A summary of Module 6: Particles and Medical Physics of the OCR AS/A-Level Physics A (H156/H556) course. Comprehensive set of notes made throughout the teaching/learning, re-learning, revision and exam periods of the course. After revisiting this part of the course and devising these notes, this st...
OCR Physics A, Module 5 : Newtonian World and Astrophysics (H156/H556) notes/summary
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Module 6 : Particles and Medical Physics
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Module 6 : Particles and Medical Physics
Module 6 – Particles and Medical Physics
Capacitors (6.1)
Capacitors (6.1.1)
Capacitance
• Ratio of change in electric charge to corresponding change in potential difference
• Charge stored per unit p.d. across a capacitor
• Capacitor: device that can store charge: two metal plates separated by dielectric (insulator)
• Symbol: C
• Unit: F (Farads)
𝑸
• 𝑪 = 𝑽 so: 1 F = 1 CV-1
• 𝑸 ∝ 𝑽, with constant of proportionality = C
• How a capacitor retains its charge / becomes charged:
o Electron flow from negative to positive, opposite to conventional current
o So, e– (and other negatively-charged particles) deposited onto first plate
o This means that this plate becomes negatively charged
o e– (and other negatively-charged particles) removed from second plate
o This means that this plate becomes positively charged
o The attraction between the positive and negative plates causes an electric field
Total capacitance in series
• 1/CT = 1/C1 + 1/C2 + 1/C3 + … + 1/Cn
𝟏
• ∴ total capacitance = ∑𝒏𝟏 𝑪
𝒏
• Derivation (not required but useful to help understand):
o Kirchhoff’s 2nd law => V1 + V2 = VT and Q1 = Q2 = QT where Q1 is charge on capacitor
#1 with p.d. V1 and capacitance C1
o ∴ if two capacitors placed in series, then their total p.d. is V1 + V2
o Q = CV so V = Q/C so as VT = V1 + V2, VT = Q1/C1 + Q2/C2 = QT/CT
o As Q1 = Q2 = QT the whole equation can be divided by this so 1/C1 + 1/C2 = 1/CT
o Similar derivation for parallel
Total capacitance in parallel
• CT = C1 + C2 + C3 + … + Cn
• ∴ total capacitance = ∑𝒏𝟏 𝑪𝒏
Series circuits
• Capacitor and resistor connected in series with voltmeter connected to capacitor in parallel
• Disconnect wire at capacitor which connected to voltmeter at t = 0
• V0 (initial p.d.) is 0 and capacitor will charge
• V-t or Q-t graphs:
o Begin at origin and increase at a decreasing rate, trending at the end
o As time increases, graph continues to have the same y-value, equal to e.m.f.
o This is not an asymptote, but rather theoretical
o It is an asymptote
Parallel circuits
• Capacitor, resistor and voltmeter all connected in parallel
• Disconnect wire at positive terminal when t = 0, DON’T just turn off the power
• V0 (initial p.d.) is maximum (v. close to e.m.f.) and capacitor will discharge
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, Module 6 : Particles and Medical Physics
Energy (6.1.2)
Energy stored by capacitor
• V-Q graphs for a capacitor linear and go through the origin
• Q = CV so when capacitance is constant Q α V and gradient is 1/C
• Energy stored by capacitor is area under graph
𝟏 𝟏 𝑸𝟐 𝟏
• As area under graph is a triangle as the graph is linear, 𝑾 = 𝑸𝑽 = = 𝑽𝟐 𝑪
𝟐 𝟐 𝑪 𝟐
Uses of capacitors
• Capacitors can be used as a storage of energy in devices such as mobile phones and laptops
Charging and discharging capacitors (6.1.3)
Charging capacitors
• Series circuit used like one explained above in 6.1.1
• Reason for charge:
o When t = 0, p.d. on capacitor, Vc = 0, so ε = IR + 0 = IR => I is large
o t increases, Vc increases, so Vc > 0 and ε = IR + Vc at decreasing rate, so I is decreasing
o Theoretically, when graph stops increasing, gradient stops increasing therefore rate
of flow charge is 0 so ε = Vc and I = 0
• V-t or Q-t or I-t graph:
o Begin at origin and increase at a decreasing rate, trending at the end
o As time increases, y-value diverges to a limit equal to e.m.f.
o Therefore, y = e.m.f. is an asymptote of this graph
o Gradient is decreasing and not constant due to the current decreasing and it also
being the rate of flow of charge, so charge increases at a decreasing rate; similar to
explanation below for discharging a capacitor
𝒕 𝒕 𝒕
• 𝑽 = 𝑽𝟎 (𝟏 − 𝒆−𝑹𝑪 ) ; 𝑸 = 𝑸𝟎 (𝟏 − 𝒆−𝑹𝑪 ) ; 𝑰 = 𝑰𝟎 (𝟏 − 𝒆−𝑹𝑪 ) for capacitor-resistor circuits
• As t increases in equal intervals, charge increases by same proportion each time
• This time interval is known as the time constant, τ, and 𝝉 = 𝑹𝑪
• Time constant: produce of capacitance and resistance for capacitor-resistor circuit
• If time constant inputted into one of the above equations, then we have a special case:
𝑹𝑪
𝑽𝟎
o 𝑽 = 𝑽𝟎 (𝟏 − 𝒆−𝑹𝑪 ) = 𝑽𝟎 (𝟏 − 𝒆−𝟏 ) = 𝑽𝟎 − 𝒆
o If you want to calculate 2 time constants after t = 0, then replace t with 2RC and
𝑽
after simplification you will get 𝑽 = 𝑽𝟎 − 𝟐𝟎
𝒆
𝑽
o Therefore, after x number of time constants, 𝑽 = 𝑽𝟎 − 𝒆𝒙𝟎
Discharging capacitors
• Parallel circuit like one explained above in 6.1.1
• Reason for discharge:
o Electrons flow from negative terminal of power supply one of the capacitor plates
▪ Capacitor must always be placed in this way otherwise it will be damaged
o This plate becomes negatively charged, so electrons begin to repel plate
o Hence, they travel back on themselves and now go towards resistor
o Travel through resistor and go towards other plate, which is now positively charged
o Over time, first plate loses negative charge and second plate loses positive charge
o As the plates become more neutral, p.d. between plates decreases
o Hence plate discharges, and when plates are neutral, p.d. = 0, hence Q = 0
o Therefore, capacitor discharges over time
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, Module 6 : Particles and Medical Physics
• V-t or Q-t or I-t graph:
o Exponential decay with y-intercept being V0 / Q0 / I0 where V0 = e.m.f.
o x-axis is asymptote (as 1/ex can never equal 0) so always V/Q/I reading, but v. small
o Why the gradient is decreasing and not constant:
▪ Initially there is a large current due to large p.d. across metal plates
▪ Capacitor is discharging, so current drops, and p.d. drops too as V = IR
▪ Charge drops quickly at first due to large current
▪ Q = It and so I is the rate of flow of charge
▪ As the rate of flow of charge decreases, charge decreases at decreasing rate
▪ Q = CV so p.d. across plates directly proportional to charge on plates
▪ Therefore, V-t graph and Q-t graph have the same shape
▪ NB: I-t graphs also have the same shape in this context
𝒕 𝒕 𝒕
• 𝑽 = 𝑽𝟎 𝒆−𝑹𝑪 and 𝑸 = 𝑸𝟎 𝒆−𝑹𝑪 and 𝑰 = 𝑰𝟎 𝒆−𝑹𝑪 for capacitor-resistor circuits
• Rate of removal of charge is proportional to the amount of charge remaining
• As t increases in τ, charge decreases by same proportion each time (≈ by 0.37 of initial V/Q/I)
• If time constant inputted into one of the above equations, then we have a special case:
𝑹𝑪
𝑽
o 𝑽 = 𝑽𝟎 𝒆−𝑹𝑪 = 𝑽𝟎 𝒆−𝟏 = 𝒆𝟎
o If you want to calculate 2 time constants after t = 0, then replace t with 2RC and
𝑽
after simplification you will get 𝑽 = 𝟐𝟎
𝒆
𝑽𝟎
o Therefore, after x number of time constants, 𝑽 = 𝒆𝒙
• Finding resistance or capacitance experimentally:
𝒕
o 𝑽 = 𝑽𝟎 𝒆−𝑹𝑪
𝒕
𝒕
o 𝐥𝐧 𝑽 = 𝐥𝐧(𝑽𝟎 𝒆−𝑹𝑪 ) = 𝐥𝐧 𝑽𝟎 − 𝑹𝑪
o lnV-t graph:
▪ y-intercept = lnV0 so V0 = ey-intercept
▪ gradient = –1/RC
• rearrange this equation to find missing variable
• accurate measurement of gradient can be found by inputting data of
V and t into spreadsheet then working out lnV and plotting lnV by t
Electric fields (6.2)
Point and spherical charges (6.2.1)
Electric field
• Region created by charged objects where object with charge experiences a force
• Always from positive to negative
• Examples of electric fields:
o Electrons and protons both charged so create electric fields and affect each other
o Positive and negative plates of capacitor create electric field between them
• Electric field lines (always drawn from positive to negative)
o Up to the edge of plates
▪ Electric field uniform so lines parallel to each other, perpendicular to plate
o At edge of plates
▪ Electric field not uniform here as horizontal distance from plates increases
▪ As you go further out, electric field weakens
▪ At edge and beyond, lines curve outwards more and more
• A uniformly charged sphere can be modelled as a point charge at its centre
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