Slide 1 / 39 Slide 2 / 39 P Physics & M apacitance and ielectrics 20151205 www.njctl.org Slide 3 / 39 apacitors capacitor is any two conductors seperated by an insulator, such as air or another material. ach conductor has equal magnitude, but opposite charge so the net charge across the two conductors is zero. Slide 4 / 39 apacitors To charge a capacitor you can simply connect each conductor to the opposite end of each battery. Once the plates have achieved a magnitude of Q the battery can be disconnected. There will be a fixed potential across the conductors equal to the potential difference of the battery. In a circuit a capacitor is represented by: In previous sections we discussed that both the electric field and the potential difference are both directly proportional to the charge Q. y doubling the charge we double both the electric field and potential difference. apacitance is the ratio of charge to potential difference, therefore it is independent of the charge and potential. Slide 5 / 39 alculating apacitance The simplest version of a capacitor is the parallel plate capacitor, which has two conducting plates of area. Slide 6 / 39 alculating apacitance apacitance is the ratio of charge to potential difference across the plates, which can be written as: Using Gauss's law we can find that the electric field between a parallel plate capacitor is: Since the electric field is uniform and the plates are separated by a distance d, the electric potential is equal to: y plugging in what we just solved for v, we will see that the capacitance is only dependent on the area of the plate and their separation. (apacitance of a parallel plate capacitor in a vacuum)
Slide 7 / 39 Unit of apacitance apacitance is the ratio of charge to the potential difference. Slide 8 / 39 1 The plates of a parallel plate capacitor are 3x10 3 m apart and they are each 2cm squares. What is the capacitance of the capacitor? So the unit for capacitance is coulomb/volt better known as a Farad, which was named in honor of the nglish physicist Michael Faraday. Farad is denoted by a capital F. 1 F = 1 Farad = 1 /V = 1 coulomb/volt 5.9x10 11 F 4x10 15 F 1.2pF 3.6µF 2.4pF Slide 9 / 39 Slide 10 / 39 2 parallel plate capacitor has a charge of 6µF. The electric field present between the plates is 3x10 7 N/ and the plate's separation is 4cm. What is the capacitance of the capacitor? 5nF 12pF 5µF Q Q Spherical apacitor so, lectric Field for point charge If the electric field is the same for a point charge then so is the potential difference. 5MF 5pF Slide 11 / 39 Slide 12 / 39 ylindrical apacitor rb ra L efore we found the electric potential for a cylinder to be equal to: The equation for capacitance is: The charge on the cylinder is represented as: x z y 2 The potential difference across the battery results in both of the two capacitors to begin charging. The first plate of will acquire a positive charge equal to Q which will displace the charge on the second plate making it negative and the first plate of 2 positive, which in turn will make the second plate of 2 to become negatively charged. This is not an immediate change in the charge, it slowly builds up, but in the end the charge is equal for each capacitor. The capacitance per unit of length is:
Slide 13 / 39 Slide 14 / 39 When we discussed ircuits the last two years you could always replace any combination of resistors with one that has the equivalent resistance. For resistance in series we would just add their values to find the net resistance and just draw a new circuit only using the one resistor. We can do the same for apacitors, but instead of just adding their values like the resistors, when they are in a series circuit we have to take their reciprocals. Slide 15 / 39 3 Three capacitors are connected in series,, 2, and 3. Their capacitance's are 4µF, 3µF, and 6µF respectively. Slide 16 / 39 4 Three capacitors are connected in series,, 2, and 3. Their capacitance's are 2µF, 7µF, and 13µF respectively. 13µF 1.33µF µf 7µF 2 22µF 7µF 20µF 1.39µF 2 2.78µF 3 2.6µF 3 x 2 y Slide 17 / 39 The voltage across each capacitor is equal to the potential difference of the battery because in a parallel circuit the voltage is the same, but in this case the charge is the sum of those on the capacitors. The charges on each of the capacitors are: Slide 18 / 39 When we discussed ircuits the last two years you could always replace any combination of resistors with one that has the equivalent resistance. For resistance in parallel we would add their reciprocals to find the net resistance and just draw a new circuit only using the one resistor. We can do the same for apacitors, but instead of adding their reciprocals like the resistors, when they are in a parallel circuit we just have to add their values. therefore:
Slide 19 / 39 5 Three capacitors are connected in parallel, 1, 2, 3. Their capacitance's are 3μF, 4μF, and 2μF Slide 20 / 39 6 Three capacitors are connected in parallel, 1, 2, 3. Their capacitance's are 12μF, 5μF, and 7μF 2µF 23.8µF 1.1µF 9µF 3µF 2 3 24µF 2µF 8µF 2 3 2.5µF 32.4µF Slide 21 / 39 7 Five capacitors are placed into a circuit as shown below. The five capacitors are, 2, 3, 4, and 5. Their capacitance's are 2µF, 1µF, 3µF, 2µF, and 1µF respectively. What is the net capacitance of the circuit? 2µF 23/4 µf 3µF 2 1 3 4 5 Slide 22 / 39 nergy Storage in a apacitor and lectricfield nergy nother way of looking at the amount of energy stored in a capacitor is to look at the electric field it produces because of the charged plates. We can say that the electric potential energy is spaced throughout the electric field therefore it has an energy density, which is denoted by a small u. The energy density is the ratio of electric potential energy to the volume between the plates. quation for apacitance: Potential ifference across the plates: 6µF 1µF Slide 23 / 39 nergy Storage in a apacitor and lectricfield nergy For a capacitor to be used in a practical manner we need to know how much electric potential energy it can store. This can be achieved by determining how much work is required to charge each of the plates. The small charges we will add up to equal the net charge will be denoted as dq and the potential difference can be given as Q/. Slide 24 / 39 nergy Storage in a apacitor and lectricfield nergy Since the work done to charge the capacitor is equal to the potential energy stored on the capacitor and since we know =Q/V the potential energy can be represented as: The work done to charge the capacitor is also the same as the amount of work required to discharge it. The uncharged capacitor has a potential energy of zero so the work to charge the capacitor is equal to the stored potential energy.
Slide 25 / 39 nergy Storage in a apacitor and lectricfield nergy The equation for the energy density is: The equation is derived from the simple case of a parallel plate capacitor, however it works for every type of capacitor in a vacuum and for every electric field also in a vacuum. Slide 26 / 39 ielectrics ielectrics are materials which are placed between the parallel plates or any other configuration for a different number of reasons. It helps to maintain the shape of the capacitor, preventing the walls from coming in contact with one another. It allows the capacitor to reach a higher potential difference then it could normally before dielectric breakdown, which is the ionization of the air around the capacitor which would result in charge leaving the capacitor. It also enables the capacitor to increase its capacitance. Slide 27 / 39 ielectrics When a dielectric is placed between the plates of a parallel plate capacitor the voltage drops to a smaller value then its original, but the charge on the capacitor remains the same. Slide 28 / 39 ielectric onstant The ielectric onstant is denoted by a capital Kappa and it is the ratio of the final capacitance to the original capacitance. Original apacitance apacitance with ielectric The new equation for apacitance is now represented as: and Q is the same,therefore > o y adding a dielectric between the plates of a capacitor with a constant charge decreases the potential difference, but also the electric field. Slide 29 / 39 Slide 30 / 39 9 capacitor is completely charged and afterwards the battery is disconnected. What will cause the potential difference across the capacitor to decrease? Increase the Plate's surface area dd a dielectric with a greater # ecrease the distance ll of the above oth and
Slide 31 / 39 10 ylindrical capacitor is filled half way with a dieclectric of #. What is the net capacitance of this configuration? Slide 32 / 39 11 spherical capacitor is filled halfway with a dielectric of #. What is the net capacitance? R r R r L Slide 33 / 39 Slide 34 / 39 ielectric's effect on lectric nergy ensity y adding a dielectric you alter the way the capacitor normally behaves. You allow it to carry a greater potential difference and also have a greater electric field passing through it without dielectric breakdown occuring. The normal electric field between the plates is which can be determined through oulomb's law for the infinite charged disk. However when the dielectric is added the net charge density along the surface of the capacitor plate is the difference of that on the plate and the one induced on the dielectric because of the effects of polarization. The net electric field is now represented as: Slide 35 / 39 Slide 36 / 39
Slide 37 / 39 Slide 38 / 39 Slide 39 / 39 12 The outer shell has a charge of Q and the inner shell has a charge of Q. Half of one side of a cylindrical capacitor of length l is filled with a dielectric of #. What is the magnitude of the electric field at a radius r, if r a < r < r b, for both parts of the capacitor? ielectric Vacuum rb ra