Download EP 229 Tutorial 2 1. Three equal charges q are placed at the corners

EP 229 Tutorial 2
1. Three equal charges q are placed at the corners of equilateral triangle of side a: Find
(a) the potential at the center
(b) the electric …eld at the center, and
(c) the potential energy of the system.
Sol.
(a) The potential is
q 1
4 "0 r
where r is the distance between the corner and center. From
p
a=2
3
= cos 30 =
r
2
a
r=p
3
Then
p
q 1
=3 3
4 "0 a
=3
(b) The sum of three electric …eld vectors at the center vanish.
(c) The potential energy of one charge is
U1 =
1 q2
q2
(1 + 1) =
2 4 "0 a
4 "0 a
The total potential energy is
U=
3q 2
4 "0 a
2. Two spheres carrying charges q and q are separated by a distance d: The spheres have equal
mass m and radius a: After being released, they collide. Find the velocity at the collision.
Sol.
The initial potential energy is
Ui =
q2
4 "0 d
and the …nal potential energy is
q2
Uf =
4 "0
2a
The di¤erence in the potential energies
Ui
is converted into kinetic energy 2
Uf =
q2
4 "0
1
2a
1
mv 2 : Then
2
s
q2
1
v=
4 "0 m 2a
1
1
d
1
d
3. In the resistance network shown, what is the condition for the potentials at a and b to be
equal?
a
R1
R2
R5
R3
R4
b
Sol.
If a and b are at the same potential, no current ‡ows in R5 and R5 can be removed without
a¤ecting anything. From
R1
R3
=
R2
R4
R1 R4 = R2 R3
This is the principle of the Wheatstone bridge to measure unknown resistance (or impedance).
4. (a) Find the current in the 3
method.
(b) Replace the 3
R:
resistor using either multiple loop method or Thevnin’s
resistor with a variable resistor R to maximize the power delivered to
2Ω
3Ω
5Ω
10 V
5V
Sol.
(a) Let us use Thevnin. The open voltage when the 3
5
7
= 2:14 V
VT h = 10
resistor is removed is
5
The resistance seen from the open terminals is
RT h = 2 k 5 =
The current in the 3
10
= 1:43
7
resistor is
i=
2:14
= 0:483 A
4:43
(b) The power becomes maximum when RT h is equal to load resistance 1.43
2
:
I
-I
I
5. Three long parallel currents, I; I; I are placed at the corners of equilateral triangle of side
a. Find the magnetic force per unit length on each current.
Sol.
The magnitude of the force between two currents is
F
I2
= 0 (N/m)
l
2 a
Parallel currents attract and antiparallel currents repel.
6. A singly charged uranium 238 (mass 3:98 10 25 kg) accelerated to 3 keV enters a magnetic
…eld of 2 T with its velocity normal to the magnetic …eld. Find the cyclotron radius.
Sol.
The velocity of 3 keV U238 ion is
v =
r
2E
=
m
= 4:91
s
2
3000 1:6
3:98 10
10
19
25
104 m/s
The cyclotron radius is
v
4:91 104 3:98 10
=
eB=m
1:6 10 19 2
= 6:11 cm
25
rc =
7. A current source I0 is suddenly connected to a parallel RL circuit as shown. Find the current
is the inductor.
S closed at t = 0
I0
R
Sol.
3
L
The initial current in L is 0 because inductor resists any change in the current. At t = 1;
iL = I0 : Then
iL (t) = I0 1 e t=
where
= L=R is the time constant. The current in the resistor is
iR (t) = I0
Homework: Show that
Z
0
1
iL (t) = I0 e
t=
1
R [i (t)]2 = LI02
2
that is, the same amount of energy as delivered to an inductor has to be dissipated. This
holds independent of the resistance.
8. Omitted.
4