Download Recitation on Electric Fields Solution

Electricity and Magnetism: PHY-204
Fall Semester 2014
Recitation on Electric Fields
Solution
1. A line of positive charge is formed into a semicircle of radius R = 60.0 cm as shown in
Fig. (1). The charge per unit length along the semicircle is described by the expression
λ = λ0 cos θ. The total charge on the semicircle is 12.0 µC. Calculate the total force
on a charge of 3.00 µC placed at the center of curvature P.
FIG. 1
Answer
The magnitude of electric field at point P due to one segment of bended line of positive
charge having a charge dq,
1 dq
4πε0 R2
1 λdl
1 (λ0 cos θ)(Rdθ)
=
=
,
2
4πε0 R
4πε0
R2
dE =
where dl = Rdθ is the arc length and λ is the charge per unit length.
θ
dE
Date: 8 September, 2014
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Electricity and Magnetism: PHY-204
Fall Semester 2014
−→
Figure shows the electric field contribution dE at point P due to a single segment of
charge at the top of semicircle. Because of the the symmetry of the situation, the
horizontal component of electric field dEx = dE sin θ will cancel out. Therefore, only
vertical component will contribute
dEy = dE cos θ
1 (λ0 cos θ)(Rdθ)
1 λ0 cos2 θdθ
=
cos
θ
=
·
4πε0
R2
4πε0
R
To obtain the total electric field at P , integrate this expression over the limits −π/2
to π/2,
∫
λ0
Ey =
4πε0 R
π/2
cos2 θdθ
−π/2
( ∫ π/2
)
∫
1
λ0
1 π/2
=
dθ +
cos 2θdθ
4πε0 R 2 −π/2
2 −π/2
( )
π
λ0
λ0
=
=
·
4πε0 R 2
8ε0 R
∵ cos2 θ =
1 + cos 2θ
2
Lets find λ0 ,
∫
Q=
∫
dQ =
λdl
∫
π/2
=
−π/2
(λ0 cos θ)(Rdθ)
∫
= λ0 R
π/2
−π/2
cos θdθ = 2λ0 R
12µC = 2λ0 (60 cm)
∴ λ0 = 10.0µC/m
Therefore, the force on a charge 3 µC placed at the center of curvature P , is
−
→
F = −Fy ĵ = −qEy ĵ
λ0
= −(3.00µC)
ĵ
8ε0 R
= −(3.00µC)
10.0µC/m
= 0.706(−ĵ).
8(8.85 × 10−12 )0.600 m
2. A charged cork ball of mass 1.00 g is suspended on a light string in the presence of a
⃗ = (3î + 5ĵ) × 105 N/C, the ball
uniform electric field as shown in Fig. (2). When E
is in equilibrium at θ = 37◦ . Find
Date: 8 September, 2014
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Electricity and Magnetism: PHY-204
Fall Semester 2014
(a) the charge on the ball,
(b) the tension in the string.
FIG. 2
Answer
(a) Free body diagram for a charged ball suspended in the presence of a uniform
electric field in given below;
Let us sum force components to find,
∑
Fx = qEx − T sin θ = 0
qEx
,
and
⇒T =
sin θ
∑
Fy = qEy + T cos θ − mg = 0.
(1)
(2)
Substituting T in Eq(2), yields
(
)
qEx
qEy +
cos θ − mg = 0
sin θ
q(Ex cot θ + Ey ) = mg
q =
Date: 8 September, 2014
mg
(1.00 × 10−3 )(9.8)
=
= 1.09 × 10−8 C.
Ex cot θ + Ey
(3 cot 37◦ + 5) × 105
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Electricity and Magnetism: PHY-204
Fall Semester 2014
(b) By substituting values in relation for T , derived in (a)
qEx
sin θ
(1.09 × 10−8 )(3 × 105 )
=
= 5.44 × 10−3 N.
sin 37◦
T =
3. A negatively charged particle −q is placed at the center of a uniformly charged ring,
where the ring has a total positive charge Q as shown in Fig. (3). The particle,
confined to move along the x-axis, is moved a small distance x along the axis (where
x << a) and released. Show that the particle oscillates in simple harmonic motion
with a frequency given by
(
)1/2
1 kqQ
f=
.
2π ma3
FIG. 3
Answer
First we need to calculate the electric field due to the ring at a point P lying a distance
x from its center along the central axis perpendicular to the plane of the ring.
(a)
(b)
−
→
Figure (a) shows the electric field contribution d E at P due to a single segment of
charge at the top of the ring. This field vector can be resolved into components dEx
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Electricity and Magnetism: PHY-204
Fall Semester 2014
parallel to the axis of the ring and dE⊥ perpendicular to the axis. Figure (b) shows the
electric field contributions from two segments on opposite sides of the ring. Because of
the symmetry of the situation, the perpendicular components of the field cancel and
only the parallel components will contribute.
dEx = dE cos θ
1
dq
1 dQ
cos
θ
=
=
cos θ.
4πε0 r2
4πε0 (x2 + a2 )
From the geometry of the figure, cos θ = x/r = x/(x2 + a2 )1/2 . So we have
dEx =
=
=
=
⇒ Ex =
=
=
1
xdQ
2
4πε0 (x + a2 )3/2
x(λadθ)
1
4πε0 (x2 + a2 )3/2
1
x(λadθ)
4πε0 (x2 + a2 )3/2
1
xQ
adθ
Q
∵λ=
2
2
3/2
4πε0 2π
a(x + a )
2πa
∫ 2π
xQ
1
dθ
2
4πε0 2π(x + a2 )3/2 0
1
xQ
4πε0 a3 (1 + xa22 )3/2
1 xQ
for x << a.
4πε0 a3
The magnitude of the force experienced by a charge −q placed along the axis of the
ring is
F = −qEx
qxQ
= −
4πε a3
( 0
)
qQ
= −
x.
4πε0 a3
This expression for the force is in the form of Hooke’s law, with an effective spring
constant of k =
qQ
·
4πε
√0 a3
Since ω = 2πf =
k
,
m
we have
1
f=
2π
Date: 8 September, 2014
√
qQ
.
4πε0 ma3
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Electricity and Magnetism: PHY-204
Fall Semester 2014
4. Identical thin rods of length 2a carry equal charges +Q uniformly distributed along
their lengths. The rods lie along the x axis with their centers separated by a distance
b > 2a (Fig. (4)). Show that the magnitude of the force exerted by the left rod on the
right one is
(
F =
) (
)
kQ2
b2
ln 2
.
4a2
b − 4a2
FIG. 4
Answer
Two identical positively charged rods of linear charge density λ having equal lengths
2a lying along x axis. The magnitude of electrostatic repulsion due to one segment of
left rod having a charge dq1 , exerted on one segment of right rod having a charge dq2
is,
dF =
kdq1 dq2
,
r2
where
Q
dx1 ,
2a
Q
dq2 = λdx2 = dx2 ,
2a
dq1 = λdx1 =
and r = x2 − x1 is the separation between two segments as shown in figure. So we
have
dF =
Date: 8 September, 2014
kQ2 dx1 dx2
·
4a2 (x2 − x1 )2
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Electricity and Magnetism: PHY-204
Fall Semester 2014
Integrate this expression to obtain the total force
∫ ∫
∫ ∫
kQ2
dx1 dx2
F =
dF =
·
2
4a x2 x1 (x2 − x1 )2
Now
∫
x1
dx1
=
(x2 − x1 )2
∫
a
−a
dx1
,
(x2 − x1 )2
let
y = x2 − x1 ,
⇒ dy = −dx1 .
When x1 = −a, y = x2 + a and when x1 = a, y = x2 − a. By these substitutions,
integral becomes
∫
x2 −a
x2 +a
x −a
1 2
−dy
= y2
y x2 +a
1
1
=
−
x2 − a x2 + a
]
[
∫
kQ2
1
1
∴F =
−
dx2
4a2 x2
x2 − a x2 + a
[∫ b+a
]
∫ b+a
kQ2
dx2
dx2
=
−
4a2 b−a x2 − a
b−a x2 + a
b+a ]
[
b+a
kQ2
=
ln(x2 − a) − ln(x2 + a)
2
4a
b−a
b−a
[ (
)
(
)]
2
b +
a −
a
b+a+a
kQ
ln
− ln
=
4a2
b−a−a
b +
a −
a
[ (
)
(
)]
2
kQ
b
b + 2a
=
ln
− ln
2
4a
b − 2a
b
[ (
)
(
)]
2
kQ
b
b
=
ln
+ ln
4a2
b − 2a
b + 2a
(
)
2
2
→
−
b
kQ
F =
ln
î.
4a2
b2 − 4a2
5. An electric dipole in a uniform horizontal electric field is displaced slightly from its
equilibrium position as shown in Fig. (5), where θ is small. The separation of the
charges is 2a, and each of the two particles has mass m.
(a) Assuming the dipole is released from this position, show that its angular orientation exhibits simple harmonic motion with a frequency,
√
1
qE
f=
·
2π ma
Date: 8 September, 2014
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Electricity and Magnetism: PHY-204
Fall Semester 2014
(b) Suppose the masses of the two charged particles in the dipole are not the same
even though each particle continues to have charge q. Let the masses of the
particles be m1 and m2 . Show that the frequency of the oscillation in this case is
√
1 qE(m1 + m2 )
f=
·
2π
2am1 m2
FIG. 5
Answer
(a) Torque exerted by electrostatic force on positive charge τ = R ×F = −aF sin θ =
−qEa sin θ.
Torque exerted by electrostatic force on negative charge τ = −qEa sin θ. The
electrostatic forces result in a net torque τ = −2qEa sin θ. For small θ, sin ≈ θ,
we have τ = −2qEaθ.
Also, from analogy of Newton’s second law for rotational motion:
τ = Iα = I
d2 θ
,
dt2
where I moment of inertia for particles rotating about an axis depends only on
their masses and their distances from the axis:
I = ma2 + ma2 = 2ma2
d2 θ
⇒ 2ma2 2 = −2qEaθ
dt
(
)
qE
d2 θ
= −
θ,
dt2
ma
which is the standard equation characterizing simple harmonic motion, with ω 2 =
qE
.
ma
Date: 8 September, 2014
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Electricity and Magnetism: PHY-204
Fall Semester 2014
Then the frequency of oscillation is f = ω/2π, or
√
1
qE
f=
,
2π ma
which is the required result.
(b) When the masses are unequal, the axis of rotation of the dipole passes through
the center of mass of the two charges.
We call moment of inertia about this axis ICM , then by using parallel axis theorem:
I = ICM + M d2
⇒ ICM = I − M d2
2
= (m1 a2 + m2 a2 ) − (m1 + m2 )rcm
,
Here rcm is the distance of center of mass from origin which is the point equidistance from the two charges on the straight line connecting them.
O
then,
Icm =
=
=
=
=
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rcm
a
a
q
rcm =
q
m1 a − m2 a
(m1 − m2 )a
=
·
m1 + m2
(m1 + m2 )
(
)2
(m1 − m2 )a
(m1 + m2 )a − (m1 + m2 )
(m1 + m2 )
2 2
(m1 − m2 ) a
(m1 + m2 )a2 −
(m1 + m2 )
(
)
(m1 − m2 )2 2
m1 + m2 −
a
(m1 + m2 )
)
(
(m1 + m2 )2 − (m1 − m2 )2 2
a
m1 + m2
(
)
4m1 m2
a2 .
m1 + m2
2
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Electricity and Magnetism: PHY-204
Fall Semester 2014
Now
d2 θ
Icm 2 = −2qEaθ
(
) dt
4m1 m2 2 d2 θ
= −2qEaθ
a
m1 + m2
dt2
(
)
d2 θ
qE(m1 + m2 )
⇒ 2 = −
θ
dt
2am1 m2
then the frequency of oscillation is
√
1 qE(m1 + m2 )
.
f=
2π
2am1 m2
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