Electricity and Magnetism: PHY-204.
17 September, 2014
Collaborative ASSIGNMENT
Assignment 1: Charges and Fields
Solution
1. A charge q is positioned at the tip of a hollow cone. If the top half of the cone is removed and thrown away (see Fig. 1), what is the force on the charge q due to remaining part of the cone? For what angle θ is this force maximum?
q
θ θ
L/2
σ
L/2
Fig. (1)
Answer
The force on the charge q due to a segment dx of larger cone can be found as dF z
=
1
4 πε
0 qdq cos θ, x 2
F z θ
F dx
Side view of ring x
θ θ
L/2 dx r L/2 r ϕ
Top view of ring
Fig. (a)
Due Date: September. 23, 2014, 5:00 pm 1

Electricity and Magnetism: PHY-204.
17 September, 2014 where dq = σrdϕdx is the charge stored on a small horizontal ring as shown in Figure
(a). From the geometry of the figure, sin θ = r/x .
dF z
=
F z
=
=
1 σqx sin θ cos θdϕdx
∫
πε
0 x 2
σq sin θ cos θ
∫
2 π dF z
4 πε
=
σq sin θ cos θ
4 πε
0
(2 π )
(
0 ln L
− ln
0
L
2
∫
L dx dϕ
)
=
L/ 2
σq x sin 2 θ
4 ε
0 ln 2 .
F z will be maximum when sin 2 θ is maximum. Therefore, θ must be 45
◦ to maximize the force.
2. An half-infinite line has linear charge density λ . Find the electric field at a point that is “even” with the end, a distance l from it, as shown in Fig. 2. You should find that the field always points up at a 45
0 angle, independent of l .
λ
Fig. (2)
Answer
Refer to Fig (b) the electric field due to a segment ds of a half-infinite line having linear charge density λ , at a distance l from the end of line, is given by d ⃗ = dE x i + dE y
Now and dE x
= dE sin θ dE y
λds
=
4 πε
0 r 2
= dE cos θ sin θ
=
λds
4 πε
0 r 2 cos θ
Due Date: September. 23, 2014, 5:00 pm 2

Electricity and Magnetism: PHY-204.
17 September, 2014 y
θ x r
θ
λ ds s
Fig. (b)
From the geometry of figure; r = l/ cos θ, s = l tan θ
⇒ ds = l cos 2 θ dθ.
Therefore, dE x
=
λl sin θ cos 2 θ 4 πε
0 l 2 cos
2
θ =
λ
4 πε
0 l sin θ.
Likewise, dE y
=
λ
4 πε
0 l cos θ
·
∫ ∫
λ
E x
= dE x
=
4 πε
0 l
E y
=
−
λ
4 πε
0 l
=
λ
4 πε
0 l
· cos θ
0
π/ 2
=
0
π/ 2
4
λ sin
πε
0 l
θdθ
Hence
⃗
=
λ
4 πε
0 l
(ˆ j ) representing the field at 45
◦ angle from the axis. This direction is independent of the value of l .
3. A point charge q is located at the origin. Find the flux through the circular disk shown.
You are allowed to use only Gauss’s law. HINT: Consider drawing a suitable Gaussian surface conforming to the geometry and seeing how the circular disk is related to this surface.
Due Date: September. 23, 2014, 5:00 pm 3

Electricity and Magnetism: PHY-204.
17 September, 2014
θ
θ
Fig. (3)
Answer
Draw a spherical Gaussian surface that encloses the circular disk. By construction
(point charge and spherical symmetry), the flux through the curved section of the sphere (shown in bold) will be the same as through the flat conical cap of section AB .
A
θ
θ
B
The total flux through the sphere is q/ε
0
. We need to find the flux through the segment of the sphere bounded by A and B . Suppose the area of this segment is A .
θ
A
B
Then
⇒ flux through the segment flux through the sphere
=
Flux through the segment =
A
4 πl 2
Aq
4 πε
0 l 2
·
Due Date: September. 23, 2014, 5:00 pm 4

Electricity and Magnetism: PHY-204.
17 September, 2014
We need to find A . A small area segment on the surface of the sphere is l 2 sin βdβdϕ , where β is the angle from the z -axis and ϕ is the azimuthal angle. Hence
∫
θ
∫
2 π
A = l
2 sin θdθdϕ
θ =0 ϕ =0
= l
2
(2 π )(
− cos θ )
θ
0
= 2 πl
2
(1
− cos θ ) .
β
φ
θ
Hence flux through the disk = flux through the segment =
=
2 πl
2
(1
− cos θ ) q
4 πε
0 l 2 q (1
− cos θ )
·
2 ε
0
4. Four positively charged bodies, two with charge Q and two with charge q , are connected by four unstretchable strings of equal length. In the absence of external forces they assume the equilibrium configuration shown in Fig. 4. Show that tan
3
θ = q
2
/Q
2
. You could show that this relation must hold if the total force on each body, the vector sum of string tension and electrical repulsion, is zero. Hint: Draw free body diagram for each charge.
q
2
1
Q Q 3 q
4
Fig. (4)
Due Date: September. 23, 2014, 5:00 pm 5

Electricity and Magnetism: PHY-204.
17 September, 2014
Answer
We have to draw the free body diagram for each charge to find the vector sum of forces acting on it along the x -axis and y -axis.
Let’s find the vector sum of forces acts on charge 1 (leftmost in the figure) along the x -axis and y -axes. Here
⃗
’s represent string tensions and
⃗
’s represent electric forces.
the charges are labeled clockwise as shown.
F
3
F
4
F
2
θ
θ
Q
1
T
θ
T
θ x y
∑
F
1 y
= T sin θ
−
T sin θ + F
4 sin θ
−
F
2 sin θ = 0
⇒
F
2
∑
F
1 x
= F
4
=
−
T cos θ
−
T cos θ
−
F
3
+ F
4 cos θ + F
2 cos θ
=
−
2 T cos θ +
1 Q 2
4 πε
0
(2 r cos θ ) 2
+ 2 F
2 cos θ
⇒
T
=
−
2 T cos θ +
(
1 1
=
2 cos θ 4 πε
0
1
4 πε
0
(2 r
(2
Q 2 cos r
θ
Q 2 cos
) 2
θ ) 2
+ 2
+ 2
1
4 πε
0
4 πε
0 qQ r
1
2 qQ r 2 cos θ cos θ = 0
)
.
(1)
Similarly, the vector sum of forces acts on charge 2 along the x -axis and y -axis can also be found as
∑
F
2 x
= T cos θ
−
T cos θ + F
3 cos θ
−
F
1 cos θ = 0
⇒
F
1
∑
F
2 y
= F
3
= F
4
+ F
3 sin θ + F
1 sin θ
−
2 T sin θ
Due Date: September. 23, 2014, 5:00 pm 6

Electricity and Magnetism: PHY-204.
F
4
F
3
T
θ
θ q
2
θ
θ
T
F
1 x
17 September, 2014 y
=
=
⇒
T =
1 q
2
4 πε
0
(2 r sin θ ) 2
+ 2 F
1 sin θ
−
2 T sin θ
4
1
πε
1
0
2 sin θ
(2 r
4 q 2 sin
1
πε
0
θ ) 2
(2 r
+ 2 q 2 sin
4 πε
0
θ
1
) 2 qQ r 2
+ 2 sin θ
−
2 T sin θ = 0
)
1 qQ sin θ .
4 πε
0 r 2
(2)
By comparing Eq (1) and Eq (2),
1
(
1 Q 2 1 qQ
+ 2
2 cos θ 4 πε
0
(2 r cos θ ) 2 4 πε
0 r 2 cos θ
)
Q 2 cos 3 θ
=
=
⇒ tan
3
θ =
1
2 sin θ
(
1 q 2
4 πε
0
(2 r sin θ ) 2 q 2 sin
3 q 2
,
Q 2
θ
1
+ 2
4 πε
0 qQ r 2 sin θ
) which is the required result.
5. Consider a high-voltage direct current power line that consists of two parallel conductors suspended 3 meters apart. The lines are oppositely charged. If the electric field strength halfway between them is 15,000 N/C, how much excess positive charge resides on a 1km length of the positive conductor?
Answer
The magnitude of electric field at a distance x from an infinite line of charge is
λ
E =
2 πε
0 x
(as derived in class).
Due Date: September. 23, 2014, 5:00 pm 7

Electricity and Magnetism: PHY-204.
Q Q
P
17 September, 2014
3m
The electric field strength halfway between two oppositely charged conductors becomes
E at mid point
= E
+
=
λ
+ E
−
+
2 πε
0 x
1
λ
(since fields are in the same direction)
2 πε
0 x
2
15000 =
λ (2)
∵ x
1
= x
2
= 1 .
5
2 πε
0
(1 .
5)
∴
λ = 1 .
5(15000)( πε
0
) = 1 .
5(15000)(3 .
14
×
8 .
85
×
10
−
12
)
= 6
×
10
−
7
Cm
−
1
, is the linear charge density for each conductor. Therefore, the positive charge residing on 1 km length of positively charged conductor is
Q = λl = (6
×
10
−
7
)(1
×
10
3
) = 6
×
10
−
4
C .
6. The neutral hydrogen atom in its normal state behaves, in some respects, like an electric charge distribution that consists of a point charge of magnitude e surrounded by a distribution of negative charge whose density is given by ρ ( r ) =
−
Ce
−
2 r/a
0 . Here a
0 is the Bohr radius , 0 .
53
×
10
−
10 m, and C is a constant with the value required to make the total amount of negative charge exactly e . What is the net electric charge inside a sphere of radius a
0
? What is the electric field strength at this distance from the nucleus? You will get full credit even if you write the integral without solving it.
Answer
Electronic charge inside a volume of radius a
0
∫ a
0
∫ a
0
∫
2 π
∫
π
ρdV = is
Ce
−
2 r/a
0 ( r
2 radius=0 r =0 ϕ =0 θ =0 sin θdrdθdϕ ) ,
Due Date: September. 23, 2014, 5:00 pm 8

Electricity and Magnetism: PHY-204.
17 September, 2014 where ( r 2 sin θdrdθdϕ ) is the volume element. hence the electronic charge inside a
0
(∫
π
)∫ a
0
∫ a
0
C (2 π ) sin θdθ r
2 e
−
2 r/a
0 dr = C (4 π ) r
2 e
−
2 r/a
0 dr.
θ =0 r =0 r =0 is
Now
∫ r
2 e
−
2 r/a
0 dr = I (say)
I =
= r
(
2 e
−
−
2 r/a
0
2 /a
0
) a
0 r
2 e
−
2 r/a
0
2
+
2
2 /a
0 [ re
−
2 r/a
0
+ a
0
∫
(
− re
−
2 r/a
0 dr
2 /a
0
)
+
1
(2 /a
0
)
∫ integration by parts e
−
2 r/a
0 dr
]
=
= a
0 r
2 e
−
2 r/a
0
2 a
2
0 r
2 e
−
2 r/a
0
− a 2
0 re
−
2 r/a
0
2
−
− a 2
0 re
−
2 r/a
0
2
− a 2
0
2 e
−
2 r/a
0
(
−
2 /a
0
) a 3
0 e
−
2 r/a
0 .
4
Putting in the limits. (Dont confuse e with the charge in this integral).
∫
0 a
0 r
2 e
−
2 r/a
0 dr =
− a
=
− a
=
− a
2
3
0
3
0
3
0 e
(
(
−
2
2
1 e
5
4 e 2
2
−
1
4 a
2
3
0 e
1
−
2
+
)
2 e 2
−
+
4 a
4
1 e
= 0 .
08 a
3
0
2
3
0
.
e
−
2
−
1
4
+
) a
4
3
0
Hence the total charge inside the sphere of radius a
0 is
+ e due to proton
−
C (4 π
{z
.
08) a
3
} due to electron’s distributed charge
= Q enc
.
The field at this distance E a is such that
E a
(4 πa
2
0
E
) = a
=
Q enc
ε
0
+ e
−
C (4 π )(0 .
08) a 3
0
4 πa 2
0
ε
0
·
One can also find C by the condition that the total electronic charge is e .
∫
∞
C (4 π ) r
2 e
−
2 r/a
0 dr = e
Now
∫ 0
∞
0 r
2 e
−
2 r/a
0 dr = a
4
3
0
∴
C (4 π ) a
4
3
0
= e
C = e
πa 3
0
·
Due Date: September. 23, 2014, 5:00 pm 9

Electricity and Magnetism: PHY-204.
17 September, 2014
7. A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge per unit length of λ , and the cylinder has a net charge per unit length of 2 λ . From this information, use Gausss law to find
(a) the charge per unit length on the inner and outer surfaces of the cylinder
(b) the electric field outside the cylinder, a distance r from the axis.
Answer
(a) Draw the Gaussian surface S
1 shown.
Cross sectional view cylinder wire
Gaussian surface S
1
This results in a charge
−
λ on the inner surface and +3 λ on the outer surface.
(b) The field is
⃗
=
3 λ
2 ε
0
πr
Since the cylindrical symmetry dictates that the field is (i) radially outward and field is (ii) identical to a field from a single wire of charge density 3 λ .
Due Date: September. 23, 2014, 5:00 pm 10