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CK Cheung
ELECTROSTATICS
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Experimental facts:
1/ There are 2 , & only 2, kinds of electric charge
2/ Like charges repel, unlike charges attract.
3/ Fundamental charge = e- = p+ = 1.6 x 10-19 Coulomb (C).
Coulomb’s Law
(For point charges)
+Q
F
r
+Q
F
By experiment,
F
1
r2
F
F  Q1Q2
F = k (
Q1Q2
)
r2
Q1Q2
r2
where k = 9x109 Nm2/C2
vacuum)
In general,
Q1Q2
F=
4 0 r 2
1
(0 = 8.85418x10-12 C2/Nm2)
Called permittivity constant
1
CK Cheung
Properties of Coulomb Force:
1)
Obey Superposition Principle
..\..\powerpoint\couloml,s law\superposition of forces.ppt
+Q
r1
1
2)
3)
+Q
-Q
F1
F2
r2
1
A conservative force
Acts along the line joining the 2 point charges.
e.g.
2 identical point charges repel each other with a force 0.1 mN. They
are moved 5 mm further apart, and the repulsive force is reduced
to 25  N.
1.
2.
How far apart were they originally?
What was the size of the charges.
1987-I-30
+Q
5 mm
 5.3x10-10C
-Q
W
X
Z
Y
q
-Q
Three charges +Q, -Q and -Q are fixed at the corners W, X and Y
respectively of a square as shown. A fourth charge, q, is fixed at Z, after
which the charge at X experiences a NET electrostatic force indicated by
the arrow. q is equal to
A.
+Q
B
+ 2Q
C
+ 4Q
D
+
E
+2
Q
2
2
Q
2
CK Cheung
Electric Field
Q
+q
r
P
Force
transmitted
by what ?
field
concept
An electric field is said to exist at a point P if a force is exerted on a
stationary test charge (+) placed at that point.
Definition:
The electric field strength, E, at P is:
E=
F
q
NC-1
Qq
4 0 r 2
E=
q
1

Hence,
E=
Q
4 0 r 2
1

1
(for radial field only)
r2
Note:
1/ Actually, q must not carry too many charge (q
0 ), as not to affect
the charge distribution of Q.
(Why do we not define E as numerically equal to the force acting on
a charge of one coulomb?)
2/ F = qE
3/ Apply to region outside any spherical conductors (~ point charge).
3
CK Cheung
Representation of E
Use field line concept:
1)
The tangent to a line of force at any point gives the direction of E at
that point.
E shows the initial direction of motion
of a resting positive charge.
2)
The number of field lines drawn per unit cross-sectional area is
proportional to the magnitude of E
Strong E
weak E
Note:
 = E A = EAcos
Called electric flux
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CK Cheung
Properties of electric field lines
1/ They begin from +ve charge & end on –ve charge
-
+
2/ They cannot cross (otherwise, direction of E at a point is not unique)
..\..\powerpoint\E-field line property.ppt
3/ They cut equipotential surfaces (i.e. surface of equal potential
e.g. conductor) at 900
+ +
+ +
+
+
+ +
+
+
+
conductor
+
+
+
+
+
+
+
+
4/ The electric potential is decreasing along the electric field line
direction.
5
CK Cheung
1984-I-48.
Y
Z
X
The diagram shows a pattern of electric field lines in which X, Y and Z are points marked on one
of the field lines. It would be correct to say that
(1)
(2)
X is at a higher potential than Z.
a negative charge placed at Z would accelerate to the left along the tangent to the field
line at Z.
(3)
the force exerted on a charge at Y would be greater than if the charge were placed at X.
A.
(1), (2) and (3)
B.
(1) and (2) only
C.
(2) and (3) only
D.
(1) only
E.
(3) only
1989-I-30
IV
X
III
II
E
I
Y
E
A particle carrying a negative charge is free to move in a uniform electric field E. If the particle
starts with a certain velocity from point X, which of the paths shown could represent the route
which the particle would follow from X to Y ?
A.
I and II only
B.
I and IV only
C.
II and III only
D.
III and IV only
E.
I, II, III and IV
1999-II-21(AL)
Which of the following statements about electric field lines is incorrect ?
A.
They are closest where the field is strongest.
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CK Cheung
B.
They are always perpendicular to equipotential lines.
C.
They always point from high electric potential to low electric potential.
D.
Work has to be done in moving an electron along the direction of a field line.
E.
They tend to attract one another.
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CK Cheung
Relative magnitude of gravitational & electrostatic forces.
+
R
Fe =
1 Q1Q2
40 R 2
(9 x10 9 )(1.6 x10 19 ) 2
=
R2
Fg =
GMeMp
(6.67 x10 11 )(9.11x10 31 )(1.6 x10 27 )
=
R2
R2
Fe

Fg
10 39
∴ In general , Fe >> Fg
..\..\powerpoint\This demonstrates that Coulomb force ( not
gravitational.ppt
This demonstrates that Coulomb force ( not gravitational forces ) are
responsible for binding :
 Electrons to nuclei to form atoms
 Atoms to atoms to form molecules
 Molecules to molecules to form solids and liquids
8
CK Cheung
E – field for a group of point charges
..\..\powerpoint\E-field due to a system of charges.ppt
P
Calculate E-field due to individual charge at the given point as if it were
the only charge present ( i.e. superposition principle ) and then add
vectorially.
9
CK Cheung
e.g.
Refer to the following diagram, in which r >> 2a. Find, in magnitude,
EL
?
ET
T
r

-q
2a
L
+q
r
Q
1
1
(

)
EL =
2
4 (r  a)
(r  a) 2
ET =
=
Q
1
4
(( a  r ) )
2
2
4
1
2
2
Q
1
(cos Θ) x 2 =
4
2
a
1
2
(( a  r ) ) (a  r )
2
2
2
2
2
1
2
x2
3
Qa
(a  r )
2
2
2
Q 1  a 
 a   Q 1 4a
( )
=
1    1    =
2
4 r 2  r 
 r   4 r r
2
Hence, EL/ET = 2
3
2
=
2 Qa
a2 2
(
1

)
4 r 3
r2
0
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CK Cheung
e.g.
A point charge +Q is placed a distance R from an infinite conducting
thin sheet (earthed) on which it induces a charge
-Q.
a/ draw the electric field lines for this arrangement.
..\..\powerpoint\e-field lines.ppt
-Q
+Q
(note: -Q is not evenly distributed on the surface of the conducting sheet)
b/ draw the field pattern associated with point charges +Q & -Q
separated by a distance of 2R.
-Q
+Q
R
c/
compare (a) & (b) , find the attractive forces that the plane & the point charge
+Q exert on each other.
..\..\powerpoint\e-field lines2.ppt
F=
Q2
4 0 (2 R) 2
1
11
CK Cheung
Formula:
1/ Infinite conducting thin sheet
+
Uniform E =

2 0
σ = surface charge
density
2/ 2 infinite sheets
..\..\powerpoint\e-field lines3.ppt
+
+
Uniform E =

0
E=0
Uniform E =
12

0
CK Cheung
3/ Infinite conducting wall
Uniform E =
E=0

0
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CK Cheung
Electric Potential
Consider:
hill
WD = mgh = PE difference between
top & bottom
g
Similiary:
B
+q
If we carry +q from A to B, &
work must be done = WAB
 there exists potential
difference between A & B.
A
Definition:
ΔV = VB – VA =
W AB
Volts ( V )
q
Note:
1/ the charge must be in steady speed when moving it from A to B,
otherwise WAB ≠ unique value.
2/ If A →  , define VA = 0
 VB =
W B
 definition of electric potential at a point
q
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CK Cheung
Consider:
X
E
Bring +q from X to Y:
qE
Y
W XY
VY-VX =
=
q
+q
  qE  d x
X
q
Y
∴ ΔV =
-  E  dx
X
from maths.:
FApply = -qE
E=-
dV
dx
Y
-ve  E points in the
direction of decreasing
potential.
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CK Cheung
e.g.
Find the potential difference between A & B :
rA
rB
FApply
+Q
+q
B
W AB
VB – VA =
=
q
 VB – VA =
B

(
A
1 Qq
)dl
4 r 2
=
q
E
dl
B
1 Q
A 4 r 2 (- dr) =
rB
A
1 Q
 4 r
2
(- dr)
rA
Q 1 1
(  ) >0
4 rB rA
 potential decreases along the direction of E-field lines
If
r A    VA ~ 0
∴ VB =
1
Q
)
4 rB
(
hence, in general:
V=
Q
4 r
1
Note:
1. V > 0
or
V<0
2. For radial field only
3. If +Q is a sphere  treat it ~ point charge ( ∵ also radial E-field),the formula
holds for r > radius of sphere.
16
CK Cheung
Potential due to a group of charges:
P
VP =
V
i
i
=

i
1  qi
(
)
4 ri
( algebric sum )
e.g.
Q1
Q2
a
What is the electric potential at
the center of the square?
Given: Q1=+1x10-8 C
Q2= -2x10-8 C
Q3=+3x10-8 C
Q4=+2x10-8 C
Q4
V=
Q3
V
i
i

a=1m
Q1  Q2  Q3  Q4
1 1 2  3  2
(
)(10 8 )  500 V
) =
4
0.7
4
r
1
(
17
CK Cheung
Potential & E-field inside conductors
1/ thin spherical shells
+
+
+
+
E-field = 0 inside the outer
surface because:
 Current = 0
 All charges are at outer
surfaces
R
∵ E=
+
+
∴ if E
-
dV
dx
=0
dV
=0
dx
 V = constant of position
 All points are at the same potential inside the outer surface
 potential at all points inside the shell = surface potential of shell
+
+
+
+
R
+
R
+
∴ surface potential =
Q
)
4 R
1
(
For r  R
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CK Cheung
2/ Solid conducting sphere
potential =
Q
)
4 R
1
(
Hence: for Shell or sphere:
R
E=
1 Q
4 r 2
E=0
V=
V=
1 Q
4 r
1 Q
4 R
19
CK Cheung
e.g.
Find the p.d. between the two concentric spherical shells?
q
Q
R
r
Vr =
1 Q
q
+
4 R
4 r
VR =
1 q
1 Q
+
4 R
4 R
1
 Vr – VR =
q
1 1
(  )
4 r R
Independent of Q
20
CK Cheung
e.g.
Two charged conducting spheres are connected as shown. In steady
state, the charges on them are q1 & q2
σ1
q1
Long
wire
R1
R2
Prove that
 1 R2

 2 R1
σ2
q2
Proof:
V1 =
1 q1
( )
4 R1
∵ V1 = V2


&
1 q1
( )
4 R1
V2 =
=
1 q2
( )
4 R2
1 q2
( )
4 R2
q1 R1

q 2 R2
(q  R)
q1
R R2

4R12
q R2
∴ 1 
= 1 . 22 = ( 1 )( 22 )
q2
2
R2 R1
q 2 R1
2
4R 2

 1 R2

 2 R1

point action

( 
1
)
R
21
CK Cheung
E – field between 2 parallel plates
E
+
-
+
+
+
+
-
+q
+
+
+
-
X
∵ ΔV =
or
WD (qE ) X

 EX
q
q
E=
V
X
For uniform field only
potential
E-field
0
X/m
22
CK Cheung
Potential due to neighboring bodies
-q
P
+Q
R
Vp =
1
Q
)
4 R
(
r
Vp =
-
1
q
)
4 r
(
23
CK Cheung
e.g.
Sketch the V-X & E-X graphs both before & after the conducting
slab is inserted. (i.e. 4 curves in 2 graphs)
+
+
+
+
+
+
+
+
-
..\..\powerpoint\e-field & potential.ppt
V/V
E/Vm-1
X/m
X/m
24
CK Cheung
Electric Potential Energy
rA
rB
FApply
+Q
+q
B
E
A
By def.:
∵ ΔV = VB-VA =
W AB
q
∴ WAB = q (VB-VA)
from above diagram:
 WAB = q (
1 Q
1 Q
)
4 rB
4 rA
If rA =  , take VA = 0
∴
WB = qVB =
1 qQ
4 rB
= electrostatic P.E. of the system
Def:
We define the electric P.E. of a system of point charges as the work
required to assemble the system of charges by bringing them in from an
infinite distance.
In general:
U=
1 ( q1 )(  q2 )
(
)
4
r
>0 ( repulsion)
Note:
U
<0 (attraction)
25
CK Cheung
e.g.
Three charges are held fixed as shown. What is the electrical P.E. of
the system? Take q = 1x10-7 C, a = 10 cm.
( hint: use superposition principle)
2
-4q
a
a
1
+q
a
3
+2q
Soln.:
Use superposition principle:
U
= U12 + U23 + U13
=
1  ( q)( 4q) ( q)( 2q) (4q)( 2q) 



4 
a
a
a
= -9x10-3 J
Note:
U <0,  +9x10-3 J of work must be done to separate the structure.
26
CK Cheung
e.g.
A particle of charge +Q is fixed at P. A second particle of mass m and
charge –q moves at constant speed, in a circle of radius R1, centered at P.
Find the work that must be done by an external agent on the second
particle in order to increase the radius of the circle of motion to R2.
R2
-q
m
+Q
R1
At r = R1:
F=
1 Qq
mv 2
=
4 R12
R1
1 Qq
mv 2
=
8 R1
2
1 Q(q)
 P.E. =
4 R1
 K.E. =
1
∴ Total energy , E1, = K.E. + P.E. = -
Qq
8 R1
Similarly, Total energy, E2, at r = R2 = -
Qq
8 R 2
W.D. = E2 – E1 =
1
Qq 1
1
(  )
8 R1 R2
27
CK Cheung
Capacitance
Consider any pair of isolated conductors
+q
-q
V
∵ V  q
∴
q
V
= constant, called capacitance of the system,
C
or
C=
q
V
units of C :

CV-1
 Farad ( F )
 1 uF = 10 –6 F
28
CK Cheung
Calculating Capacitance
1/
Spherical shells:
-Q
+Q
r
R
Vr =
1 Q
1 Q
+
4 R
4 r
VR =
1 Q
1 Q
+
4 R
4 R
 Vr – VR =
∴ C=
Q 1 1
(  )
4 r R
Q
=
V
4
1 1
(  )
r R
Note:
1. C
depends on
2.
If
R=  
Isolated sphere  C = 4  r
3.
If
C
 r=
= 1 F,
geometrical factors
1
4
only
= 9x109
m !! ( c.f. RE
)
29
CK Cheung
2/ Parallel plate type
Area = A
+q
E
d
-q
∵ E=

q

 q = EA 
 A
Also, V = E(d)
∴ C=
q EA

V
Ed
 C= ε
A
d
30
CK Cheung
e.g.
a/ A parallel-plate capacitor whose plates have an area 1 m2 and which
are separated by 2 mm is connected across the terminals of a 100V
battery. Find:
Area=1m2
100 V
0.002m
1/ the electric field between the plates.
∵ V = Ed
∴ E=
100
 5 x10 4 NC-1
0.002
2/ the magnitude of the charge density on the plates
∵ E=

   E  (5x10 4 )(8.85x10 12 )  4.43x10 7 C m-2

3/ the charge on either plate
Q = A  4.43x10 7 (1) = 4.43x10-7 C
4/ the capacitance of the system
Q 4.43x10 7

 4.43x10 9 F
∵ C=
V
100
31
CK Cheung
b/ A conducting plate 1 mm thick is now inserted between the plates,
the battery being still connected, Find:
1/ the electric field inside the conducting plate
E=0
inside conductors
2/ the electric field in the remaining air space
∵ V = E d
 100 = E(0.002-0.001)  E = 105 Vm-1
3/ the induced surface charge density on the faces of the inserted
plates.
∵ each plate induces equal charges on the surface of the conducting plate.
∴ E=

   E  (10 5 )(8.85x10 12 )  8.85x10 7 Cm-2

Q = A  (8.85 x10 7 )(1)  8.85 x10 7 C
4/ the capacitance of the system
Q 8.85 x10 7
∵ C=

 8.85x10-9 F  C increases
V
100
C/
If the conducting slab is inserted after the
capactor is disconnected from the
100V battery, repeat part (b) above
32
CK Cheung
Capacitors in parallel
q1
V
C1
q2
C2
q1 = C1V
q2 = C2V
∴ total charge Q = q1 + q2 = ( C1+C2)V

Q
 C1  C 2
V
= CEQUIVALENT = Ceq
∴ The system is equivalent to a single capacitor of
capacitance:
Ceq = C1+C2+…………..
Note:
Charge on Ceq = q1+q2+………..
33
CK Cheung
Capacitors in series
+q
+q
-q
-q
C2
C1
+ V
Total charge
stored in the
system = q
-
∵ V = V1 + V2
=
∴
q

V
q
q
1
1

 q( 
)
C1 C 2
C1 C 2
1
1
1
( 
)
C1 C 2
= Ceq
Hence , the system is equivalent to a single capacitor
of capacitance:
1
1
1


 ..................
Ceq C1 C 2
Note:
1/ Qeq = qi (charge on any one )
2/ charge stored in each capacitor is the same
34
CK Cheung
C1=1  F
e.g.
C3=3  F
S2
C2=2  F
C4=4  F
12 V
S1
Find the charge on each capacitor when:
a/ S1 is closed
b/ S2 is also closed
solution:
a/ Given circuit is equivalent to :
C
C*
12 V
1
1
1
3


C  F
C C1 C3
4
1
1
1
4


 C*   F
*
C2 C4
3
C
∵ q = CV = (
also:
3
 F)(12) = 9x10-6
4
q* = C*V = (
C = q1 =q3
4
 F)(12) = 16x10-6
3
C = q2 =q4
35
CK Cheung
b/
C1=1  F
C2=2  F
C3=3  F
C4=4  F
12 V
Hence, given circuit is equivalent to:
C1=1  F
V1
C2=2  F
C3=3  F
V2
C4=4  F
12 V
∵
q1 + q 2 = q 3 + q 4
 V1 + 2V1 = 3V2 + 4V2
 3V1 = 7 V2------------- (1)
Also,
V1 + V2 = 12----------(2)
 V1 = 8.4 V & V2 = 3.6 V
 q1 = C1V1; q2 = C2V1 ; q3 = C3V2; q4 = C4V2
36
CK Cheung
87’ MC 28
e.g.
A
3μF
4μF
C
1200 V
450 V
1μF
2μF
D
B
If after some time, the voltage across CD suddenly jumps to 600V, which capacitor(s)
has been shorted ( i.e. become a conductor)?
1/
2 μF shorted:
 VCD = 0
2/
1 μF shorted
 charges on 4 μF & 2 μF will neutralize each other
 VCD = 0
3/
3 μF shorted:
If VCD = 600 V,  VAC also = 600 V !  impossible !
∵ Q2 = Q4 ( series) but C2  C4  V2  V4
4/
4 μF shorted:
 given circuit is equivalent to:
 answer.
3μF
600V
1200V
3μF
600V
37
CK Cheung
Energy stored in a capacitor
-q
+q
V
C
Vo
Let q = charge at any instant before the capacitor is saturated
(q & V increase, until q = Q & V = Vo
∵ WD = qV  dw = V(dq) =
Q
∴ WD =
but
C=
q
 cons tan t )
V
q
(dq )
C
q
 (dw)   ( C )dq
0
1 Q2
 WD =
2 C
or
 WD =
1
CVo2
2
 energy stored in the capacitor in the form of E-field
38
CK Cheung
e.g.
S
Qo
Vo
V
C1
C2
C1 is charged to V0 and Qo and then connected to a C2 ( uncharged ) by closing S.
1/
what is the final p.d. across the combination?
Qo = Q1 + Q2
 C1V0 = C1V + C2V
C1
 V = V0 (
)
C1  C 2
This suggest a way to measure an unknown
C2(e.g.) in terms of a known one.
2/
what is the stored energy before & after S is closed?
Initial energy = U0 =
1
C1V02
2
Final energy = U f =
1
1
1
C1V 2 + C 2V 2 = V 2 (C1  C 2 )
2
2
2
=
VC
1
(C1  C 2 )( 0 1 ) 2
2
C1  C 2
= (
C1
)U 0
C1  C2
U f< U0 the missing energy appears as heat in the connecting
wires as the charges move through them.
39
CK Cheung
e.g.
(a) Two // metal plates, each of area 900 cm2, are mounted 2 mm apart
in air. If the 2 metal plates are connected to the terminals of a 1.5 V
battery, find
1/
the capacitance of the capacitor,
C=
1.5 V
A
900 x10 4
 8.85 x10 12
d
2 x10 3
∴ C = 3.98x10-10 F
2/ the energy stored in the capacitor.
Energy =
1
1
CV 2  (3.98 x10 10 )(1.5) 2  4.48 x10 10 J
2
2
(b)
1.5 V
P
x
P is a point between the 2 plates and is at a distance x from the
lower plate. Draw sketches to show how
1/ the E-field at point P,
2/ the potential V at point P
varies with x.
1/
E=
V
1.5

 750Vm 1
x 2 x10 3
E/Vm-1
750
X/m
2x10
-3
40
CK Cheung
2/ V = Ex  linear
V/V
1.5
X/m
2x10
-3
(c) with the separation of the plates fixed at 2 mm, the top plate is
then shifted sideways so that the overlapping area of the plates is
reduced to 600 cm2. Neglecting the edge effects, what is the energy
stored in the capacitor if its plates:
2 mm
1/ remain connected to the 1.5 V battery
2/ are disconnected from the battery,
while the top plate is being moved?
In each case, why is the answer different from that in (a)(ii)? Explain
briefly.
A f (8.85 x10 12 )(600 x10 4 )
1/ Cf = ε

 2.66 x10 10 F
3
d
2 x10
1
1
Ef = C f V 2  (2.66 x10 10 )(1.5) 2  2.99 x10 10 J
2
2
Lesser energy stored in a system of lesser capacitance with same applied
voltage
41
CK Cheung
2/ ∵ disconnected from battery,  Q unchanged  Qf = Qi
Qi = CiVi = (3.98x10-10)(1.5) = Qf = 5.97x10-10 C
Ef =
1 Qi2 1 (5.97 x10 10 ) 2

 6.7 x10 10 J
10
2 C f 2 (2.66 x10 )
More energy stored in the system due to W.D. by external agent
in pulling the top plate.
spontaneous
42
CK Cheung
The RC ( dc) circuit
R
a
b
C
E
1/ charging
(S a)
At any time:
q
C
E=iR+

d
di
1 dq
(E) = R ( ) +
dt
C dt
dt
 0 = R(

- RC
di
1
(i )
) +
C
dt
di
 dt
i
i

- RC

i0


(1)
-RC ln (
t
di
 dt
i 0
i
)=t
i0
i  i0 e
hence,
t
(
)
RC
(Where: i 0 = original current at t = 0)
At t = 0,  i  i 0 & q = 0
From (1):  i 0 =
E
R
t
(
)
E
i  ( )e RC
R
43
CK Cheung
i /A
i0 
E
R
~ e-t
t/s
0
For the charges q on the capacitor:
∵ i
dq
dt
 q=
 (i)dt =
t
t
E ( )
0 ( R e RC )dt
 q = CE ( 1- e
(
t
)
RC
)
q/C
q0=CE
0
t/s
44
CK Cheung
Definition
Capacitative
= τ= RC
Time constant
At
t = τ  q = q0 ( 1- e
(
RC
)
RC
)
 q = q0 ( 1- e 1 )
 q = q0 ( 1- 0.37 )

q
 0.63
q0
 capacitor is charged up to 63 % of its equilibrium (saturated ) value
q/C
q0=CE
0.63q0
0
τ
t/s
Note:
1. if t > 3 τ  circuit can be assumed to be in steady state
2.
for calculation, put t =  for steady state
45
CK Cheung
e.g.
S
R
C
E
After how many time constant ( RC ) will the energy stored in the
capacitor reach half of its equilibrium value?
t
(
) 
1 q2
1 
RC 
∵ Ut =

CE
(
1

e
)
2 C 2C 

2
t
=
but,
(
)
1
CE 2 (1  e RC ) 2
2
U =
1
CE 2
2
 Ut = U  (1  e

(
t
)
RC 2
)
t
(
)
Ut
1
RC 2
= (1  e
) =
2
U
 t = 1.22 RC
46
CK Cheung
2/
Discharging ( S  b )
R
a
b
VC=
E
At t = 0,
VC=
i
i
Q
=
C
Q
C
C
E
VC
E
 i0
=
R
R
decreases as t
=
 i  - i0 e
(
t
)
RC
increases
 direction reversed
also,
 VC
=E e
(
t
)
RC
and
q=Q e
(
t
)
RC
VC/q/ i
t/s
47
CK Cheung
89’
26. An isolated spherical conductor of radius r is charged to a potential V.
The total electrical energy stored is
A.
V²/(80 r).
B.
V²/(40 r).
C.
V²/(20 r).
D.
V²  (20 r).
E.
V²  (40 r).
E
=
1
1
CV 2 = ( 4r) V2
2
2
V² ( 2r).
87’
29.
i /m A
10
5
t /s
0
10
20
30
40
A capacitor C is charged to a certain pd. and then discharged through a resistor R.
The variation of the current i with time t is shown in the above graph. Which of
the following is/are correct?
(1) The time constant of the circuit is about 15 s.
(2) The area under the graph is proportional to the energy stored in the capacitor.
(3) If the resistance of R is doubled, the current at t = 0 will also be doubled.
A.
(1), (2) and (3)
B.
(1) and (2) only
C.
(2) and (3) only
D.
(1) only
E.
(3) only
1/ correct
2/ E
= i2 R t = R  i 2 dt area under i2 – t curve
 incorrect
3/ i0 =
E
; if R increases  i0 decreases
R
 incorrect
91’ essay (6)--------self study
48
CK Cheung
85’ IIB
5.
GM tube
amplifier
R
C
microammeter
1V
50  s
Figure 2
Figure 3
Figure 2 shows a simplified diagram of a ratemeter. The signal
from the GM tube is fed into an amplifier. After shaping and
amplification, the output signal is a square wave of amplitude 1 V
and duration 50 s (see Figure 3).
Using an RC circuit, a
corresponding current signal is observed in a microammeter.
When the GM tube is brought near a radioactive source, a steady
pulse train of 100 pulses per second ( T= 0.01 s )is produced
and measured. The capacitance C and resistance R are of values 0.5
F and 3 k, respectively. The output impedance of the amplifier is
small during the charging of C but large during its discharging.
(a) (i)
Calculate the time constant for the RC circuit.
τ= RC = (3000)(0.5x10-6) = 1.5x10-3 s
(ii)
A student claims that the capacitor is almost completely
discharged during the interval between successive pulses.
Justify this claim.
50 μs
t
T=0.01s
t= 0.01 - 5x10-5 = 9.95x10-3s >> τ capacitor is almost
completely discharged.
49
CK Cheung
(b) Calculate the current through the microammeter.
In 1 pulse, q stored = CV = (0.5x10-6)(1) C
Current = q x frequency = (0.5x10-6)(100) = 5x10-5 A
(c) The original radioactive source is replaced by a weaker one
such that, on average, only one pulse per second is produced by
the GM tube. In order to have a steady reading on the
microammeter, should the time constant of the RC circuit be
increased or decreased? Explain your answer. (8 marks)
τ should be increased.
Whenτ >> interval between pulses  p.d. across the capacitor remains
more or less constant & thus gives a steady current.
50
CK Cheung
e.g. 81’ (4)
4. A student is asked to perform an experiment to demonstrate that the
charge on a capacitor is proportional to the p.d. across its terminals.
The circuit to be used is shown in Figure 2.
A
10 V
C
S
to oscilloscope y-input;
timebase off;
y-sensitivity set to
d.c. 2 V/cm
Figure 2
The principle of the experiment is to close switch S and charge the
capacitor linearly by continuously adjusting the variable resistor to
keep the current in the circuit constant at its initial value. The p.d.
across the capacitor will therefore increase linearly with time, and a
graph of the deflection of the oscilloscope spot in the y-direction
against time will be a straight line.
(a) The student decides that it would be most convenient to start with
the variable resistor set to its maximum value of 100 k and to
use an ammeter which he can maintain at full scale deflection
when the switch is closed. What should be the full scale
deflection of the meter?
i=
V
10

 10  4 A
R 100 x10 3
(b) Having selected his meter, the student considers the deflection of
the oscilloscope spot. He decides that he will be able to time a
deflection rate of 0.1 cm/s satisfactorily.
he choose?
What value of C should
Voltage changing rate = (2 V cm-1)(0.1 cm s-1 )
= 0.2 V s-1
∵ V=
Q it
V
i
10 4



 0.2 =

C C
t C
C
C = 5x10-4 F
51
CK Cheung
(c) What will be the value of the variable resistor 10 s after closing
the switch S?
After 10 s.
p.d.
R=
p.d. across C = ( 0.2 V s-1) (10) = 2 V
across R = 10 – 2 = 8 V
8
= 8 x 104
10  4
Ω
52
CK Cheung
86’ IIB 4
4.
V
S
R
12 V
C
A
Figure 3
A student performs an experiment in which a 500 F capacitor C is to
be charged at a constant rate, using the circuit in Figure 3. After
switch S is closed, the microammeter shows an initial current reading
of 120 A. By continuously adjusting the variable resistance R, the
charging current is maintained at this magnitude throughout the
subsequent charging process.
V /V
R
12
10
8
6
4
2
0
50
100
150
200
250
300
time / s
Figure 4
53
CK Cheung
At the same time, the student also notes down, from the voltmeter V,
the variation of the p.d. across R (VR) as a function of time. The
observed variation is shown in Figure 4. The student then draws the
conclusion that the capacitor is leaking. In answering the following
questions, you may assume that a leaking capacitor can be regarded
as a capacitor with a high resistance R0 (the leakage resistance)
connected in parallel.
R0
(a) Sketch on Figure 4, the expected variation of VR with time if the
capacitor is NOT leaking.
V /V
R
∵V=VR+
12
Q
C
it
V = VR + C
10
it
 VR= 12 - C
8
 st.
line
6
4
2
0
50
100
150
200
250
300
time / s
Figure 4
54
CK Cheung
(b) Explain the difference between the two curves, and indicate the
reason why the fact that VR approaches a constant value after 200
s suggests a leaking capacitor.
Since the capacitor is charged at a constant rate, the voltage
across the capacitor will also increase at a constant rate. So, the
voltage across the variable resistor will decrease at a constant
rate. This supports the curve for a not leaking capacitor to be a st.
line.
The reason for the voltage across the resistor drops to a constant
value is that when the capacitor is charged up (i.e saturated when
t ~ 200 s), a steady current will pass through the leaking resistor,
 voltage across the variable resistor is then steady.
(c) (i)
Sketch below the variation of charge Q stored in capacitor C
as a function of charging time t. (No numerical values to be
calculated.)
Q
T/s
200s
(ii)
Calculate the final charge stored on capacitor C.
Q = CV = (500x10-6)(12-2.4) = 4.8x10-3C
(iii) Calculate the leakage resistance R0.
(12 marks)
V=iR

(12-2.4) = (120x10-6)(R)

R = 80000 Ω
55