PHYS 1444 Lecture #5

1. PHYS 1444 Lecture #5 Tuesday June 19, 2012 Dr. Andrew Brandt • Short review • Chapter 24 • Capacitors and Capacitance PHYS 1444 Dr. Andrew Brandt

2. Coulomb’s Law – The Formula Formula A vector quantity. Newtons • Direction of electric (Coulomb) force (Newtons) is always along the line joining the two objects. • Unit of charge is called Coulomb, C, in SI. • Elementary charge, the smallest charge, is that of an electron: -e where PHYS 1444 Dr. Andrew Brandt

3. Vector Problems • Calculate magnitude of vectors (Ex. force using Coulomb’s Law) • Split vectors into x and y components and add these separately, using diagram to help determine sign • Calculate magnitude of resultant |F|=(Fx2+Fy2) • Use = tan-1(Fy/Fx) to get angle PHYS 1444 Dr. Andrew Brandt

4. Gauss’ Law • The precise relation between flux and the enclosed charge is given by Gauss’ Law • e0 is the permittivity of free space in the Coulomb’s law • A few important points on Gauss’ Law • Freedom to choose surface • Distribution of charges inside surface does not matter only total charge • Charges outside the surface do not contribute to Qencl. PHYS 1444 Dr. Andrew Brandt

5. Example 22-3: Spherical conductor. A thin spherical shell of radius r0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere? *q3 Figure 22-11. Cross-sectional drawing of a thin spherical shell of radius r0 carrying a net charge Q uniformly distributed. A1 and A2 represent two gaussian surfaces we use to determine Example 22–3. Solution: a. The gaussian surface A1, outside the shell, encloses the charge Q. We know the field must be radial, so E = Q/(4πε0r2). b. The gaussian surface A2, inside the shell, encloses no charge; therefore the field must be zero. c. All the excess charge on a conductor resides on its surface, so these answers hold for a solid sphere as well. Key to these questions is how much charge is enclosed PHYS 1444 Dr. Andrew Brandt

6. Example 22-4: Solid sphere of charge. An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0). *q4 Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4πε0r2). b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r03 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4πε0r03). PHYS 1444 Dr. Andrew Brandt

7. Example 22-5: Nonuniformly charged solid sphere. Suppose the charge density of a solid sphere is given by ρE = αr2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere. Solution: a. Consider the sphere to be made of a series of spherical shells, each of radius r and thickness dr. The volume of each is dV = 4πr2dr. To find the total charge: Q = ∫ρE dV = 4παr05/5, giving α = 5Q/4πr05. b. The charge enclosed in a sphere of radius r will be Qr5/r05. Gauss’s law then gives E = Qr3/4πε0r05. PHYS 1444 Dr. Andrew Brandt

8. Electric Potential Energy • Concept of energy is very useful solving mechanical problems • Conservation of energy makes solving complex problems easier. • Defined for conservative forces (independent of path) PHYS 1444 Dr. Andrew Brandt

9. What are the differences between the electric potential and the electric field? • Electric potential (U/q) • Simply add the potential from each of the charges to obtain the total potential from multiple charges, since potential is a scalar quantity • Electric field (F/q) • Need vector sums to obtain the total field from multiple charges • Potential for a positive charge is large near a positive charge and decreases to 0 at large distances. • Potential for the negative charge is small (large magnitude but negative) near the charge and increases with distance to 0 Properties of the Electric Potential PHYS 1444 Dr. Andrew Brandt

10. What is a capacitor? • A device that can store electric charge without letting the charge flow • What does it consist of? • Usually consists of two oppositely charged conducting objects (plates or sheets) placed near each other without touching • Why can’t they touch each other? • The charges will neutralize each other • Can you give some examples? • Camera flash, surge protectors, computer keyboard, binary circuits… • How is a capacitor different than a battery? • Battery provides potential difference by storing energy (usually chemical energy) while the capacitor stores charge but very little energy. Capacitors (or Condensers) PHYS 1444 Dr. Andrew Brandt

11. Capacitors • A simple capacitor consists of a pair of parallel plates of area A separated by a distance d. • A cylindrical capacitors are essentially parallel plates wrapped around as a cylinder. • Symbols for a capacitor and a battery: • Capacitor -||- • Battery (+) -|i- (-) Circuit Diagram PHYS 1444 Dr. Andrew Brandt

12. What do you think will happen if a battery is connected (voltage is applied) to a capacitor? • The capacitor gets charged quickly, one plate positive and the other negative with an equal amount. of charge • Each battery terminal, the wires and the plates are conductors. What does this mean? • All conductors are at the same potential. • the full battery voltage is applied across the capacitor plates. • So for a given capacitor, the amount of charge stored in the capacitor is proportional to the potential difference Vba between the plates. How would you write this formula? • C is a proportionality constant, called capacitance of the device. • What is the unit? Capacitors C is a property of a capacitor so does not depend on Q or V. PHYS 1444 Dr. Andrew Brandt Normally use mF or pF. C/V or Farad (F)

13. C can be determined analytically for capacitors w/ simple geometry and air in between. • Let’s consider a parallel plate capacitor. • Plates have area A each and separated by d. • d is smaller than the length, so E is uniform. • For parallel plates E=s/e0, where s is the surface charge density. • E and V are related • Since we take the integral from the lower potential point a to the higher potential point b along the field line, we obtain • So from the formula: • What do you notice? Determination of Capacitance C only depends on the area (A) and the separation (d) of the plates and the permittivity of the medium between them. PHYS 1444 Dr. Andrew Brandt

14. Example 24 – 1 Capacitor calculations: (a) Calculate the capacitance of a capacitor whose plates are 20 cm x 3.0 cm and are separated by a 1.0 mm air gap. (b) What is the charge on each plate if the capacitor is connected to a 12 V battery? (c) What is the electric field between the plates? (d) Estimate the area of the plates needed to achieve a capacitance of 1F, given the same air gap. (a) Using the formula for a parallel plate capacitor, we obtain (b) From Q=CV, the charge on each plate is PHYS 1444 Dr. Andrew Brandt

15. Example 24 – 1 (c) Using the formula for the electric field in two parallel plates Or, since we can obtain (d) Solving the capacitance formula for A, we obtain Solve for A About 40% the area of Arlington (256km2). PHYS 1444 Dr. Andrew Brandt

16. Spherical capacitor: A spherical capacitor consists of two thin concentric spherical conducting shells, of radius ra and rb, as in the figure. The inner shell carries a uniformly distributed charge Q on its surface and the outer shell an equal but opposite charge –Q. Determine the capacitance of this configuration. Example 24 – 3 Using Gauss’ law, the electric field outside a uniformly charged conducting sphere is So the potential difference between a and b is Thus capacitance is PHYS 1444 Dr. Andrew Brandt

17. Capacitor Cont’d • A single isolated conductor can be said to have a capacitance, C. • C can still be defined as the ratio of the charge to absolute potential V on the conductor. • So Q=CV. • The potential of a single conducting sphere of radius rb can be obtained as where • So its capacitance is • Although it has capacitance, a single conductor is not considered to be a capacitor, as a second nearby charge is required to store charge PHYS 1444 Dr. Andrew Brandt

18. Capacitors in Series or Parallel • Capacitors are used in many electric circuits • What is an electric circuit? • A closed path of conductors, usually wires connecting capacitors and other electrical devices, in which • charges can flow • there is a voltage source such as a battery • Capacitors can be connected in various ways. • In parallel and in Series or in combination PHYS 1444 Dr. Andrew Brandt

19. Capacitors in Parallel • Parallel arrangement provides the same voltage across all the capacitors. • Left hand plates are at Va and right hand plates are at Vb • So each capacitor plate acquires charges given by the formula • Q1=C1V, Q2=C2V, and Q3=C3V • The total charge Q that must leave battery is then • Q=Q1+Q2+Q3=V(C1+C2+C3) • Consider that the three capacitors behave like a single “equivalent” one • Q=CeqV= V(C1+C2+C3) • Thus the equivalent capacitance in parallel is For capacitors in parallel the capacitance is the sum of the individual capacitors

20. Series arrangement is more “interesting” • When battery is connected, +Q flows to the left plate of C1 and –Q flows to the right plate of C3 • This induces opposite sign charges on the other plates. • Since the capacitor in the middle is originally neutral, charges get induced to neutralize the induced charges Capacitors in Series • So the charge on each capacitor is the same value, Q. (Same charge) • Consider that the three capacitors behave like an equivalent one • Q=CeqV  V=Q/Ceq • The total voltage V across the three capacitors in series must be equal to the sum of the voltages across each capacitor. • V=V1+V2+V3=(Q/C1+Q/C2+Q/C3) • Putting all these together, we obtain: • V=Q/Ceq=Q(1/C1+1/C2+1/C3) • Thus the equivalent capacitance is PHYS 1444 Dr. Andrew Brandt The total capacitance is smaller than the smallest C!!!