Download Capacitance – Dr Taleatu`s Lecture note

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…
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 and the vectors E and dA 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 & dS 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 = ….. 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 uF) and a battery (V=20V). Thickness and
face area of the lower plate (copper material)
Capacitor are 6.0mm and 2.4x10-2m2 respectively.
What depth would electrons attain within the plate
face as the capacitor becomes charged? Density of
conduction electrons of Cu is 8.49x1028es/m3.
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
= ……… 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 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 1a,
q1= C1V….
Total charge q= q1 + q2 + q3
= C1V +C2V+C3V
Ceq= C1+C2+C3 = 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,
V1 ……, V2….
For n capacitors, ……. 2.6
Energy stored in an electric field
In attempt to transfer electrons from one plate of a
capacitor to othe 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 required will be
dW=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 CV2 ….. 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 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 //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 E0 b/w plates without a
dielectric, we enclose the charge +q on the top plate
with a Gaussian sfc and then apply Gauss’ law;
……..
E0=……. 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 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 E0
by a factor k,
E= E0/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 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 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.dA
where dA 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 dA,
then J is also uniform and //e/ to dA.
i=………..
J=i/A S.I units is A/m2 …..
1.4
Where A is the total sfc area
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, nAL, 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=(nAL)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,
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/m3.
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 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 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 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 1a
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=iR 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, even a pn jxn diode (see fig 1b) 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 2nd law as
a=F/m = eE/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 free time). Thus, at any instant,
vd=aτ = eEτ /m ………….2.8
Substitute for vd from eqn 1.7
vd=J/ne = eEτ /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 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
dU=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 dU/dt;
dU/dt = P = iV ………………… 3.2
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 0.75A. (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
= 0.75A/125 = 6.0 x 10-3A
b
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 = dW/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 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 the thermal energy
that appears in the resistor.
dW = Ԑdq = Ԑ idt
Ԑ idt = i2Rdt (amount of energy in the resistor)
Ԑ = iR
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 the cct in fig ii above, to determine the
change in potential about point a clockwise directions
Ԑ - iR = 0
≡ i = Ԑ/R
Now, resolving anticlockwisely;
- Ԑ + iR = 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 –iR; in the opposite directn it is +iR
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 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- iR = 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 - iR1 – iR2- iR3 = 0 (clockwisely)
i = Ԑ/R1+R2+R3
= 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- iR = 0
now,
≡ i = Ԑ/R+r
≈ 12/ 4+2 = 2A
Vb - Va = ԐR/(R+r)
≈ 8V
OR,
If we wish to avoid the battery, we resolve anticlockwisely passing
thru the resistor R implying moving in opposite drn to current.
Va + iR = Vb
Vb - Va = iR == 2(4) = 8V
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 source in case of a battery) to the charge carriers.
The rate of this energy transfer to the charge carrier is
P = iV
= i(Ԑ - iV) = iԐ - i2r
Example:
In the cct below determine the current I given that Ԑ1 = 4.4V, Ԑ2
= 2.1V, r1 = 2.3 Ω, r2 = 1.8 Ω amd R = 5.5 Ω.
Solution
Applying loop rule from point a in a counterclockwise drn
-Ԑ1 + ir1 + iR + ir2 + Ԑ2= 0
- 4.4 + (2.3 + 5.5 + 1.8)I + 2.1 = 0
≡ i = 0.24 A
In clockwise drn,
Ԑ1 - Ԑ2 - ir1 - iR - ir2 = 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) 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,
Ԑ1 – i1R1 + i3R3 = 0
if clockwisely; - i3R3 + i1R1 - Ԑ1 = 0
For the loop bdcb, and also counterclockwisely,
-i3R3 – i2R2 - Ԑ2 = 0
if clockwisely; Ԑ2 + i2R2 + i3R3 = 0
For the big loop badcb counterclockwisely,
Ԑ1 – i1R1 – i2R2 - Ԑ2 = 0
if clockwisely; Ԑ2 + i2R2 + i1R1 - Ԑ1 = 0
Then, for the conservation of charge, i.e, applying jxn rule,
i1 + i3 = i2
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
i1 = V/R1;
i2 = V/R2, and i3 = V/R3
Applying jxn rule at point a,
i = i1 + i2 + i3
1
V V V
1
1
1
 

V  
   V.
R1 R2 R3
Req
 R1 R2 R3 
1 1
1
1
 
 
Req  R1 R2 R3 
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 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 -iR – 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
(∞) 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;

q
Vc    1  e t / RC
C

……………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 V0
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 eqn and write for q through the capacitor
q = q0 e-t/RC (discharging a capacitor) ……..12
From i= dq/dt,
i= -(q0/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
= q0 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 Vo
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
i in the cct is to be 1.0mA? Take E1=2.0V, E2=3.0V and
r1=r2=3.0Ω. (b) what is the rate at which thermal energy
appears in R
soln
2. An ideal battery of e.m.f=12V is connected to a
network of resistances as shown below. If R1=6Ω,
R2=12Ω, R3=4Ω, R4=3Ω and R5=5Ω. What is the p.d
across resistance 5?
soln
3. A 100W lightbulb is plugged into a standard 120V
outlet. (a) How much does it cost per 31-day month to
leave the light turned on continuously? assume electrical
energy cost $0.06 kWh. (b) what is the resistance of the
bulb? (c) what is the current in the bulb?
soln
4. A 10V 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