Capacitors
graphs of V, Q and I for charging a capacitor and why
(I-t): exponential decay (because current decreases as charge and voltage of capacitor increases so there's more resistance. Decreases from I₀ and tends towards 0) (Q-t): exponential growth (because charge is increasing when charged. Tends towards Q₀) (V-t): exponential growth (because V is proportional to Q, until voltage of capacitor V = voltage of charging source V₀. Tends towards V₀) -when voltage of capacitor is max, no more charge can flow, so current is 0 current = gradient of Q-t graph
graph of capacitor charging at constant current (I-t) and (Q-t)
(I-t): rectangle where area = Q (Q-t): straight line where gradient = I Same properties for curved graphs
Why does your hair stand up when you are about to be struck by lightening
-a potential difference builds between the body and the cloud as the cloud builds charge -since hair is an insulator, it acts as a dielectric (negative electrons move slightly to positive ground, positive ions attracted to negative cloud) -the electric field induced in the hair causes the strands to repel each other and stand up
definition of capacitance
-amount of charge a capacitor can store per unit P.D across it C = Q/V -constant for a capacitor (units = F (Farads) = 1CV⁻¹)
why is a capacitor not a suitable source for powering a cordless telephone
-capacitor would be impossibly large to fit in phone -capacitor would need recharging very frequently/capacitor could only power the phone for a short time -capacitor voltage/current supplied would fall continuously while in use
A parallel-plate capacitor is fully charged and left connected to the power supply. A dielectric is then inserted between the plates. What happens to the charge stored, electric field strength, and energy stored?
-charge increases -electric field strength stays the same (E = V/d and V remains the same, so does d) -energy stored increases (because C increases)
effect on voltage of doubling charge of capacitor
-charge is proportional to voltage (Q = CV) -so doubling charge of capacitor will double voltage
A parallel-plate capacitor is fully charged and then disconnected from the power supply. A dielectric is then inserted between the plates. What happens to the charge stored, electric field strength, and energy stored?
-charge stays the same -electric field strength decreases (because original field from charges is slightly cancelled out from polarised dielectric which creates its own field in opposite direction Q = CV since Q is the same and C has increased, V must decrease, so field strength decreases) -energy stored decreases (because E = 1/2CV², C increases but V decreases by the same factor, so energy stored decreases)
1 farad per metre
-charge stored per unit potential difference per unit distance between two parallel plates -for plates with a spacing of 1m at 1V, Q = 8.85x10⁻¹² C (relative permittivity)
Explain why the rate of change of pd between capacitor plates decreases as the capacitor discharges
-current decreases as it discharges -so charge is lost more slowly so P.D falls more slowly -because Q = CV and C is constant so V ∝ Q
what is a capacitor
-device designed to store charge -two parallel metal plated (separated by a dielectric for higher performance) -charges held in place by electric field
discharging capacitor through fixed resistor - what happens to current
-discharge current starts at max and decreases to 0 -because potential difference across capacitor decreases as capacitor is discharged -so current = PD/fixed resistor resistance
why is only 50% of energy supplied by battery stored in capacitor
-during charging, battery forces charge Q through pd V round the circuit -so it transfers energy QV to the circuit -half of this energy is stored in the capacitor (E = 1/2QV) -the other half is wasted due to resistance in circuit, transferred to surroundings when charge flows through circuit -resistance is due to charge of capacitor's plates repelling electrons
electric field strength and energy stored in a thundercloud
-electric field strength of E = V/d -for a thundercloud with constant charge, energy stored = 1/2QV or 1/2QEd -when winds force cloud to rise by Δd, energy increases by 1/2QEΔd -because work is done to overcome electrical attraction of cloud and ground by force of wind
Dielectric
-electrically insulating material that becomes polarized in an electric field -may contain polar molecules which rotate to become polarised
what happens when a capacitor is charged by a battery
-electrons move through battery from one plate to other -battery forces electrons from plate connected to positive of battery to plate connected to negative of battery -each plate gain an equal and opposite charge -plate losing electrons becomes positively charged, and plate gaining electrons becomes negatively charged -so a Potential Difference builds
why do 3 capacitors in series get charged the same irrespective of their capacitance? And how does the source voltage get divided between them?
-electrons transferred through battery from plate on one end to plate on other end of series capacitors (the ones connected to the cell) -electrons are transferred from on side to the other, causing electrons on plate facing plate gaining electrons to be repelled to other plate of connected capacitor, leaving a positive charge on the plate they have left -so the charge is effectively shifted along from one plate to another, so the amount of charge stored in each capacitor is the same -since the charge is the same in each one, but they have different capacitances, the voltage is divided between each of the capacitors (because V = Q/C) -the voltage across each capacitor must sum to match the source voltage -the capacitors in series equation creates a theoretical single capacitor that performs the same as all of the others in series. The charge stored in that capacitor can be calculated, which is the same in each of the actual capacitors
finding time constant from V-t discharge graph
-find initial voltage -find 37% of initial voltage -find time matched with this voltage -this time is the time constant (because time constant = time taken by a discharging capacitor to decrease to 37% of the initial voltage, current, or charge)
A 2F, 3F and 5F capacitor are connected in series, along with a 50v battery. Find the potential difference across the 2F capacitor, and the energy stored in it.
-find total capacitance: (1/2 + 1/3 + 1/5)⁻¹ = 0.967 F -find charge using this: 50 x 0.967 = 48.39 C -this is the charged stored in each capacitor -find the voltage across the 2F capacitor: 48.39/2 = 24.20 V -find the energy stored: 1/2 CV² 1/2 (2)(24.2)² = 388.8 J
why is the discharge curve for a capacitor exponential (prove mathematically)
-for a small decrease in charge: ΔQ = -IΔt (negative because charge decreases over time) -substitute I = Q/RC in I position: ΔQ = -QΔt/RC -rearrange ΔQ/Δt = -Q/RC -for small intervals of time (Δt -> 0): dQ/dt = -Q/RC -so the rate of change of charge is proportional to the charge -which means it is an exponential
effect of increasing R on graphs of discharging capacitor
-if R is increased, time constant increases -so the discharge takes more time, so the graph is stretched longer parallel to the x-axis
The flash tube in a camera produces a flash of light when a 180 μF capacitor is discharged across the tube. The capacitor in the circuit in the diagram above is replaced by a capacitor of greater capacitance. Discuss the effect of this change on the photograph image of a moving object.
-image would be less sharp (or blurred) because the discharge would last longer and the image would be photographed as it is moving -image would be brighter because the capacitor stores more energy and therefore produces more light
A 200μF capacitor is in series with a 2200μF capacitor and they are charged until the 200μF capacitor stores 30μC. What is the charge on the other capacitor? Calculate the P.D across each capacitor
-in the series circuit, the two capacitors' plates that are connected together are isolated from the power supply, so they will have the same charge (in equilibrium) -this means the charge on both capacitors in series is identical -since we know the charge on each one, P.D can be calculated for each one 30/200 = 0.15 V 30/2200 = 0.014 V
how to achieve the largest possible capacitance
-make area A as large as possible -make plate spacing as small as possible -fill gap between plates with dielectric which has a relative permittivity as large as possible (easily achieved by rolling two strips of aluminium foil separated by layer of dielectric)
permittivity
-measure of how difficult it is to generate an electric field in a medium -increased permittivity = increased charge needed to create an electric field permittivity of free space (ϵ₀ = 8.85x10⁻¹² Fm⁻¹)
A capacitor is made from two parallel plates with a polythene sheet between them. The capacitor is charged with a battery. When the polythene sheet is pulled out from between the plates, there is an increase in the energy stored by the capacitor. Explain why this happens.
-polar dielectric molecules align with positive charged end towards negative plate -so work is done on capacitor separating charged surface from negative plate which it is attracted to
how are dielectrics with polar molecules polarised
-polar molecules are already polarised, but are randomly orientated when dielectric is uncharged -molecules turn when placed between charged plates so negative side of molecule aligns with positive plate and vice versa for other side -so surface of dielectric facing positive plate gains negative charge and surface facing negative plate gains positive charge
relative permittivity of dielectric substance (ϵᵣ) (dielectric constant)
-ratio of charge stored by parallel-plate capacitor with dielectric to charge without dielectric between plates at same P.D ϵᵣ = 1 with no dielectric ϵᵣ = Q/Q₀ Q = charge stored by capacitor when space between plates is completely filled with a dielectric Q₀ = charge stored at same P.D when space is completely empty ϵᵣ = ϵ₁/ϵ₀ ϵ₁ = permittivity of material 1 ϵ₀ = permittivity of free space ϵᵣ = C/C₀ C = capacitance of capacitor when space between plates is completely filled by dielectric C₀ = capacitance of capacitor when space is completely empty
graphs of current and charge against time
-same shape curve (exponential decay from starting value) -because they are proportional, as: current = PD/fixed resistor resistance since V = Q/C: I = Q/RC
where are capacitors used
-smoothing circuits to smooth unwanted variations in voltage -back-up power supplies -timing circuits (capacitor discharge through fixed resistor) -filter circuits (to remove unwanted frequencies)
A 470μF capacitor is charged using a 10V battery. It is then disconnected from the battery, and connected to an uncharged 220μF capacitor. Calculate the voltage across the capacitors once the current has stopped flowing
-the capacitance can be calculated as if they are in parallel -voltage is the same at all points in a parallel circuit, so when the battery is disconnected, the two capacitors will gain the same voltage -since charge is always conserved: initial charge stored in 470μF capacitor = 470 x 10 = 4700 -find total capacitance of both capacitors: 470 + 220 = 690μF -calculate the potential difference: 4700 / 690 = 6.8 V
half life of a charged capacitor
-time taken for charge of capacitor to half: Q = Q₀e⁻ᵗ/ᴿᶜ so: Q₀/2 = Q₀e⁻ᵗ/ᴿᶜ 1/2 = e⁻ᵗ/ᴿᶜ 2⁻¹ = e⁻ᵗ/ᴿᶜ -ln2 = -t/RC (half life) t = RCln2
time constant (𝜏) (symbol = tau)
-time taken to by a discharging capacitor to decrease to 37% its initial voltage, current, or charge -time taken to charge a capacitor to 63% of the voltage source -time constant = RC (unit = seconds) -at time t = RC, charge falls to 0.37 (e⁻¹) times its initial value (units derived from R = V/A, C = Q/V -> VQ/VA = Q/A = t)
-how to measure energy stored in capacitor
-use joulemeter when charging: -battery and fixed resistor on input, capacitor on output -measure energy transferred (energy stored = work done to force charge to plate) or when discharging (with manual joulemeter): -capacitor on input, bulb on output -record capacitor P.D before discharging -record joules before discharge starts -record joules after -difference in energy = 1/2CV² -so capacitance can be calculated
investigating capacitor discharge (to measure RC and compare with actual value)
-using an oscilliscope or by recording values manually: -record PD and current of discharging capacitor connected to fixed resistor at regular intervals of time -calculate RC using capacitance and resistor value -plot graph of V against t -time taken for voltage to decrease to 0.37 times (1/e) the initial value can be measured from graph and compared with calculated value RC
purpose of dielectrics
-when polarised, the positive side of dielectric attracts more electrons from battery onto negative plate -and negative side of dielectric repels electrons from positive plate to force them through the battery -so a dielectric increases charge stored in a capacitor for any given PD across the capacitor terminals -increasing capacitance of capacitor
how is a dielectric polarised
-when uncharged, molecules are uncharged -when charged, electrons of molecules pull slightly towards positive plate -so each molecule of dielectric become polarised -so surface of dielectric facing positive plate gains negative charge and surface facing negative plate gains positive charge (from electrons moving away from that side)
how is energy stored in a capacitor
-work done when electrons forced from one plate to the other -electric potential energy builds (because plates are unlike charges so charges are forced together against their electrostatic repulsion)
A 6.0nF capacitor is in parallel with a 10nF capacitor. The voltage across the 6.0nF capacitor is 36V. What is the voltage across the other capacitor?
36 V because voltage is the same in parallel across each branch
capacitance of a capacitor, and how it is found
C (F) - Farads charge stored per unit potential difference C = Q/V -found by plotting graph of Q against V when charging a capacitor at constant current (by adjusting resistance of variable resistor in series with capacitor) gradient = capacitance
capacitance of capacitor with dielectric
C = Aϵ₀ϵᵣ/d A = effective surface area of each plate ϵᵣ = relative permittivity of dielectric d = spacing between plates ϵ₀ = permittivity of free space
energy stored in charged capacitor (and how is this derived)
E = 1/2QV -found from finding area under line of V against Q graph E = 1/2CV² E = 1/2 Q²/C -found by substituting Q = CV into top equation
equation for discharge curve (P.D and current)
I = I₀e⁻ᵗ/ᴿᶜ V = V₀e⁻ᵗ/ᴿᶜ I₀ = initial current (capacitor does't have to be fully charged) I = current of capacitor at time t V₀ = initial P.D (capacitor doesn't have to be fully charged) V = potential difference across capacitor at time t (P.D, current, and charge curves for discharging are all exponential decay)
how to find initial current when charging + P.D at any instant
I₀ = V₀/R (source voltage/R) -assuming capacitor is initially uncharged (at any instant, V₀ = IR + Q/C (source PD = resistor PD + capacitor PD)
charged stored by a capacitor
Q so positive plate has charge +Q and negative plate has charge -Q
equations for graphs of V, Q, and I for charging a capacitor
Q = Q₀(1 - e⁻ᵗ/ᴿᶜ) V = V₀(1 - e⁻ᵗ/ᴿᶜ) I = I₀e⁻ᵗ/ᴿᶜ -found by substituting I equation here into equations from previous card
equation for discharge curve (charge)
Q = Q₀e⁻ᵗ/ᴿᶜ Q₀ = initial charge of capacitor (does not have to be fully charged) t = time since discharging began C = capacitance of capacitor Q = charge of capacitor at time t R = resistance of fixed resistor used to limit current
energy stored in capacitor vs battery
battery: E = QV so line is parallel to x-axis capacitor: E = 1/2QV so line forms triangle with x-axis
-finding time constant using table of time and charge values
lnQ = - t/RC + lnQ₀ -find lnQ for each Q value -plot graph of lnQ against t: gradient = -1/RC y-intercept = Q₀
equations for total capacitance when capacitors are in series or parallel
series: 1/C₁ + 1/C₂ = 1/Cₜₒₜₐₗ parallel: C₁ + C₂ = Cₜₒₜₐₗ (remember that charge is always conserved when solving these questions) (can be used to find energy stored in capacitors by finding two of their: P.D, capacitance, charge stored)
