance is the ability to store a charge. C Capacitance (farads) q Charge (Coulombs) V Potential (Volts)
Chapter 21Electric Fields A 3 x 102 pF capacitor has a potential difference of 30 volts across it. What is the charge on the capacitor? C = 3 x 102 pF = 3 x 10-10 F q = V = 30 volts q = CV = (3 x 10-10F)(30V) = 9 x 10-9C