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Physics Capacitors
Physics Autumn Y13
| Question | Answer |
|---|---|
| A mystery capacitor can store 3.0J of energy when connected to a 10V supply. How much energy can it store when connected to a 5.0V supply? | 1.5J incorrect as the supply would would change the charge it could store as well as the voltage across it. 0.75J correct |
| How capacitors work | Battery causes oppositely charged plates with uniform electric field between them. PD across capacitor equal to emf of battery but in opposite direction. Once at same pd no more current flow |
| How is a capacitor made | 2 metal plates separated by insulating dielectric (insulating, and doesnt let electric fields easily transfer energy?) then rolled together. Capacitance given by C = permittivity x area / separation |
| Uses of capacitors | Used to store small amounts of energy. Used to protect other components in a circuit - reduces current in branches. |
| What does it mean for a capacitor to have a capacitance of one farad | If the potential across it rises by 1V when a charge of 1C is placed on it. |
| Explaining capacitor charging | Electrons flow from A to B. Equal numbers removed and deposited and this leaves equal magnitude of charge on plates. Q = CV so as charge increases pd increases across capacitor. Process stops when pd = emf (K2L). Hard equation. |
| Explaining capacitor discharging | Electrons move ACW through resistor. magnitude of charge on both plates decreases until zero charge on plates Q = CV and V is zero. Energy stored in capacitor dissipated as thermal energy in resistor. Easy equation. Exponential decay. |
| Permittivity units | c^2/N/m^2 = F/m |
| 'Full scale deflection' | Once the capacitor is fully charged |
| Capacitors in parallel | V the same. QT = Q1 + Q2 etc Plug in Q = CV Cancel V |
| Capacitors in series | Q the same so pd split VT = V1 + V2 + V3 V = Q/C |
| Why work done on capacitor when charging | Electrons must have work done on them to exert the force needed (because must overcome attractive/repulsive force) |
| What does the area under a pd charge graph represent | The work done in charging = energy stored in capacitor. W = 1/2 QV Normal EMF cell does W = QV so dont mess up. |
| Switch opened from fully charged 6V capacitor. Voltmeter then reads 4V once switch closed again | Energy dissipated = energy stored at 6V - energy stored at 4V |
| What the xs mean in the main equation | V, I, or Q C and R constant |
| Time constant (lower case tau) | Resistance x capacitance Equal to the time taken for a discharging capacitor for the x to decrease to 37% of its initial value (when t = RC in the formula) |
| Capacitance definition | Charge per unit p.d. |
| Capacitance of a charged sphere (also stores charge if isolated - single plate capacitor) | Ratio of the charge is stores to the pd at its surface C = Q/V V of point charge from electric fields. Simplifies to C = 4*pi*permittivity*resistance Earth is a capacitor but low capacitance |
| Capacitance between parallel plates depends on | Area of plates, separation of plates, material between plates (C proportional to A/d and constant of proportionality depends on material between plates) |
| Relative permittivity | Permittivity = (relative) x (of free space). So for vacuum is one, then higher the higher the permittivity is (permittivity always higher than of free space). |
| Perfect vacuum efficacy as a dielectric | The worst possible dielectric because electric fields formed so easily. Has a relative permittivity of 1, while everything else is higher. |
| Capacitor banks | Store large amounts of energy and can release it very quickly. In parallel for high charge, in series for high pd Use the time constant RC. Changing R to change how quickly it charges (max speed) and discharges (necessary speed) |
| Rectification | Changing AC into DC. Transformers only work with AC (changing fields) so allows for the high pd of long distance transfer. Capacitors allow for smoothing the output voltage. |
| Smoothing | Distance between high and low is the 'ripple voltage'. |
| What is the force between 2 metal plates | Electric force. |
| 12V cell, when the capacitor is halfway through charging and is at 6.6V, what pd across resistor? | 5.4V |
| How to get the equation deltaQ/deltat = -Q/RC | Q against t graph. Gradient = -I I = -deltaQ/deltat I = V/R V= Q/C I = Q/CR Q/CR = -deltaQ/deltat |
| How to use the equation deltaQ/deltat = -Q/RC to model capacitor discharge | deltaQ = -Q*deltat/CR So find deltaQ for one value of Q, then subtract from Q to get a new value of Q, then repeat process (using spreadsheet). Allows us to see how Q changes over time |
| Potential difference | Joules per coulomb not per electron |
Created by:
Pyrogearos2