7: Fields & Their Consequences Flashcards
Concept of a Force Field
Region in which a body experiences a non-contact force
Force fields arise from ____
The interaction of mass, of static charge and between moving charges
Similarities & Differences between Gravitational & Electrostatic Forces (1:4, 1:1)
- Similarities:
• Both have inverse-square laws
• Use field lines
• Use of potential concept
• Equipotential surfaces - Differences:
• Masses always attract, but charges may attract or repel
Gravity
Universal attractive force between all matter
Magnitude of Force between Two Point Masses
F = G m₁ m₂ / r² where G is the gravitational constant
g
Force per unit mass as defined by g = F / m
Magnitude of g in Radial Field
g = G M / r²
Representation of Gravitational Field by Gravitational Field Lines
https://qph.fs.quoracdn.net/main-qimg-1e788e321185dc777fa96ced46727c30
Gravitational Potential
The work that needs to be done to move a unit mass from infinity to the point
The gravitational potential at an ____ from the mass will be ____
Infinite distance, zero
Gravitational Potential Difference
The energy needed to move a unit mass between two points with different gravitational potentials
Work Done in Moving Mass m
ΔW = m ΔV
Gravitational Equipotential Surfaces
They contain all the points with the same gravitational potential
No ____ when moving along ____
Work is done, an equipotential surface
V in Radial Field
V = - GM / r
Significance of Negative Sign in Gravitational Potential
Gravitational potential is 0 at ∞ so work needs to be done against the field to reach ∞
Graphical Representation of Variation of g with r
https://www.a-levelphysicstutor.com/images/fields/g-graph01.jpg
Graphical Representation of Variation of V with r
https://www.cyberphysics.co.uk/Q&A/KS5/gravitation/q5a.png
Relation of V & g
g = - ΔV / Δr
Area Under a Graph of g against r
ΔV
Orbital period and speed related to ____
Radius of circular orbit
Derivation of T² ∝ r³ (5)
- Force between two masses: F = G m₁ m₂ / r²
- Centripetal force: F = m₂ v² / r
- m₂ v² / r = G m₁ m₂ / r²
⇒ v = √(G m₁ / r) - v = 2 π r / T (distance / time)
- 2 π r / T = √(G m₁ / r)
⇒ T² / 4 π² r² = r / G m₁
⇒ T² = (4 π² / G m₁) r³
∴ T² ∝ r³
Escape Velocity
The minimum speed an object needs to leave a gravitational field and not fall back due to gravitational attraction
Escape Velocity Formula (3)
- Kinetic energy lost = Gravitational potential energy gained
- ½ m v² = G M m / r
- v = √(2 G M / r)
Energy Considerations for an Orbiting Satellite (3)
- An orbiting satellite has kinetic and potential energy – its total energy is always constant
- In a circular orbit, orbital speed and distance above the mass are constant so kinetic and potential energy are both constant
- In an elliptical orbit, a satellite will speed up as its orbital radius decreases so kinetic energy increases as potential energy decreases
Synchronous Orbit
When an orbiting object has an orbital period equal to the rotational period of the object it is orbiting
Geostationary Orbit (4)
- Geostationary satellites are always above the same point on Earth
- Their orbit is in the plane of the equator
- Their orbital period is the same as Earth’s rotational period (synchronous orbit and same angular speed)
- Given their orbital period is 24 hours, orbital radius can be calculated as ~42 000 km
Low Orbits (3)
- Low orbiting satellites are cheaper to launch and require less powerful transmitters so they’re useful for communications
- Imaging satellites have low orbits for imaging and monitoring weather
- As the planet and the satellite rotate at different angular speeds, the satellite doesn’t stay over the same point on Earth so it can cover the whole of the surface
Force between Point Charges in a Vacuum
F = (1 / 4 π ε₀) ((Q₁ Q₂) / r²)
Permittivity of Free Space
ε₀ is the opposition offered against the formation of an electric field in a vacuum
Comparison of Magnitude of Gravitational and Electrostatic Forces between Subatomic Particles
The distance between the particles is small, but the mass of the particles is also small so gravitational forces are small whereas electrostatic forces are big
Representation of Electric Fields by Electric Field Lines (2)
Uniform field: https://th.bing.com/th/id/OIP.cgxeo3AL1zJpHRVbgAeLzQHaFj
Radial field: https://i.ytimg.com/vi/c8H1j9LWjls/maxresdefault.jpg
Electric Field Strength
Force per unit charge
E
Force per unit charge defined by E = F / Q
Magnitude of E in a Uniform Field
E = V / d
Derivation from Work Done Moving Charge between Plates (3)
- E = F / Q and E = ΔV / d
- F / Q = ΔV / d
- F d = Q ΔV
Trajectory of Moving Charged Particle Entering a Uniform Electric Field Initially at Right Angles (3)
- A charged particle feels a constant force parallel to the electric field lines
- If it is positively / negatively charged, the force is the same / opposite direction to the field lines
- This causes the particle to accelerate at right angles to its original motion so it follows a parabolic path
Magnitude of E in Radial Field
E = (1 / 4 π ε₀) (Q / r²)
Absolute Electric Potential
The electric potential energy that a unit positive charge would have at that point in an electric field
Absolute Electric Potential is Zero ____
At an infinite distance from the charge creating the field
Electric Potential Difference
The energy needed to move a unit positive charge between two points in an electric field
Work Done in Moving Charge Q
ΔW = Q ΔV
Electric Equipotential Surfaces
They contain all the points with the same absolute electric potential
No ____ Moving Charge along an ____
Work is done, equipotential surface
Magnitude of Electric V in a Radial Field
V = (1 / 4 π ε₀) (Q / r)
Graphical Representation of Variation of E with r
https://entrancecorner.s3.amazonaws.com/media/uploads/2018/06/10/3321-a.PNG
Graphical Representation of Variation of E & V with r
https://www.shaalaa.com/images/_4:4e26160e991b4d31bc8a40237a4a8441.png
V Related to E
E = ΔV / Δr
ΔV is Area under Graph of ____
E against r
Definition of Capacitance
C = Q / V
Dielectric Action in a Capacitor
C = A ε₀ εᵣ / d
Relative Permittivity & Dielectric Constant
The ratio of the permittivity of a material to the permittivity of free space
Polar Molecules (2, 1:2)
- A dielectric is made up of lots of polar molecules (with a positive and negative end)
- When no charge is stored by a capacitor, no electric field is generated, so the molecules are aligned randomly
- When charge is applied to the plates of a capacitor, an electric field is generated between them:
• The negative ends of the molecules are attracted to the positively charged plate, and vice versa
• This causes the molecules to rotate and align themselves anti-parallel to the electric field generated between the plates
Area under Graph of Charge against PD
Energy stored
Capacitor Energy Stored Equation
E = 1/2 Q V = 1/2 C V² = 1/2 Q² / V
Graphs of Q, V, I against Time for Charging
https://www.scienceandmathsrevision.co.uk/wp-content/uploads/2019/05/cap-discharge_2-1024x339.jpg
Graphs of Q, V, I against Time for Discharging
https://www.scienceandmathsrevision.co.uk/wp-content/uploads/2019/05/cap-discharge_3-1024x339.jpg
Capacitor Discharge Equations
Q = Q₀ e^(-t / (R C)), V = V₀ e^(-t / (R C)), I = I₀ e^(-t / (R C))
Capacitor Charge Equations
Q = Q₀(1 - e^(-t / (R C))), V = V₀(1 - e^(-t / (R C))), I = I₀ e^(-t / (R C))
Time Constant
R C
Time to Halve
T_(1/2) = 0.69 R C
Force on a Current-Carrying Wire in a Magnetic Field
F = B I l when field is perpendicular to current
Fleming’s Left-Hand Rule (3)
- The first finger points in the direction of the uniform magnetic field
- The second finger points in the direction of conventional current
- The thumb points in the direction of the force
Magnetic Flux Density
B, the force on one metre of wire carrying a current of one amp at right angles to the magnetic field
Tesla
One tesla is the magnetic flux density that produces a force of one newton in a wire with length one metre with one amp
Required Practical 9
Investigation of the charge and discharge of capacitors
Required Practical 9: Charging Method (3)
https://www.inchcalculator.com/wp-content/uploads/2019/02/resistor-capacitor-circuit.png
1. Fully discharge the capacitor by connecting its two ends with a wire
2. Ensuring that it is broken, set up the circuit as shown in the diagram, connecting a data logger in parallel with the capacitor
3. Set the data logger to recording voltage, record the resistance of the resistor and complete the circuit
Required Practical 9: Discharging Method (3)
https://www.inchcalculator.com/wp-content/uploads/2019/02/resistor-capacitor-circuit.png
1. Charge the capacitor using a 6V battery
2. Ensuring it is broken, set up the circuit as shown in the diagram, connecting a data logger in parallel with the capacitor
3. Set the data logger to recording voltage, record the resitance of the resistor and complete the circuit
Required Practical 10
Investigate how the force on a wire varies with flux density, current and length of wire using a top pan balance
Required Practical 10 Method (4)
https://media.sciencephoto.com/image/c0021045/800wm/C0021045-Conductor_in_magnetic_field_experiment.jpg
1. Set up the apparatus as shown in the diagram with a variable resistor in series with the multimeter
2. Zero the balance, which the cradle is resting on
3. Turn on the power supply and record the reading on the balance and current
4. Repeat step 3, adjusting the variable resistor, number of magnets on the cradle or length of the wire in the cradle
Newton’s Law of Gravitation
The force of attraction between two point-masses is proportional to the product of the masses and inversely proportional to square of distance between them
Force on Charged Particles Moving in Magnetic Field
F = B Q v when the field is perpendicular to velocity
Direction of Force on Positive & Negative Charged Particles
To use Fleming’s left-hand rule for charged particles, use you middle finger as the direction of motion of a positive charge. If the particle carries a negative charge, point your finger the other way
Circular Path of Particles (5)
- By Fleming’s left-hand rule, the force on a moving charge travelling perpendicular to a magnetic field is always perpendicular to its direction of travel
- This satisfies the condition for circular motion
- Force due to magnetic field, F = B Q v, and centripetal force, F = m v² / r
- Radius of circular motion, r = m v / B Q, and frequency of rotation, f = B Q / (2 π m)
- So, increasing a particle’s velocity means its circular path has a larger radius, but it takes the same time to complete the path
Cyclotron (5)
- Cyclotrons are used to produce radioactive tracers or high-energy beams of radiation
- A cyclotron is made up of two hollow semi-circular electrodes with a uniform magnetic field applied perpendicular to the plane of the electrodes and an alternating pd applied between the electrodes
- Textbook Page 335
- Particles follow a semi-circular path in an electrode then, when they leave the electrode, they are accelerated in the gap. In the other electrode, their circular path has a greater radius as they are faster. When they leave this electrode, the pd has been reversed, repeating the process and causing the particles to spiral outwards increasingly fast before exiting the cyclotron
- Since the frequency of the motion doesn’t depend on radius, the particle spends the same amount of time in each electrode so the alternating pd has a fixed frequency
Magnetic Flux
Φ = B A where B is normal to A
Flux Linkage at an Angle (3)
- https://www.jobilize.com/ocw/mirror/col12074/m58477/CNX_UPhysics_30_03_FChanges.jpg
- The component of the magnetic field perpendicular to the area of the coil is B cos θ
- So, flux linkage, N Φ = B A N cos θ
Faraday’s Law
Induced emf is directly proportional to the rate of change of flux linkage
Magnitude of Induced EMF =
Rate of change of flux linkage
Magnitude of Induced EMF Equation
ε = N ΔΦ / Δt
Straight Conductor Moving in a Magnetic Field (4)
- Textbook Page 344
- Lenz’s law says that the induced emf will produce a force that opposes the motion of the conductor
- Fleming’s left-hand rule can be used to find the direction of the emf
- If the conductor has length l and moves with constant velocity v:
ε = N ΔΦ / Δt
= B ΔA / Δt
= B l Δd / Δt
= B l v
Electromagnetic Induction (3)
- If there is a relative motion between a conductor and magnetic field, the electrons in the conductor will experience a force and accumulate at one end
- This induces an emf across the rod
- If the rod is part of a complete circuit, a current will be induced
EMF Induced in a Coil Rotating Uniformly in a Magnetic Field
ε = B A N ω sin ω t
Alternating Currents & Voltages
Currents and voltages that vary sinusoidally with time
Root Mean Square Current & Voltage
The average current and voltage from an alternating power supply is the root mean square - dividing the peak value by √2
Alternating Current Equations
Iᵣₘₛ = I₀ / √2
Vᵣₘₛ = V₀ / √2
Peak-to-Peak Current & Voltage
Double the peak current / voltage or the distance between the peak and trough on the waveform
Mains Electricity Root Mean Square Voltage
230 V
Transformer Equation
Nₛ / Nₚ = Vₛ / Vₚ
Transformer Efficiency Equation
Iₛ Vₛ / (Iₚ Vₚ)
Eddy Currents (4)
- Eddy currents are looping currents induced by the changing magnetic flux in the core
- They create a field that acts against the field that induced them, reducing its strength
- They also dissipate energy by generating heat
- These effects can be reduced by having layers of the core separated by thin layers of insulator so a current can’t flow
Causes of Inefficiencies in a Transformer (3)
- Heat is generated by resistance in the coils. To reduce this, wires with low resistance can be used
- Energy is needed to magnetise the core and this is wasted as it heats the core. A magnetically soft material, that magnetises easily, should be used
- The coils should be as close as possible to increase the amount of magnetic flux (created by the primary coil) cut by the secondary coil
Transmission of Electrical Power at a High Voltage
Electrical power is transmitted at a high voltage, in the national grid, to produce a low current in the power lines so the power dissipated is reduced
