Download Individual ASSIGNMENT Assignment 6: Select Circuit Elements

Electricity and Magnetism: PHY-204
Fall Semester 2015
Individual ASSIGNMENT
Assignment 6: Select Circuit Elements
This homework must be solved individually.
Solution
1. A variable air capacitor used in a radio tuning circuit is made of N semicircular
plates each of radius R and positioned a distance d from its neighbors, to which it is
electrically connected. As shown in Fig. 1, a second identical set of plates is enmeshed
with the first set. Each plate in the second set is halfway between two plates of the
first set. The second set can rotate as a unit. Determine the capacitance as a function
of the angle of rotation θ, where θ = 0 corresponds to the maximum capacitance.
Fig. (1)
Answer
Recall first that capacitance of a parallel plate capacitor is given by
C=
ε0 A
d
where A is the area of each plate and d is the distance between two plates. This
problem looks scary at first. But it really isn’t! It is just the case of a large number
of parallel plates such that the area of adjacent plates inside capacitor is adjustable.
More clearly, as we increase the value of θ between 0 and π, the overlap between the
adjacent plates decreases. In fact, when θ = 0 we have the case of a large number of
parallel plate capacitors with the shape of plates being that of a semi circle. When
θ = π there is no area of overlap of the adjacent plates and the capacitance is zero.
But before we compute the actual value of capacitance as a function of angle θ we find
out what is the number of parallel plate capacitors formed by the two sets of N plates
each. Let us tale N = 3 for convenience and count the number of capacitors formed.
Date: 14 December, 2015, 5 : 00 pm
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Electricity and Magnetism: PHY-204
Fall Semester 2015
1
2
3
+
-
4
5
6
Here dotted lines show the set of moveable plates, while solid lines show the set of
fixed plates. Clearly the number of capacitors formed is 5. It is easy to check for other
values of N and conclude that the total number of capacitors formed by two sets of N
plates is 2N − 1. So the total capacitance Ctotal in terms of capacitance of one parallel
plate capacitance C is
Ctotal = (2N − 1)C
We add these capacitances because the resulting capacitors form a parallel network.
Now we compute C as a function of θ. The distance between the plates is d/2 and
area is a function of θ. So
C=
2ε0 A(θ)
·
d
To find A(θ) observe the following diagram of a semi-circular plate.
θ
A
R
The shaded region in this figure shows A(θ). Using the fact that area of a semi-circle
of radius R is πR2 /2, we obtain,
(
)
πR2 π − θ
A(θ) =
2
π
We have multiplied by
π−θ
π
since this the ratio of angle subtended by shaded region to
the entire angle subtended by the semi-circle.. Plugging this value in Ctotal we obtain,
(
(
))
2ε0 πR2 π − θ
Ctotal = (2N − 1)
d 2
π
2
(2N − 1)ε0 (π − θ)R
·
=
d
Date: 14 December, 2015, 5 : 00 pm
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Electricity and Magnetism: PHY-204
Fall Semester 2015
2. The circuit in Fig.(2) consists of two identical, parallel metal plates connected by
identical metal springs to a 100-V battery. With the switch open, the plates are
uncharged, are separated by a distance d = 8.00 mm, and have a capacitance C =
2.00 µF. When the switch is closed, the distance between the plates decreases by a
factor of 0.500.
(a) How much charge collects on each plate?
(b) What is the spring constant for each spring?
HINT : Derive the relation for force exerted on each plate using F =
C=
Aε0
,
x
dU
dx
and capacitance
where x is the arbitrary plate separation.
Fig. (2)
Answer
(a)
d = 8.00 mm
d
new d′ = = 4.00 mm
2
C0 = 2.00 µF
A
C0 d
C0 = ε0
⇒ A=
d
ε0
ε
A
ε
A
0
0
=2
= 2C0
new C ′ =
d′
d
Q = C ′ △V = 2C0 △V = 400 × 10−6 = 4.00 × 10−4 C.
(b) When the switch was open, there was no potential energy associated with the
electric field. In steady state, the voltage across capacitor is 100 V. So Ue =
Date: 14 December, 2015, 5 : 00 pm
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Electricity and Magnetism: PHY-204
1 ′ 2
CV
2
Fall Semester 2015
= C0 V 2 = 2 × 10−6 (100)2 = 2.00 × 10−2 J. Each spring stretches by an
amount △x =
0.4
2
= 0.2 mm. Hence for each spring its potential energy (elastic)
must decrease by 1.00 × 10−2 J. So
1
k(△x)2
2
2 × 10−2
= 5.00 × 105 N/m.
k =
(0.2 × 10−3 )2
1.00 × 10−2 =
This calculation is by virtue of conservation of energy.
3. A parallel-plate capacitor consists of a fixed plate and a movable plate that is allowed
to slide in the direction parallel to the plates. Let x be the distance of overlap, as
shown in Fig. (3). The separation between the plates is fixed.
(Movable)
(Fixed )
x
Fig. (3)
(a) Assume that the plates are electrically isolated, so that their charges ±Q are
constant. In terms of Q and the (variable) capacitance C, derive an expression
for the leftward force on the movable plate. HINT : Consider how the energy of
the system changes with x.
(b) Now assume that the plates are connected to a battery, so that the potential
difference ϕ is held constant. In terms of ϕ and the capacitance C, derive an
expression for the force.
(c) If the movable plate is held in place by an opposing force, then either of the above
two setups could be the relevant one, because nothing is moving. So the forces
in (a) and (b) should be equal. Verify that this is the case.
Answer
I’ll discuss this in the class.
4. A conducting sphere of radius R initially has a uniform surface-charge density
σ)0 .
(
t
Beginning at t = 0, this charge is drained off over a period t0 such that σ = σ0 1−
.
t0
Date: 14 December, 2015, 5 : 00 pm
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Electricity and Magnetism: PHY-204
Fall Semester 2015
R
σ
Ic
(a) Find the magnitude of the displacement current Id just outside the surface of the
sphere.
(b) Calculate the current Ic carried off by the wire. Compare the displacement current
Id to Ic .
Answer
Area Vector
Id
R
σ
Ic
Dashed line is an open
area whose edge is
the red circle. This is
similar to an inflated
balloon with an open
neck.
We use Maxwell-Ampere’s Law in this question. We choose a very small circular path
encircling the wire carrying the conduction current. This path is shown in red in the
corresponding diagram. For part (a) we consider the spherical surface just outside
the conducting sphere as the open surface corresponding to this path. Note that Ic
through this spherical surface (shown as a dashed circle in the diagram above) is zero.
(a)
d
Id = ε0
dt
∫
⃗ dA
⃗
E.
Open Area
= ε0
Date: 14 December, 2015, 5 : 00 pm
dE
(4πR2 ).
dt
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Electricity and Magnetism: PHY-204
Fall Semester 2015
dE
. Using Gauss’s Law for Electric Charges we find the electric
dt
field just outside the sphere in terms of the electric charge density:
∫
⃗ dA
⃗ = Qenc
E.
ε0
We need to find
Closed Area
E(4πR2 ) =
Q
ε0
Q
4πε0 r2
σ
=
,
ε0
=
from which we obtain
dE
1 dσ
=
dt
ε0 dt
(
)
1
d
t
=
(σ0 )
1−
ε0
dt
t
(
)0
1
1
(σ0 ) −
=
ε0
t0
σ0
= −
.
ε0 t0
(4πR2 )σ0
(4πR2 )σ0
and |Id | =
. The negative sign of the displacet0
t0
ment current shows that it is directed towards (into) the charged sphere whose
Hence Id = −
area is pointing outwards.
2
Ιd = - 4π R σ0
t0
2
Ιc = + 4π R σ0
t0
(b)
Ic = −
dQ
dt
(as current comes from charge depletion)
)
(
d
t
2
= −(4πR )σ0
1−
dt
t0
2
(4πR )σ0
= +
.
t0
Date: 14 December, 2015, 5 : 00 pm
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Electricity and Magnetism: PHY-204
Fall Semester 2015
Therefore, |Ic | = |Id | and Ic = −Id . Note that this was expected from the
Maxwell-Ampere’s Law. Both currents equal the path integral of the magnetic
field around the small circular red path, and therefore must be equal in magnitude. Their signs are perfectly consistent with the definition of the area vectors.
Think about this. If in confusion, come and talk to me.
5. The induced emf in an inductor of inductance L varies with time according to εind (t) =
−2At, where A is a positive constant.
(a) If there is no current through the inductor at t = 0, calculate the current as a
function of time for t > 0.
(b) Is the current increasing or decreasing when t > 0?
(c) Discuss how your answer to part b is consistent with the sign of εind and what
you know about the behavior of inductors?
Answer
(a)
εind (t) = −2At, A > 0
di
εind (t) = −L
dt
di
−2At
2At
=
=
dt
−L
L
∫ t
2A
i(t) =
t′ dt′ + i(t = 0)
L 0
A
At2
= t2 + 0 =
·
L
L
(b) The current is increasing with time.
(c) The current increases with time, so the induced emf acts in a direction to arrest
this increase. So it grows in the ‘negative’ direction with respect to time. This
is in line with the behavior of inductors.
6. The toroid in the figure below has 200 rectangular windings, and the toroid redii
are Rin = 160 mm and Rout = 240 mm. The height of each winding is h = 20
mm, such that the rectangular cross section of each winding is (Rout − Rin ) × h =
(80 mm) × (20 mm). What is the inductance of the toroid?
Date: 14 December, 2015, 5 : 00 pm
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Electricity and Magnetism: PHY-204
Fall Semester 2015
Rout
Rin
h
Answer
d
|ε| = N
dt
∫
⃗ · dA
⃗
B
Now the magnetic field inside the toroid depends upon the distance from the center
r. A coil is shown in the cross section view.
r
Rin
h
Rext
⃗ is constant at a given r.
From symmetry |B|
(B)(2πr) = µ0 N i
µ0 N i
B =
·
2πr
Date: 14 December, 2015, 5 : 00 pm
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Electricity and Magnetism: PHY-204
Fall Semester 2015
Hence over the area inside the loop shown, we have
∫
∫
µ0 N i Rout 1
⃗
⃗
dr
B · dA =
2π r=Rin r
(
)
µ0 N ih
Rout
=
ln
2π
Rin
(
)
2
µ0 N h
Rout di
ln
|ε| =
2π
Rin dt
(
)
2
µ0 N h
Rout
⇒ L =
ln
2π
Rin
( )
−7
4π × 10 × (200)2 × 0.02
80
=
ln
2π
20
−4
= 2.22 × 10 H.
Note that L has units of Henry (H). 1 H= 1 VsA−1 and it depends on N 2 .
Date: 14 December, 2015, 5 : 00 pm
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