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FIELDS

FIELDS.  V. Gradient of slope =.  x. Electrical field. +ve charge moving towards –ve charge. Higher potential. +. V. x. Lower Potential. Increases if V bigger and  x smaller. - Potential gradient = Electrical Field Strength. +. -. +. Non-uniform fields. Uniform field.

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FIELDS

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  1. FIELDS

  2. V Gradient of slope = x Electrical field +ve charge moving towards –ve charge Higher potential + V x Lower Potential Increases if V bigger and x smaller - Potential gradient = Electrical Field Strength

  3. + - + Non-uniform fields Uniform field Field Strength Line patterns (line to indicate direction of force felt by a +ve charge) - +

  4. Field strength lines Equipotential, 90 to field strength lines 800V 0V 200V 400V 600V Lines of equipotential – no work done on charge Equipotentials in a uniform field will be equally spaced

  5. Describing a uniform field Electric Force • Force between two charges • May be attractive (-ve) or repulsive (+ve) Symbol = F Units = N

  6. E = F q E = V d distance between charges (plates) Describing a uniform field Electric Field Strength • Force per unit +ve charge Important due to repulsive/attractive nature of force Symbol = E Unit = N C-1 or V m-1

  7. Describing a uniform field Electric Potential Energy • Energy of a +ve charged particle (q) due its position in an electric field Symbol = EPE Units = C V = J EPE = qV

  8. EPE = qV 10V + + 0V EPE = 0 Acceleration due to electric field  potential energy =  kinetic energy qV = ½mv2

  9. Important point: K.E. gained is the same for both particles Velocity is different due to difference in mass Electron velocity quickly approaches the speed of light - relativistic. Velocity equation only valid when v << c

  10. The electron volt (eV) A measure of energy qV = ½mv2 1 eV = Energy gained when an electron is subjected to a potential difference of 1V 1 eV = 1.6 x 10-19C x 1V = 1.6 x 10-19 J

  11. Millikan’s Oil Drop Experiment to measure charge on an electron

  12. Forces experienced by oil drop? + F (electric attraction) = qE - Oil drop (charge q) F (weight) = mg - If the oil drop hovers then… mg = qE

  13. The cunning bit... Add or remove electrons Then ionise oil drop (using radioactive source) Re-adjust the voltage (and therefore the size of E) to make oil drop hover again. Change in voltage proportional to charge on an electron

  14. Kepler’s Laws

  15. Kepler 2

  16. Kepler 3

  17. Apple and moon experiment by Newton Question - how does gravity extend into space? Newton calculated... Acceleration of Moon towards Earth Acceleration of an apple towards Earth

  18. Conclusion... Angular acceleration of moon towards earth (due to circular motion) Strength of earth’s gravity at 60RE = Gravity changes at a rate of inverse square of distance This extended gravitational force out into the universe - an amazing result (!)

  19. F = -Gm1m2 r2 Gravitational force, F F = gravitational force, N G = gravitational constant, 6.67 x 10-11 N m2 kg-2 m1 = mass of first object, kg m2 = mass of second object, kg r = distance between the two objects, m

  20. This is the gravitational force between the sun and earth Earth orbiting the sun

  21. 90 to velocity of object Needed for circular motion Gm1m2 m2v2 = r r2 Requirement for an object to orbit Fgravitational = Fcentripetal N.B. m1 = mass at centre of orbit, m2 = mass of satellite

  22. Very important Gm1m2 m242r = r2 T2 T2 42r3 = Gm1 v = 2r Remembering T (Kepler III)

  23. The satellite must be travelling fast enough for its orbit radius (Kepler III) • Not faster enough - orbit will collapse • Too fast - will overcome gravitational forces and escape

  24. Relative to the earth it doesn’t move T = Orbits N-S (over the poles), the earth rotates and so it looks at a different place each orbit. T = Types of orbit Geostationary 24 hours Polar 90 minutes

  25. Gravitational field strength, g - better known as gravity If there is a force there is an acceleration F = ma If the force is due to gravitational forces then acceleration is acceleration due to gravity F = mg

  26. F = -GMm r2 = -GM mr2 r2 g = F units N kg-1 m Or In words, gravitational force per unit mass acting at a point Know g = -GMm Then

  27. Remember: work = force x distance (in direction of force) = mg x h Uniform Field (near surface of Earth) = mgh Gravitational Potential Energy Stored ability to do work - something else has done work to get the object to that point

  28. mg(h + 3Δh) mg(h + 2Δh) mg(h + Δh) Δh mgh h h=0 Change in GPE

  29. Area = mg Δh = work done Δh h1 h2 Work done moving an object by Δh (near the earth’s surface, g  constant) Force (mg) /N Height /m

  30. V = mgh = gh Symbol = V m Unit = J kg-1 Gravitational potential ...“The work done to move unit mass from infinity to that point”... Don’t forgeth  r

  31. Gravitational potential is the total work, against the gravitational force, for 1kg to go from a point where g = 0 to the point in question where g = x N kg-1. g = 0 N kg-1 at r =  g = 9.8 N kg-1 at r = 6.4 x 106m

  32. Space calculations GPE = 0 GPE = - x A convention... Earth only calculations GPE = x GPE = 0

  33. r /m Δr 0 r    = m g dr r Force (mg) / N Work done moving an object from  to r (Δr) Area = work done N.B. g isn’t constant (non-uniform field)

  34.  g dr V = r    V = - GM dr = GM = - GM r r r2 r r Gravitational potential - work done on unit mass i.e. m = 1kg

  35. g = - dV dr Another interesting point...   g dr V = r Can be rearranged to ... The gradient of gravitational potential is gravitational field strength.

  36. F = - GMm Gravitational force r2 g = - GM Gravitational field strength r2 V = - GM Gravitational potential r

  37. Energy conservation Etotal = Ekinetic + Epotential mgh = Etotal - ½mv2 gh = constant - ½v2 V  -½v2

  38. r r -½v2 V The gradient of either graph is g

  39. Energy gained if falling into hole Escape velocity V = 0, r =  V = -62.5MJ kg-1 Energy required to get out of hole

  40. v   -2V V is -ve Stationary object (v = 0), at V = 0 ½mv2 + mV = ETotal  0 + 0 = 0 Nudge object into well, ETotal = 0 K.E. increases as P.E. become more -ve ½mv2 + mV = 0

  41. At Earth’s surface V = -62.5MJ kg-1, a 1kg mass will hit the ground at ~11km s-1 if nudged into well. Conversely... A 1kg mass launched at 11km s-1 will just make it to V = 0, the brim of the potential well. 11km s-1 = escape velocity

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