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

Tuesday, June 19, Coulomb’s Law – The 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 Formula PHYS 1444 Dr. Andrew Brandt

Tuesday, June 19, 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|=  (F x 2 +F y 2 ) Use  = tan -1 (F y /F x ) to get angle PHYS 1444 Dr. Andrew Brandt

Tuesday, June 19, Gauss’ Law The precise relation between flux and the enclosed charge is given by Gauss’ Law  0 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 Q encl. PHYS 1444 Dr. Andrew Brandt

Example 22-3: Spherical conductor. A thin spherical shell of radius r 0 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 Cross-sectional drawing of a thin spherical shell of radius r 0 carrying a net charge Q uniformly distributed. A 1 and A 2 represent two gaussian surfaces we use to determine Example 22–3. Solution: a. The gaussian surface A 1, outside the shell, encloses the charge Q. We know the field must be radial, so E = Q/(4 πε 0 r 2 ). b. The gaussian surface A 2, 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 5 Tuesday, June 19, 2012PHYS 1444 Dr. Andrew Brandt

Example 22-4: Solid sphere of charge. An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r 0. Determine the electric field (a) outside the sphere (r > r 0 ) and (b) inside the sphere (r < r 0 ). *q4 Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4 πε 0 r 2 ). b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr 3 /r 0 3 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4 πε 0 r 0 3 ). Tuesday, June 19, PHYS 1444 Dr. Andrew Brandt

Example 22-5: Nonuniformly charged solid sphere. Suppose the charge density of a solid sphere is given by ρ E = α r 2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r 0. (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 = 4 π r 2 dr. To find the total charge: Q = ∫ ρ E dV = 4 πα r 0 5 /5, giving α = 5Q/4 π r 0 5. b. The charge enclosed in a sphere of radius r will be Qr 5 /r 0 5. Gauss’s law then gives E = Qr 3 /4 πε 0 r 0 5. Tuesday, June 19, PHYS 1444 Dr. Andrew Brandt

Tuesday, June 19, 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

Tuesday, June 19, 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

Tuesday, June 19, PHYS 1444 Dr. Andrew Brandt Capacitors (or Condensers) 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.

11 Capacitors A simple capacitor consists of a pair of parallel plates of area A separated by a distance d.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 Tuesday, June 19, 2012PHYS 1444 Dr. Andrew Brandt

Tuesday, June 19, PHYS 1444 Dr. Andrew Brandt 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 V ba 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/VorFarad (F) C is a property of a capacitor so does not depend on Q or V. Normally use  F or pF.

Tuesday, June 19, PHYS 1444 Dr. Andrew Brandt Determination of Capacitance 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=  0, where  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? C only depends on the area (A) and the separation (d) of the plates and the permittivity of the medium between them.

Tuesday, June 19, PHYS 1444 Dr. Andrew Brandt 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

Tuesday, June 19, PHYS 1444 Dr. Andrew Brandt Example 24 – 1 (c) Using the formula for the electric field in two parallel plates (d) Solving the capacitance formula for A, we obtain Or, sincewe can obtain Solve for A About 40% the area of Arlington (256km 2 ).

Tuesday, June 19, PHYS 1444 Dr. Andrew Brandt Example 24 – 3 Spherical capacitor: A spherical capacitor consists of two thin concentric spherical conducting shells, of radius ra ra and r b, 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. 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

17 PHYS 1444 Dr. Andrew Brandt 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 r b can be obtained as 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 where Tuesday, June 19, 2012

18 PHYS 1444 Dr. Andrew Brandt 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 paralleland in Series or in combination

19 Capacitors in Parallel Parallel arrangement provides the same voltage across all the capacitors. –Left hand plates are at Va Va and right hand plates are at VbVb –So each capacitor plate acquires charges given by the formula Q 1 =C 1 V, Q 2 =C 2 V, and Q 3 =C 3 V The total charge Q that must leave battery is then  Q=Q 1 +Q 2 +Q 3 =V(C 1 +C 2 +C 3 ) Consider that the three capacitors behave like a single “equivalent” one  Q=C eq V= V(C 1 +C 2 +C 3 ) Thus the equivalent capacitance in parallel is For capacitors in parallel the capacitance is the sum of the individual capacitors

20 PHYS 1444 Dr. Andrew Brandt Capacitors in Series Series arrangement is more “interesting”  When battery is connected, +Q flows to the left plate of C1 C1 and –Q flows to the right plate of C3C3  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  So the charge on each capacitor is the same value, Q. ( Same charge ) Consider that the three capacitors behave like an equivalent one  Q=C eq V  V=Q/C eq The total voltage V across the three capacitors in series must be equal to the sum of the voltages across each capacitor.  V=V 1 +V 2 +V 3 =(Q/C 1 +Q/C 2 +Q/C 3 ) Putting all these together, we obtain: V=Q/C eq =Q(1/C 1 +1/C 2 +1/C 3 ) Thus the equivalent capacitance is The total capacitance is smaller than the smallest C!!!