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Physics paper 2 > fields and their concequences > Flashcards

fields and their concequences Flashcards

1
Q

Newton’s Law of Gravity

A

there exists an attractive force between two point masses which is proportional to the product of their masses and inversely proportional to their separation squared

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2
Q

concept of a force field

A

the region in which a body experiences a force

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3
Q

representation of gravitational field lines (radial and uniform fields)

A

field lines show direction of force on a point mass in a gravitational field - vectors.
Uniform field - straight lines of equal distance apart
radial field - straight lines point inwards - as lines get further apart the field becomes weaker

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4
Q

equipotential

A
  • lines joining points of equal potential
  • no work is done moving an object through a line of equipotential
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5
Q

gravitational field strength

A

the force acting per unit mass on a mass in a gravitational field
radial field - g is inversely proportional to r²
uniform field - g is inversely proportional to r

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6
Q

Gravitational field strength units

A

Nkg-1 (vector)

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7
Q

gravitational potential

A

work done per unit mass in moving from a small mass from infinity to that point

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8
Q

units of gravitational potential

A

Jkg-1 (scalar)

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9
Q

gravitational potential at infinity

A

zero

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10
Q

gravitational potential difference between two points

A

work done per unit mass in moving a small mass from one point to the other

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11
Q

graph of g against r

A

curve that reaches a maximum at the surface of an object

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12
Q

area under graph of gravitational field strength against r

A

work done moving a unit mass between the two points

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13
Q

graphical variation of V with r

A

r shaped curve

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14
Q

gradient of graph of gravitational potential against r

A

gradient = -g
g = - ∆V/ ∆r

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15
Q

derivation of Kepler’s Law

A
  1. Gravitational force = centripetal force
  2. GMm/r² = mv²/r or GMm/r² = mrω²
  3. GM/r = v² where v=2πr/t
  4. re-arrange to get T² = 4π²/GM x r³
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16
Q

energy considerations of an orbiting satellite

A
  • total satellite energy = KE + GPE = 0.5mv²
  • GMm/r = GMm/2R -GMm/r
    Circular orbit - KE and GPE are both constant.
    Elliptical orbit - KE and GPE increase and decrease inversely with each other in cycles
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17
Q

features of a geosynchronous orbit- geostationary satellite

A
  1. Orbits over the equator
  2. period is 24 hours so maintains a fixed position relative to the surface of Earth
  3. offers uninterrupted communication between transmitter and receiver without steering required
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18
Q

low orbit satellites

A
  • 180 - 2000 km above Earth
  • cheaper to launch and require less powerful transmitters
  • high speed and close proximity so need several working together- each can scan whole earth
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19
Q

Escape velocity of an object from a planet

A

loss of KE = gain in GPE
1/2mv²= GMm/r ∴ v = √2GM/r

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20
Q

field line rules

A
  • start at surface/ point leaving surface at 90 degrees
  • arrow shows direction of force lines do not cross
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21
Q

gravitational field strength and density

A

gravitational field strength is proportional to density

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22
Q

coulomb’s Law

A

-there exists a force between two point charges that is proportional to the product of their charges and inversely proportional to their separation squared
- Attractive with un-like charges (negative)
- Repulsive with like charges (positive)

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23
Q

representations of electric field lines

A
  • field lines show direction of force on a point positive charge in an electric field
  • lines always start/finish on a surface/charge, do not cross and leave a surface at 90 degrees
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24
Q

electric field strength

A

force acting per unit positive charge on a positive charge when placed in an electric field (vector)

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25
electric field strength units
NC-1 or Vm-1
26
electric potential
work done per unit charge in moving a small positive charge from infinity to point in field (scalar)
27
units of electric potential
JC-1 or V
28
electric potential at infinity
zero
29
sign of V
repulsive force = positive as Q is positive attractive force = negative as Q is negative
30
electrical potential difference
work done per unit positive charge in moving a small positive charge between two points in a field (scalar)
31
energy needed to move a charge across a p.d
W=Q∆V Derivation for uniform field: E = F/Q = ∆V/d ∴ fd=Q∆V so W=Q∆V
32
graphs of E against r
- radial field around positively charged hollow sphere - uniform fields
33
graphs of V against r
- radial field around positively charges hollow sphere - uniform field
34
gradient of tangent to graph of V against r
electric field strength at point
35
area under graph of E against r
electric potential difference
36
relationship between E and r for a uniform field
E is inversely proportional to r
37
relationship between E and r for a radial field
E is inversely proportional to r²
38
path of charged particle in a uniform electric field
- path is parabolic because magnitude of force is constant and always in the same direction - Positive charge: accelerates in same direction as field lines at right angles to original motion following a parabola. Negative charge: accelerates in opposite direction to field lines at right angles to original motion following a parabola.
39
Path of charged particle in a uniform electric field in 3D
Same parabolic motion as in 2D
40
speed of charged particle accelerated across a potential difference
loss of electric potential energy = gain in kinetic energy
41
equipotentials
flat planes (uniform field) or spherical surfaces (radial field) joining points of equal potential. No work done when travelling along a lone of equipotential because ∆V=0
42
similarities between electric and gravitational fields
- field strengths are both inversely proportional to separation squared - potentials are both inversely proportional to separation - field strengths for radial fields are inversely proportional to r² and inversely proportional to r for uniform fields
43
differences between electric and gravitational fields
masses always attract whereas charges may attract or repel
44
accelerating charges
loss of electrical potential energy = gain in kinetic energy
45
shuttling ball forces
T in string, force - along field line, weight down
46
motion of electron entering uniform field in same direction as field lines
Force acts against initial velocity so electron decelerates
47
capacitance definition
charge stored per unit potential difference
48
capacitance process
when a capacitor is connected to a power source, positive and negative charge builds up on opposite plates due to the flow of electrons in the circuit, creating a uniform field. As electrons cannot flow across the capacitor, they instead go the other way opposing the initial current so when capacitor is fully charges, I = 0
49
permittivity
how difficult it is to generate an electric field in a material
50
relative permittivity
the ratio of the permittivity of a material to the permittivity of free space
51
how does permittivity affect capacitance
when a charge is applied to a capacitor and thus its dielectric, the polar molecules in the dielectric rotate according to the charge of each plate. this alignment opposes the applied electric field of the capacitor thus reducing the overall electric field lowering the p.d needed to charge the capacitor which increases the capacitance
52
constant charging current
constant I achieved by constantly decreasing R
53
capacitance graph: Q against V
- directly proportional - capacitance = gradient - area under line = energy stored by capacitor
54
derivation of E = 1/2 QV
1. ∆W=Q∆V -work done moving charge= energy stored on capacitor - as more charge is transferred, pd across plates ↑ so more energy gained. Total energy ≠ final QV as Q and V are changing throughout. 2. energy stored=total work done in charging= charge x average voltage - if 'charged' from 0 to V, the average pd across plates = o.5V, as p.d ↑ linearly with charge. 3. energy stored =work done= 1/2QV OR VαQ ∴ as V and Q are both changing and E=VQ, E=average Q x average V =0.5QV
55
energy stored/voltage graphs
E = 1/2 CV² Energy α V²
56
time constant definition
1) Discharging = Time taken for voltage /charge /current to fall to 1/e (0.37) of its initial value. 2) Charging = time taken for charge/voltage to increase to (1- 1/e) of max value (0.63) Time constant = RC = Resistance x current
57
finding time constant from a graph of V against t
1. calculate initial voltage 2. locate value on graph of 0.37 3. time taken = RC
58
finding time constant from a graph of ln(V) against time
1. ln[V] - ln [Vo] = - 1/RC 2. ln[V] = -1/RC t + ln[Vo] 3. in form y=mx+c : RC = -1/gradient
59
graphical representations of charging a capacitor against time
V = V0 (1 - e^-t/RC)
60
graphical representation of discharging a capacitor against time
V = V0 (1 - e^-t/RC)
61
time taken to halve
T1/2 = 0.69RC = ln2 RC by calculation

Physics paper 2 (3 decks)

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