PHY 102 Capacitance Capacitor is a device to

  • Slides: 37
Download presentation
PHY 102 Capacitance Capacitor is a device to store electrical charge… Symbol is… When

PHY 102 Capacitance Capacitor is a device to store electrical charge… Symbol is… When a capacitor is charged, its plates have charges of equal magnitude but opposite signs: +q and –q. B/w the two plates, there is a p. d V. The charge q and p. d V are related by q=CV ……………. . 1. 1 C is …. a measure of how much q is required by the plates to produce certain p. d b/w them. C depends only on the geometry of plates and NOT on q or p. d. S. I unit if coulomb/volt called farad F.

To charge a capacitor, it is placed in an electric cct wit a battery…

To charge a capacitor, it is placed in an electric cct wit a battery… Once charged, it can be assumed that a capacitor can retain the charges indefinitely until it is put into a cct when it can be discharged…. . To calculate C, once we know its geometry, we can i. Assume a charge q on the plates ii. Calculate the electric field E btw plates by Gauss law. iii. Find the p. d V b/w the plates. From E btw the plates and q on either plates, Gauss Law is …… q is the charge enclosed by a Gaussian sfc and …. . is

The net electric flux thru that sfc. Assume E has a uniform magnitude E

The net electric flux thru that sfc. Assume E has a uniform magnitude E and the vectors E and d. A are //e/ q=…. 1. 3 …. is permittivity constant = 8. 85 x 10 -12 F/m The p. d V b/w the plates and E are related by ………. 1. 4 choosing a path that follows the electric fielsd line, the vector E & d. S will have opposite direction, …… 1. 5 For a //e/-plate capacitor having considerably large plates that are so close together, E can be assumed constant through out the region b/w the plates, the

Charge q on the +ve plate enclosed by Gaussian sfc is given by q

Charge q on the +ve plate enclosed by Gaussian sfc is given by q = …. . 1. 6 From eqn 1. 5, V= …… Substitute for V=Ed in eqn 1. 1 and equate it to eqn 1. 6, we have Ced=…. C=…. 1. 7 This implies that C depends on the geometrical factors A (area of plate) and d (separation b/w the plates). Ex: Fig a below comprises and uncharged capacitor (C=0. 20 u. F) and a battery (V=20 V). Thickness and face area of the lower plate (copper material)

Capacitor are 6. 0 mm and 2. 4 x 10 -2 m 2 respectively.

Capacitor are 6. 0 mm and 2. 4 x 10 -2 m 2 respectively. What depth would electrons attain within the plate face as the capacitor becomes charged? Density of conduction electrons of Cu is 8. 49 x 1028 es/m 3. Solution …… Cylinderical capacitor A cylinderical capacitor of length L formed by two coaxial cylinders of radii a&b is shown below. If we assume L>>b and neglect the fringing of the E that occurs at the end of the cylinders, each plate contains a charge of magnitude q as ……

q= …. . Where A is the area of curved part of Gaussian sfc

q= …. . Where A is the area of curved part of Gaussian sfc = ……… 1. 8 = ……. 2. 1 This indicates that the C of cylinderical capacitor depends only on geometrical factos (L, b & a) Spherical capacitor Consider a central x/section of a capacitor that consists of two concentric spherical shells of radii a&b. As a Gaussian sfc, q = …. . E = ……. 2. 2 Eqn 1. 5 yields V=……… 2. 2

Substitute for V=q/C C = …. . 2. 3 In case of a single

Substitute for V=q/C C = …. . 2. 3 In case of a single isolated spherical conductor of radius R, if we let b→…. And substitute R for a in eqn 2. 2, C= ……. . 2. 4 Arrangement of Capacitors In a cct, combination of capacitors can be replaced wt an equivalent capacitor. Capacitors in //e/ The capacitors are directly wired together at one plate and directly wired together at the other plate. P. d applied across the two groups of wired-together plates is the same. Thus, the total charge stored on the capacitors is the sum of the charges on all capacitors. Equivalent capacitor has the same total charge and the same p. d as the actual capacitors. , From fig 1 a, q 1= C 1 V…. Total charge q= q 1 + q 2 + q 3 = C 1 V +C 2 V+C 3 V Ceq= C 1+C 2+C 3 = q/V For n capacitors; Ceq= ……. 2. 5 Capacitors in series As shown in the fig below, capacitors are wired serially, one after the other, and that a p. d is applied across the two ends of the series. This applied p. d is equal to the sum of the p. ds across all capacitors which produce identical charge on them. This implies that an equivalent capacitor will have the same

charge q and the same total p. d as the actual series capacitors. Thus,

charge q and the same total p. d as the actual series capacitors. Thus, V 1 ……, V 2…. For n capacitors, ……. 2. 6 Energy stored in an electric field In attempt to transfer electrons from one plate of a capacitor to other, electric fields build up in the space b/w the plates and its direction tends to oppose further transfer. In other words, work required to charge a capacitor is stored in form of electric energy U in the electric field b/w the plates. Thus, for a charge q’ being transferred from one plate to the other, the p. d v’ b/w the plate is given by

V’=q’/C If an extra increment of charge dq’ is xferrd, the increment of work

V’=q’/C If an extra increment of charge dq’ is xferrd, the increment of work required will be d. W=V’dq’ = ……. . 2. 7 Work reqd to bring the total capacitor charge up to a final value q is W=……. … 2. 8 This work is stored as p. e U in the capacitor so that U =…… 2. 9 Alternatively, U = ½ CV 2 …. . 3. 0 eqns 2. 9 & 3. 0 do not depend on any geometry of the capacitor. Energy density Consider a //e/ plate capacitor, the E b/w the plates

is the same, energy density, u, which is given by the total p. e

is the same, energy density, u, which is given by the total p. e per unit vol b/w the plates is also uniform: u=……. 3. 1 Given that V=Ed, u=……. . 3. 2 Capacitor with a dielectric If the space b/w the plates of a capacitor is filled with a dielectric (an insulating mat’l like oil, plastic etc), the C will be increased by a numerical factor called dielectric constant k of the insulating mat’l. thus, the magnitude E produced by a point charge inside a dielectric is given by E=…… The E just outside an isolated conductor immersed in a

dielectric is E=…. 3. 3 Dielectrics and Gauss’ law The fig below shows a

dielectric is E=…. 3. 3 Dielectrics and Gauss’ law The fig below shows a //e/ plate capacitor of plate area A, both with and without dielectric. It is assumed that the charge q on the plates is the same in both situations. It is also noted that the field b/w the plate induces charge on the faces of the dielectric. To find the electric field E 0 b/w plates without a dielectric, we enclose the charge +q on the top plate with a Gaussian sfc and then apply Gauss’ law; ……. . E 0=……. 3. 4 With presence of dielectric, the E b/w the plates (and

within the dielectric) can be determined using the same Gaussian sfc. Since the sfc

within the dielectric) can be determined using the same Gaussian sfc. Since the sfc encloses two types of charge; +q on the top plate and –q (induced charge) on the top face of the dielectric, the charge on the conducting plate is said to be free bcos it can move if we change the electric potential of the plate whereas the induced charge is not free. The net charge enclosed by the Gaussian sfc is therefore q-q’. Gauss’ law gives …… Thus, E=…. 3. 5 The effect of dielectric is to weaken the original field E 0 by a factor k, E= E 0/k …. . q-q’ = q/k ……. 3. 6 In Gauss’ law: ……. . 3. 7

Current and Resistance Consider a section of conductor in the fig below. If charge

Current and Resistance Consider a section of conductor in the fig below. If charge dq passes thru a hypothetical plane (such as aa’) in time dt, the current I thru that plance is defined as i= dq/dt ………………. 1. 1 Thus, charge that passes thru the plane aa’, bb’ and cc’ in a time interval is q= ……. . 1. 2 Under steady-state condns, the current is the same for planes aa’, bb’ and cc’ and for all planes that pass completely thru the conductor. This implies that the charge is conserved.

Current density J This can be used to describe the flow of charge thru

Current density J This can be used to describe the flow of charge thru a x/section of the conductor at a particular point. For each element of the x/section, the magnitude of J is equal to the current per unit area thru that elt. i=J. d. A where d. A is the area vector of the elt └ar to the elt. Total current i= …. 1. 3 If the current is uniform across the sfc and //e/ to d. A, then J is also uniform and //e/ to d. A. i=………. . J=i/A S. I units is A/m 2 …. . 1. 4 Where A is the total sfc area

Drift speed When no current is passed across a conductor, its conduction es move

Drift speed When no current is passed across a conductor, its conduction es move randomly with no net motion in any drn. But when it has current thru it, these es still move randomly but with lesser speed (drift speed vd) in drn opposite that of length L (as shown in the fig below) and assume all charge carriers move with the same drift speed and that the current density J is uniform across the wire’s x/sectional area A. The number of charge carriers in the wire is, n. AL, where n is the number of charge carriers per unit vol. The total charge of carriers in the length L each with charge e is q=(n. AL)e ………. 1. 5 Because the carriers all move along the wire with

speed vd, this total charge moves any x/sectiona of wire in the time interval,

speed vd, this total charge moves any x/sectiona of wire in the time interval, t= ……. … 1. 6 current I is the time rate of charge xfer across a x/section i=q/t = …… Write for J=i/A …. . J=nevd …………. 1. 7 Here, ne is carrier charge density. S. I unit is coulomb/m 3. Resistance and Resistivity Resistance b/w any two points of a conductor is determined by applying a p. d V b/w the points and measuring the current I that results;

R=V/i …… 1. 8 S. I unit is Volt/ampere = ohm Ω A conductor

R=V/i …… 1. 8 S. I unit is Volt/ampere = ohm Ω A conductor whose fxn in a cct is to provide a specific resistance is called a resistor. Symbol is …. In general, properties of mat’ls are often useful. The electric field E at a point and current density at that point are related to resistivity of the mat’ls as ρ= E/J …. 1. 9. S. I unit is Ωm. In vector form, E= ρJ …. 2. 0 Eqns 1. 9 & 2. 0 hold only for isotropic mat’l (mat’l whose electrical ppties are the same in all drns). Conductivity of mat’l is simply the reciprocal of resistivity, ……………. 2. 1 S. I unit is (Ωm)-1

In terms of current density and electric field, J=……. 2. 2 Consider the length

In terms of current density and electric field, J=……. 2. 2 Consider the length and x/sectional area of a wire as L and A rsply. If a p. d V exists b/w its end and the current density is uniform thruout the wire, the electric field will be constant for all points within the wire. E=V/L and J=i/A ρ= E/J …… Put V/i =R; ρ= RA/l …. . 2. 3 Most physical ppties of mat’ls vary with temp. for linear relation b/w ρ and temp; ρ- ρ0= ρ0 α (T- To) …. . 2. 4 To is a selected reference temp and ρ0 is resistivity at

that temp. α is temp coeff of resistivity. Ohm’s law In conducting devices, a

that temp. α is temp coeff of resistivity. Ohm’s law In conducting devices, a p. d V is applied across the device. The resulting current i is measured as V is varied in both magnitude and polarity. From fig 1 a below, a plot of i vs V is a straight line passing thru the origin, so the ratio of i/V (wch is the slope of the straight line) is the same for all values of V. This means, the resistance R=V/i of the device is indpdt of the magnitude and polarity of the applied p. d. thus, ohm’s law asserts that current thru a device is always directly proportional to the p. d applied to the device. The eqn V=i. R is defining eqn for resistance and it applies to all conducting devices whether they obey

ohm’s law or not. For instance, if p. d and i thru any device,

ohm’s law or not. For instance, if p. d and i thru any device, even a pn jxn diode (see fig 1 b) are measured, the resistance at that value of V is R=V/i ……… 2. 5 In general, for conducting mat’ls, ohm’s law is related by E= ρJ ……. . 2. 6 This shows that ohm’s law obeyed when the resistivity of the mat’l is indpdt of the magnitude and drn of the applied electric field. Consider an electron of mass m placed in an electric field E, the e will experience an accln given by Newton’s 2 nd law as a=F/m = e. E/m ……. 2. 7 During the random motion, the es make collisions and

acquire a drift speed vd=aτ (τ is average time b/w collisions, i. e, mean

acquire a drift speed vd=aτ (τ is average time b/w collisions, i. e, mean free time). Thus, at any instant, vd=aτ = e. Eτ /m …………. 2. 8 Substitute for vd from eqn 1. 7 vd=J/ne = e. Eτ /m E= ( ……. 2. 9 Compare this expression (2. 9) with eqn 2. 6, we find that ρ= …………. . 3. 0 Power in electric circuits consider a cct shown in the fig below, the battery maintains a p. d V across its terminals and across the terminals with a greater potential at terminal a than at terminal b. Bcos there is an external conducting

path b/w the two terminals of the battery and bcos the p. ds set

path b/w the two terminals of the battery and bcos the p. ds set up by the battery are maintained, a steady current I is produced in the cct, directed from terminal a to b. The amount of charge dq that moves b/w those terminals in time interval dt is equal to idt. This charge dq moves thru a decrease in potential of magnitude V, and thus its electric p. e decreases by amount d. U=dq V = idt V ………… 3. 1 The principle of conservation of energy indicates that the decrease in electric p. e from a to b is accompanied by a xfer of energy to some other form. The power associated with that xfer is the rate of xfer d. U/dt; d. U/dt = P = i. V ………………… 3. 2

Ex An electrical cable consists of 125 strands of fine wire, each having 2.

Ex An electrical cable consists of 125 strands of fine wire, each having 2. 65µΩ resistance. The same p. d is applied b/w the ends of all strands and results in a total current of 0. 75 A. (a) What is the current in each strand? (b) what is the applied p. d? and ©what is the resistance of the cable? Soln a. current in each strand i = total i/# of strands = 0. 75 A/125 = 6. 0 x 10 -3 A b

Therefore, the amount of work done on the charge dq to pass thru any

Therefore, the amount of work done on the charge dq to pass thru any cross-section of a cct in any time interval dt is dw. The e. m. f of this device is given as Ԑ = d. W/dq ………………. 1 Therefore, the e. m. f of an e. m. f device is the work done per unit charge in moving the charge from its lower potential terminal to its higher potential end. S. I unit is volt. Thus, an ideal e. m. f device has no internal resistance to the internal movement of charge from terminal to terminal. This implies that the p. d b/w the terminals is equal to its e. m. f. However, a real e. m. f device (e. g a battery) has internal resistance to the internal movement of charge. When it is not connected to a cct, i. e, when no current flows thru it, the p. d b/w its terminals is equal to its e. m. f but the moment it has current thru it, the p. d b/w its terminals differs from its e. m. f, the device transfers energy to the charge carriers passing thu it and then to other devices in the cct.

CURRENT IN A SINGLE-LOOP CIRCUIT There are two equivalent ways to calculate the current

CURRENT IN A SINGLE-LOOP CIRCUIT There are two equivalent ways to calculate the current in the singleloop of cct below (Fig ii). One method is based on energy conservation and the other on the concepts of potential. The connecting wires in the cct are assumed to have negligible resistance. i. Energy Method Recall that in an ideal device, there is negligible or no internal resistance, therefore, the amount of work done in moving the charge must equal thermal energy that appears in the resistor. d. W = Ԑdq = Ԑ idt = i 2 Rdt (amount of energy in the resistor) Ԑ = i. R ii. Potential Method In a system of loop ccts, the p. ds at any point can be added algebraically. Thus, the algebraic sum of the changes in potentials encountered in a complete traversal of any loop

of cct must be zero. This is a statement of Kirchhoff’s loop rule. Consider

of cct must be zero. This is a statement of Kirchhoff’s loop rule. Consider the cct in fig ii above, to determine the change in potential about point a clockwise directions Ԑ - i. R = 0 ≡ i = Ԑ/R Now, resolving anticlockwisely; - Ԑ + i. R = 0 ≡ i = Ԑ/R Therefore, for complex or multiloop ccts, there are two criteria for finding the p. d around the loop; i. Resistance Rule: For a move thru a resistance in the direction of current, the change in potentials is –i. R; in the opposite directn it is +i. R ii. E. M. F rule: For a move thru an ideal e. m. f device in the direction of the e. m. f arrow, the change in potential is +Ԑ; in the opposite direction it is -Ԑ. Internal Resistance Consider a cct in fig iii below, it consists of a real battery with an internal resistance r, wired to an external resistor.

The int resistance of the battery is the electrical resistance of the conducting mat’ls

The int resistance of the battery is the electrical resistance of the conducting mat’ls of the battery, it is an irremovable feature of the battery. Applying loop rule in a clockwise drn at point a, the changes in potential gives Ԑ - ir- i. R = 0 ≡ i = Ԑ/R+r Resistances in Series In the fig iv below, three resistances are connected in series to an ideal battery with an e. m. f Ԑ. Applying the loop rule, Ԑ - i. R 1 – i. R 2 - i. R 3 = 0 (clockwisely) i = Ԑ/R 1+R 2+R 3 = Ԑ/Req Potential Difference b/w Two Points The p. d b/w two points a & b in the cct in fig v below can be determined by estimating the changes in potential passing thru d battery. Va + Ԑ - ir = Vb Vb - Va = Ԑ - ir

but for i across the entire cct, Ԑ - ir- i. R = 0

but for i across the entire cct, Ԑ - ir- i. R = 0 now, Vb - Va = ԐR/(R+r) ≈ 8 V ≡ i = Ԑ/R+r ≈ 12/ 4+2 = 2 A OR, If we wish to avoid the battery, we resolve anticlockwisely passing thru the resistor R implying moving in opposite drn to current. Va + i. R = Vb Vb - Va = i. R == 2(4) = 8 V Grounding a Circuit This means to connect a cct to the conducting path to earth’s surface. The potential at the grounding point in the cct is zero. Consider grounding the cct in fig v above in the fllg ways: Power, Potential & E. M. F When an e. m. f device works on the charge carriers to

establish a current i, the device transfers energy from its source (e. g chemical

establish a current i, the device transfers energy from its source (e. g chemical source in case of a battery) to the charge carriers. The rate of this energy transfer to the charge carrier is P = i. V = i(Ԑ - i. V) = iԐ - i 2 r Example: In the cct below determine the current I given that Ԑ 1 = 4. 4 V, Ԑ 2 = 2. 1 V, r 1 = 2. 3 Ω, r 2 = 1. 8 Ω amd R = 5. 5 Ω. Solution Applying loop rule from point a in a counterclockwise drn -Ԑ 1 + ir 1 + i. R + ir 2 + Ԑ 2= 0 - 4. 4 + (2. 3 + 5. 5 + 1. 8)I + 2. 1 = 0 ≡ i = 0. 24 A In clockwise drn, Ԑ 1 - Ԑ 2 - ir 1 - i. R - ir 2 = 0 ≡ i = 0. 24 A Multiloop Circuits Consider a cct containing more than one loop in fig vi. To

solve this kind of complex cct we apply both conservation of energy (loop rule)

solve this kind of complex cct we apply both conservation of energy (loop rule) and conservation of charge (junction rule). By kirchhoff’s jxn rule, the sum of the currents entering any jxn must be equal to the sum of the currents leaving that jxn. Firstly, considering loop badb starting from jxn b and going conterclockwisely, Ԑ 1 – i 1 R 1 + i 3 R 3 = 0 if clockwisely; - i 3 R 3 + i 1 R 1 - Ԑ 1 = 0 For the loop bdcb, and also counterclockwisely, -i 3 R 3 – i 2 R 2 - Ԑ 2 = 0 if clockwisely; Ԑ 2 + i 2 R 2 + i 3 R 3 = 0 For the big loop badcb counterclockwisely, Ԑ 1 – i 1 R 1 – i 2 R 2 - Ԑ 2 = 0 if clockwisely; Ԑ 2 + i 2 R 2 + i 1 R 1 - Ԑ 1 = 0 Then, for the conservation of charge, i. e, applying jxn rule, i 1 + i 3 = i 2

Resistances in Parralel Two or more resistors connected in //e/ to a battery of

Resistances in Parralel Two or more resistors connected in //e/ to a battery of emf Ԑ implying that the applied p. d across then is maintained, but producing a current thru each. This shows the p. d is the same but the current varies. In the cct below (fig vii) current in each resistor is given by i 1 = V/R 1; i 2 = V/R 2, and i 3 = V/R 3 Applying jxn rule at point a, i = i 1 + i 2 + i 3 RC Circuit This type of cct allows us to examine the charging process with time. It consists of a capacitor, an ideal batery of emf Ԑ

and a resistance R. Consider the cct in fig viii below. As soon as

and a resistance R. Consider the cct in fig viii below. As soon as the cct is complete, charge begins to flow b/w a capacitor plate and a battery terminal on each side of the capacitor. This current increases the charge q on the plates and the p. d Vc (=q/C) across the capacitor. When the p. d equals the p. d or emf of the battery (ideal), the current is zero and the capacitor is fully charged. Applying loop rule in a clockwise drn from a terminal of the battery, Ԑ -i. R – Vc = 0 where Vc = q/C (p. d across the capacitor) substitute for i = dq/dt and rearrange; Ԑ = R dq/dt + q/C (charging eqn) ………. 6 Now, assuming at the initial stage, the capacitor is uncharged, i. e, q=0 and time t=0; charge q is obtained by solving eqn 6 q = C Ԑ [1 - e-t/RC] …. (Charging a capacitor) ……. 7 This implies when t=0, e-t/RC goes to unity and q sums to zero (q=0).

But as charging continues and lets assume the time t gets to infinity (∞)

But as charging continues and lets assume the time t gets to infinity (∞) with the process, then e-t/RC goes to zero. Thus full charge on the capacitor is given by …. . q=CԐ …. . 8 The current i(t) charging the capacitor is the derivative of q(t) i=dq/dt = (Ԑ/R) e-t/RC …………… 9 To find p. d across the capacitor, Vc =q/C and compare with eqn 7; …………… 10 Therefore, at initial stage when t=0, Vc=0; and as it becomes fully charged at t~∞, Vc = Ԑ. The product RC is called capacitive time constant of the cct. It is represented as Ʈ=RC. To now examine how a fully charged capacitor discharge with time, Ata new time t=0 after it has been fully charged to potential V 0 equal to the Ԑ of the battery, and now take out the battery so that the capacitor discharge through its resistance R, then Ԑ=0 and eqn 6 above now becomes

0 = R dq/dt + q/C …… (discharging eqn) ……… 11 To solve this

0 = R dq/dt + q/C …… (discharging eqn) ……… 11 To solve this eqn and write for q through the capacitor q = q 0 e-t/RC (discharging a capacitor) ……. . 12 From i= dq/dt, i= -(q 0/RC) e-t/RC …………. . 13 This implies that the current also decreases exponentially with time, at a rate set by the capacitive time constant Ʈ=RC. Then, at time t= Ʈ, the capacitor charge q reduces to = q 0 e-1 (or ~37% of the initial value) Now, to estimate the initial current io, apply loop rule to the cct at time t=0, note that the capacitor’s initial potential V o is connected across R, io = Vo/R = q/C. 1/R = qo/RC …………………. . 14

Examples 1. In the fig below, what value must R have if the current

Examples 1. In the fig below, what value must R have if the current i in the cct is to be 1. 0 m. A? Take E 1=2. 0 V, E 2=3. 0 V and r 1=r 2=3. 0Ω. (b) what is the rate at which thermal energy appears in R soln 2. An ideal battery of e. m. f=12 V is connected to a network of resistances as shown below. If R 1=6Ω, R 2=12Ω, R 3=4Ω, R 4=3Ω and R 5=5Ω. What is the p. d across resistance 5? soln

3. A 100 W lightbulb is plugged into a standard 120 V outlet. (a)

3. A 100 W lightbulb is plugged into a standard 120 V outlet. (a) How much does it cost per 31 -day month to leave the light turned on continuously? assume electrical energy cost $0. 06 k. Wh. (b) what is the resistance of the bulb? (c) what is the current in the bulb? soln 4. A 10 V battery is connected to a series of n capacitors each of capacitance 2. 0µF. If the total energy stored in the capacitors is 25µJ, what is n? soln