PHYSICS FOR ENGINEERING
Capacitor AND CAPACITANCE
APRIL 2019 PHYSICS FOR ENGINEERING LECTURE Instructor: Engr. Claudio Cajes
REPUBLIC OF THE PHILIPPINES CEBU TECHNOLOGICAL UNIVERSITY - MAIN CAMPUS MJ. CUENCO AVE., COR. R. PALMA ST., CEBU CITY BACHELOR OF SCIENCE IN MECHANICAL ENGINEERING CAPACITOR & CAPACITANCE 88
March 2019
PHYSICS FOR ENGINEERING
Faughn, J.S., Serway, R.A., Vuille, C., & Bennet, C.A.(2006). Capacitance. Serwayβs College Physics (Thompson Asian Edition). pp-545-546 Pacific Grove, United States of America: Brooks/Cole. CAPACITANCE Capacitor is a device used in a variety of electric circuits β for example, to tune the frequency of radio receivers, eliminate sparking in automobile ignition systems, or store short-term energy for rapid release in electronic flash units. Used in electric circuit, the plates are connected to the positive and negative terminals of a battery or some other voltage source. When this connection is made, electrons are pulled off one of the plates, leaving it with a charge of +Q, and are transferred through the battery to the other plate, leaving it with a charge of βQ, as shown in the figure. The transfer of charge stops when the potential difference across the plates equals the potential difference of the battery. A charged capacitor is a device that stores energy that can be reclaimed when needed for a specification application.
Figure 1. A parallel-plate capacitor consists of two parallel plates, each area A, separated by a distance d. The plates carry equal and opposite charges.
The capacitance C of a capacitor is the ratio of the magnitude of the charge on either conductor (plate) to the magnitude of the potential difference between the conductors (plates): π
πΆ = βπ , Equation 1.0 πΆ
SI units: farad (F) = coulomb per volt (π) The quantities Q and βV are always taken to be positive when used in Equation 1.0. For example, if a 3.0 Β΅F capacitor is connected to a 12-V battery, the magnitude of the charge on eaach plate of capacitor is π = πΆβπ = (3.0 Γ 10β6 πΉ)(12 π) = 36ππΉ
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PHYSICS FOR ENGINEERING
From the Equation 1.0, we see that a large capacitance is needed to store large amount of charge for given applied voltage. The farad is a very large unit of capacitance. In practice, most typical capacitors have capacitances ranging from microfarads ((1ππΉ = 1 Γ 10β6 πΉ) to picofarads (1ππΉ = 1 Γ 10β12 πΉ). THE PARALLEL-PLATE CAPACITOR The capacitance of a device depends on the geometric arrangement of the conductors. The capacitance of a parallel-plate capacitor with plates separated by air (see firgure 1.0). can be easily calculated from three facts. First, recall that π the magnitude of the elctric field between two plates is given by πΈ = π , where Ο 0
is the magnitude of the charge per unit area on each plate. Second, we found earlier in this chapter that the potential difference between two plates is βπ = πΈπ, where d is the distance between the plates. Third, the charge on one plate is given by π = ππ΄, where A is the area of the plate. Substituting these three facts into the definition of capacitance gives the desired result: π ππ΄ ππ΄ πΆ= = = π βπ πΈπ ( ) π π0 π΄
Canceling the charge per unit area, Ο, yields: πΆ = π0 π , where A is the area of one plates, d is the distance between the plates, and π0 is the permittivity of the free space. π΄ From the equation πΆ = π0 π , we see that plates with larger area can store more charge. The same is true for a small plate separation d, because then the positive charges on one plate exert a stronger force on the negative charges on the other plate, allowing more chrage to be held on the plates.
Figure 2. (a) The electric field between the plates of parallel-plate capacitor is uniform near the center, but non-uniform near the edges. (b) Electric field pattern of two oppositely charged conducting parallel plates. Small pieces of thread on an oil surface align with electric field.
CAPACITOR & CAPACITANCE March 2019
PHYSICS FOR ENGINEERING
Figure 2 show the electric field lines of a more realistic parallel-plate capacitor. The electric field is very nearly constant in the center between the p[lates, but becomes less so when approaching the edges. For most purposes, however, the field may be taken as constant throughout the region between the plates. One practical device that uses a capacitor is the flash attachment on a camera. A battery is used to charge the capacitor, and stored charge is then released when the shutter-release button is pressed to take a picture. The stored charge delivered to a flash tube very quickly, illuminating the subject at the instant more light is needed. Computers make use of capacitors in many ways. For example, one type of computer keyboard has capacitors at the bases of its keys, as in Figure 3. Each key is connected to a movable plate, hich represents one side of the capacitor; the fixed plate on the bottom of the keyboard represents the other side of the capoacitor. When a key is pressed, the capacitor spacing decreases, causing an increase in capacitance. External electronic circuits recognize each key by the change in its capacitance when it is pressed.
Capacitors are useful for storing a large amount of charge that needs to be delivered quickly. A good example on the forefront of fusion research is electrostatic confinement. In this role, capacitors discharge their electrons through a grid. The negatively charged electrons in the grid draw prositively charfged particales to them and therefore to each other, causing some particles to fuse and release energy in the process. Figure 3. When the key of one type of keyboard is pressed, the capacitance of a parallel-plate capacitor increases as the plate spacing decreases. The substance labeled βdielectricβ is an insulating material.
APPLICATION Camera flash attachment Computer keyboard Electrostatic Confinement
CAPACITOR & CAPACITANCE March 2019
PHYSICS FOR ENGINEERING
A PARALLEL-PLATE CAPACITOR Objective: To calculate fundamental physical properties of a parallel-plate capacitor. Problem: A paralell-plate capacitor has an area π΄ = 2.00 Γ 10β4 π2 and a plate separation π = 1.00 Γ 10β3 π . (a) Find its capacitance. (b) How much charge is on the positive plate if the capacitor is connected to a 3.00 β π battery? (c) Calculate the charge density on the positive plate, assuming the density is uniform, and (d) the magnitude of the electric field between the plates. Stategy: Parts (a) and (b) can be solved by substituting into the basic equations for capacitance. In Part (c), use the fact that the voltage difference equals the electric field times the distance. Solution: (a) Find the capacitance. πΆ2
A
2.00 Γ10β4 π2
C = π0 π = (8.85 Γ 10β12 Nβπ2 ) ( 1.00Γ10β3 π ) C = 1.77 Γ 10β12 πΉ = 1.77 ππΉ (b) Find the charge on the positive plate after the capacitor is connected to a 3.00-V battery. π
πΆ = βπ β π = πΆβπ = (1.77 Γ 10β12 πΉ) Γ (3.00 π) = 5.31 Γ 10β12 πΆ (c) Calculate the charge density on the positive plate. Charged density is divoded by the area: π
5.31Γ10β12 πΆ
πΆ
π = π΄ = 2.00 Γ10β4 π2 = 2.66 Γ 10β8 π2 (d) Calculate the magnitude of the electric field between the plates. π
πΈ=π = 0
πΆ
2.66Γ10β8 2 π
πΆ2 8.85Γ10β12 Nβπ2
π
= 3.01 Γ 103 πΆ
Remarks: The answer to part (d) could also have been obtained from the lectric potential, which is βπ = πΈπ for a parallel-plate capacitor. ------------------------------------------------------------------------------------------------Exercise Problem: Two plates, each of area 3.00 Γ 10β4 π2 , are used to construct a parallel-plate capacitor with capacitance 1.00 ππΉ, (a) Find the necessary separation distance. (b) If the positive plate is to hold a charge of 5.00 Γ 10β12 πΆ, find the charge density.
CAPACITOR & CAPACITANCE March 2019
PHYSICS FOR ENGINEERING
(c) Find the electric field between the plates. (d) what voltage battery should be attached to the plate to obtain the preceeding results? πΆ
π
Answers: (a) 2.66 Γ 10β3 π (b) 1.67 Γ 10β8 π2 (c) 1.89 Γ 103 πΆ
(d) 5.00 π
------------------------------------------------------------------------------------------------Symbols for Circuit Elements and Circuits The symbol that is commonly used to represent a capacitor in a circuit is, or sometimes
. Donβt confuse either of these symbols with the circuit
symbol, which is used to designate a battery (or any other source of direct current). The positive terminal of the battery is the higher potential and is represented by the longer vertical line in the battery symbol. ------------------------------------------------------------------------------------------------CAPACITORS WITH DIELECTRICS A dielectric is an insulating material, such as rubber, plastic, or waxed paper. When a dielectric is inserted between the plates of a capacitor, the capacitance increases. If the dielectric completely fills the space between the plates, the capacitance is multiplied by the factor k, called the dielectric constant. The following experiment illustrates the effect of dielectric in a capacitor. Consider a parallerl-plate capacitor of charge π0 and capacitance πΆ0 in the absence of a dielectric. The potential difference across the capacitor plates can be measured, and is given by βπ0 =
π0 πΆ0
. Because the capacitor is not connected
to an external circuit, there is no pathway for charge to leave or be added to the plates. If a dielectric is now inserted between the plates as in Figure 4, the voltage across the plates is reduced by the factor k to the value, βπ =
βπ0 πΎ
Because k>1, βπ is less than βπ0. Because the charge π0 on the capacitor doesnβt change, we conclude that the capacitance in the presence of the dielectric must change to the value, π
πΆ = βπ0 =
π0 βπ0 πΎ
=
ππ0 βπ0
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PHYSICS FOR ENGINEERING
Figure 4. (a) With air between the plates, the voltage across the capacitor is βπ0 , the capacitance is πΆ0 , and the charges is π0 . (b) With a dielectric between the plates, the charges remains at π0 , but the voltage and capacitance change. According to this result, the capacitance is multiplied by the factor k when the dielectric fills the region between the plates. For a parallelplate capacitor, where tha capacitance in the absnece of a dielectric is πΆ0 =
π0 π΄ π
, we can express the capacitance is the presence of a
dielectric as, πΆ0 =
π0 π΄ π
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CAPACITOR & CAPACITANCE March 2019
PHYSICS FOR ENGINEERING
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