Lecture 8
iclicker A device has a charge q=10 nc and a potential V=100V, what s its capacitance? A: 0.1 nf B: 1nF C: 10nF D: 10 10 F E: 1F
iclicker A device has a charge q=10 nc and a potential V=100V, what s its capacitance? A: 0.1 nf B: 1nF C: 10nF D: 10 10 F E: 1F q = 10nC = 10 8 C V = 100V C = q/v = 10 8 /100 = 10 10 =0.1nF
Parallel Plate Capacitor (4) Remember the definition of capacitance so the capacitance of a parallel plate capacitor is Variables: A is the area of each plate d is the distance between the plates Note that the capacitance depends only on the geometrical factors and not on the amount of charge or the voltage across the capacitor. 12
Example: Capacitance, Charge, and Electrons Question: A storage capacitor on a random access memory (RAM) chip has a capacitance of 55 nf. If the capacitor is charged to 5.3 V, how many excess electrons are on the negative plate? Answer: Idea: We can find the number of excess electrons on the negative plate if we know the total charge q on the plate. Then, the number of electrons n=q/e, where e is the electron charge in coulomb. Second idea: The charge q of the plate is related to the voltage V to which the capacitor is charged: q=cv. 15
Cylindrical Capacitor (1) Consider a capacitor constructed of two collinear conducting cylinders of length L Example: coax cable The inner cylinder has radius r 1 and the outer cylinder has radius r 2 Both cylinders have charge per unit length λ with the inner cylinder having positive charge and the outer cylinder having negative charge 16
Cylindrical Capacitor (2) We apply Gauss Law to get the electric field between the two cylinders using a Gaussian surface with radius r and length L as illustrated by the red lines which we can rewrite to get an expression for the electric field between the two cylinders r1< r < r2 17
Cylindrical Capacitor (3) As we did for the parallel plate capacitor, we define the voltage difference across the two cylinders to be V = V 1 V 2 Thus, the capacitance of a cylindrical capacitor is Note that C depends on geometrical factors only. 18
Spherical Capacitor (1) Consider a spherical capacitor formed by two concentric conducting spheres with radii r 1 and r 2 19
Spherical Capacitor (2) Let s assume that the inner sphere has charge +q and the outer sphere has charge q The electric field is perpendicular to the surface of both spheres and points radially outward 20
Spherical Capacitor (3) To calculate the electric field, we use a Gaussian surface consisting of a concentric sphere of radius r such that r 1 < r < r 2 The electric field is always perpendicular to the Gaussian surface so which reduces to makes sense! 21
Spherical Capacitor (4) To get the electric potential we follow a method similar to the one we used for the cylindrical capacitor and integrate from the negatively charged sphere to the positively charged sphere Using the definition of capacitance we find The capacitance of a spherical capacitor is then 22
Capacitance of an Isolated Sphere We obtain the capacitance of a single conducting sphere by taking our result for a spherical capacitor and moving the outer spherical conductor infinitely far away Using our result for a spherical capacitor with r 2 = and r 1 = R we find meaning V = q/4πε 0 R (we already knew that!) 23
iclicker q A metal ball of radius R has a charge q. Charge is changed q -> - 2q. How did it s capacitance change? A: C->2 C0 B: C-> C0 C: C-> C0/2 D: C->4 C0 E: C->8 C0 14
iclicker q A metal ball of radius R has a charge q. Charge is changed q -> - 2q. How does it s capacitance changed? A: C->2 C0 B: C-> C0 C: C-> C0/2 D: C->4 C0 E: C->8 C0 15
Energy Stored in Capacitors Capacitors store electric energy U = 1 2 qv (V created by q s, selfinteraction) q = CV U = 1 2 CV 2 q 2 or We want small voltage, large energy: large C U = 1 2 C 37
Energy Density in Capacitors (1) We define the energy density, u, as the electric potential energy per unit volume Taking the ideal case of a parallel plate capacitor that has no fringe field, the volume between the plates is the area of each plate times the distance between the plates, Ad Inserting our formula for the capacitance of a parallel plate capacitor we find 38
Energy Density in Capacitors (2) Recognizing that V/d is the magnitude of the electric field, E, we obtain an expression for the electric potential energy density for parallel plate capacitor This result, which we derived for the parallel plate capacitor, is in fact completely general. This equation holds for all electric fields produced in any way The formula gives the quantity of electric field energy per unit volume. 39
Example: Isolated Conducting Sphere (1) An isolated conducting sphere whose radius R is 6.85 cm has a charge of q=1.25 nc. Question 1: How much potential energy is stored in the electric field of the charged conductor? Answer: Key Idea: An isolated sphere has a capacitance of C=4πε 0 R (see previous lecture). The energy U stored in a capacitor depends on the charge and the capacitance according to and substituting C=4πε 0 R gives 40
Example: Isolated Conducting Sphere (2) An isolated conducting sphere whose radius R is 6.85 cm has a charge of q = 1.25 nc. Question 2: What is the field energy density at the surface of the sphere? Answer: Key Idea: The energy density u depends on the magnitude of the electric field E according to q so we must first find the E field at the surface of the sphere. Recall: 41
What is the total energy in E-field? U tot = Z 1 R Z 1 R udv = 1 4 2 0E 2 r 2 dr = Z 1 2 1 q 2 2 0 4 0 r 4 r2 dr = 1 2 1 R q 2 4 0 R = 2 qv Yes! dv = d sin d r 2 dr =4 r 2 dr 21
Example: Thundercloud (1) Suppose a thundercloud with horizontal dimensions of 2.0 km by 3.0 km hovers over a flat area, at an altitude of 500 m and carries a charge of 160 C. Question 1: What is the potential difference between the cloud and the ground? Question 2: Knowing that lightning strikes require electric field strengths of approximately 2.5 MV/m, are these conditions sufficient for a lightning strike? Question 3: What is the total electrical energy contained in this cloud? 42
Physics of a spark V d1 E V/d 1 E 0 e E +q -q d V E k E 1µm 1eV E V/d E spark MeV/m 23
Example: Thundercloud (2) Question 1: What is the potential difference between the cloud and the ground? Answer: We can approximate the cloud-ground system as a parallel plate capacitor whose capacitance is The charge carried by the cloud is 160 C 720 million volts V = 1 q =7.2108 2 C ++++++++++++ ++++++++++++ 43
Example: Thundercloud (3) Question 2: Knowing that lightning strikes require electric field strengths of approximately 2.5 MV/m, are these conditions sufficient for a lightning strike? Answer: We know the potential difference between the cloud and ground so we can calculate the electric field E is lower than 2.5 MV/m, so no lightning cloud to ground May have lightning to radio tower or tree. 44
Example: Thundercloud (4) Question 3: What is the total electrical energy contained in this cloud? Answer: The total energy stored in a parallel place capacitor is 45
Electric circuits 27
Circuit diagram Lines represent conductors The battery or power supply is represented by The capacitor is represented by the symbol Battery provides (a DC) potential difference V 28
Charging/Discharging a Capacitor (2) Illustrate the charging processing using a circuit diagram. This circuit has a switch (pos c) When the switch is in position c, the circuit is open (not connected). (pos a) When the switch is in position a, the battery is connected across the capacitor. Fully charged, q = CV. (pos b) When the switch is in position b, the two plates of the capacitor are connected. Electrons will move around the circuit--a current will flow--and the capacitor will discharge. c c 8
demo 30
I I - - V + + V 31
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