EE2011 Engineering Electromagnetics
Solutions to Part I: Tutorial-2
SOLUTIONS TO TUTORIAL 2: CHARGE, POTENTIAL, AND E-FIELD - I
Question 1:
Two identical small metallic spheres are placed 10 cm apart. The spheres have charges of
1.7×10-9 C and -3.3×10-9 C, respectively. Find the force between the two spheres if they are
connected through a very tiny conducting needle. You may assume there is no
accumulation of charges on the tiny needle.
Solution to Question 1:
If the spheres are connected through a very thin needle, some of the charges with opposite
signs will cancelled and the net charge remained is (1.7 - 3.3) × 10-9 = -1.6 × 10-9 C.
Since the needle is very tiny and thin, so we assume that it does not accumulate any charges.
On the other hand, the two spheres are assumed to be identical and small, so this net
charge will mutually repel and evenly distributed between the two spheres as shown below:
Hence, we have the force, if neglecting the effects due to the needle, as follows:
(− 0.8 × 10 )
F=
−9 2
= 5.8 × 10 −7 N .
4πε 0 (0.1)
As the charges on the two spheres are of the same sign, so the two spheres repulse to each
other.
2
Question 2:
A uniform line charge density ρ exists on the y-axis from
y = 0 to y = ∞. Calculate the electric field due to this
charge distribution at a point P(a, 0). What will be the
electric field if the line is extended to y = -∞? If a charge
of q C is moved to the point P(a, 0), what is the force
experience by the charge if the line extends from y = -∞
to y = ∞?
Solution to Question 2:
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EE2011 Engineering Electromagnetics
Solutions to Part I: Tutorial-2
(Note: Working with vectors instead of scalars is recommended. Also, the point on source is
assumed to have the prime while the point at the observation location does not have the
prime symbol.)
From the geometry, we have the following differential elements and position vectors:
dl = yˆ dy ' ,
R = (a, 0) − (0, y ') = axˆ − y ' yˆ
R
axˆ − y ' yˆ
=
aˆ R =
.
R
a 2 + y '2
From the Coulomb’s Law, we will have the electric field due to the contribution by an
elementary charge dQ = ρ dy ' :
ρ dy '
ρ dy '
dE =
aˆ =
(axˆ − y ' yˆ )
2 R
3/ 2
4πε 0 R
4πε 0 ( a 2 + y '2 )
∞
E=∫
0
=
ρ dy ' ( axˆ − y ' yˆ )
(
4πε 0 a 2 + y '2
)
32
ρa
=
4πε 0
⎡
y ' xˆ
yˆ
+
⎢
⎢⎣ a 2 a 2 + y '2 a a 2 + y '2
y '=∞
⎤
⎥
⎥⎦
y '= 0
ρ a ⎡ xˆ
yˆ ⎤ ρ ( xˆ − yˆ )
ˆ
ˆ
0
0
.
+
−
−
=
y
x
2
4πε 0 ⎢⎣ a
4πε 0 a
a 2 ⎥⎦
If y extends from 0 to y = -∞, we have
∞
E=∫
0
ρ dy ' ( axˆ − y ' yˆ )
(
4πε 0 a 2 + y '2
)
32
0
+
ρ dy ' ( axˆ − y ' yˆ )
∫ 4πε
−∞
0
(a
2
+ y '2
)
32
2ρa ⎡
y ' xˆ
=
⎢
4πε 0 ⎢ a 2 a 2 + y '2
⎣
y '=∞
⎤
ρ ( xˆ − 0yˆ )
.
⎥
=
2πε 0 a
⎥⎦
y '=0
(Please note that we are taking the same dy’ for both segment if the superposition principle
is applied!! Once dy’ is selected, the limit should be selected to ensure the same increasing
of angle θ.)
One would expect this as the opposite dl component will generate an equivalent ycomponent in the opposite direction.
Therefore, the force, F = qE =
q ρ xˆ
, is obtained which is again a repulsive force.
2πε o a
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EE2011 Engineering Electromagnetics
Solutions to Part I: Tutorial-2
Question 3:
A uniformly charged disc of radius a carries a charge density of σ. Calculate the electric
field on the axis of the disc at a distant A away from the centre of the disc. What will be the
electric field if (i) A << a and (ii) A >> a ? Find the corresponding electric potential at A
away from the centre of the disc.
Solution to Question 3:
Approach:
Divide the ring into infinitesimally small ring area
y
elements of charges and sum up the total
x
contributions.
a
Since for each diametrically opposite area elements,
the dE sin φ component cancel out, only the zZ
component needs to be considered. Also, we have
A
dS = zˆ r ' dθ ' dr ' , and cos φ =
.
r'd '
2
A + r '2
By the Coulomb’s Law, we further have
σ r ' dθ ' dr 'cos φ
r ' Aσ dr ' dθ '
=
so that it leads to
dEz = dE cos φ =
32
2
2
4πε o ( A + r ' )
4πε o ( r 2 + A '2 )
dE
l
zˆ σ A
E( z = A) =
4πε o
dEz
a 2π
∫∫
0 0
r ' dr ' dθ '
(r ' + A )
2
2
32
r '= a
1
zˆ σ A ⎡ −1 ⎤
zˆ σ A ⎡ 1
=
=
−
⎢
⎢
⎥
2ε o ⎣ r '2 + A 2 ⎦ r '=0 2ε o ⎣ A
a2 + A2
⎤
⎥.
⎦
⎧
⎡
⎤
⎪
⎢
⎥
⎪ zˆ σ ⎢
A
⎥ σ zˆ
,
≈
A << a;
⎪ 2ε ⎢1 −
2 ⎥
⎪ o ⎢ a 1 + ⎛ A ⎞ ⎥ 2ε o
⎜ ⎟ ⎥
⎪
⎢⎣
⎝a⎠ ⎦
⎪
Therefore, we have E ( z = A ) = ⎨
⎡
⎤
⎪
⎢
⎥
⎪
2
ˆ
σ
z
A
⎢
⎥ (π a )σ zˆ
⎪
1
, A >> a.
−
≈
2
2 ⎥
⎪ 2ε o ⎢
4
πε
A
o
a
⎢ A 1+ ⎛ ⎞ ⎥
⎪
⎜ ⎟ ⎥
⎢
⎪⎩
⎝A⎠ ⎦
⎣
Apparently, the upper line solution is obtained based on the first order approximation, and
physically, the solution is the same as the one obtained based on the field next to an
infinitely large sheet with the charge density of σ. As for the lower line solution, the field is
obtained due to a point source. Both are agreeable with those in physical situation.
Approach I:
Using the potential definition itself, we derive as follows:
a 2π
r '= a
σ r ' dr ' dθ '
σ ⎡ 2
σ ⎡ 2 2 ⎤
=
A + r '2 ⎤ =
A +a −A .
V = ∫ dV = ∫ ∫
2
2
⎦ r '=0 2ε o ⎣
⎦
2ε o ⎣
0 0 4πε o A + r '
Approach II
But, we could have if we use the field expression to derive the potential:
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EE2011 Engineering Electromagnetics
Solutions to Part I: Tutorial-2
⎡
⎤
A
σ ⎡ 2 2 ⎤A
σ ⎡ 2 2 ⎤
1
−
d
=
a +A −A =
a +A −A .
A
⎢
⎥
⎦ ∞ 2ε o ⎣
⎦
2ε o ⎣
a 2 + A2 ⎦
∞
⎣
And in both cases, it is assumed that by default, V = 0 and E = 0 at infinity.
A
A
σ
2ε o
∞
V = − ∫ E • dl = − ∫
Question 4:
Two infinite non-conducting sheets of charge are parallel to each other. The sheet on the
left has a uniform surface charge density of σ1, and the one on the right has a uniform
charge density of σ2. Without loss of any generality, you could assume both σ1 and σ2 are
positive.
(a)
Determine the electric fields at points
i.
to the left of the two sheets.
ii.
between the two sheets.
to the right of the two sheets.
iii.
(b)
Deduce your solution if σ1=σ and σ2=-σ.
Solution to Question 4:
From lecture, we know that the E-field due to an infinite
sheet of charge with surface charge density σ is everywhere
outward (or inward) normal to the surface if the value of σ
is positive (or negative). Its magnitude is given by E =
σ
2ε o
(you could use Gauss law to derive this or to use the
conclusion made from an example in lecture).
It is known that sheets A and B have σ1 and σ2. Without
loss of any generality, we assume σ1 and σ2 are both
positive, so E fields are both pointing outward, but
EA =
σ1
σ
and E B = 2 .
2ε o
2ε o
For convenience, we define the x-direction toward right side as shown in the figure. Using
the superposition principle or from the Gauss’s law, we have the solutions detailed below:
(a)
The field can be calculated below:
a. To the left of the two sheets
⎛ σ
σ ⎞
1
E = − xˆ E A − xˆ E B = −xˆ ⎜ 1 + 2 ⎟ = −xˆ
(σ 1 + σ 2 ) .
2ε O
⎝ 2ε O 2ε O ⎠
b. Between the two sheets
⎛ σ
σ ⎞
1
E = + xˆ E A − xˆ E B = xˆ ⎜ 1 − 2 ⎟ = xˆ
(σ 1 − σ 2 ) .
2ε O
⎝ 2ε O 2ε O ⎠
c. To the right of the two sheets
⎛ σ
σ ⎞
1
E = xˆ E A + xˆ E B = xˆ ⎜ 1 + 2 ⎟ = xˆ
(σ 1 + σ 2 ) .
2ε O
⎝ 2ε O 2ε O ⎠
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EE2011 Engineering Electromagnetics
(b)
Solutions to Part I: Tutorial-2
When σ1=σ and σ2=-σ, we could easily deduce the solution from (a) as follows:
a. To the left of the two sheets
1
E = − xˆ
(σ 1 + σ 2 ) = 0 .
2ε O
b. Between the two sheets
1
σ
E = xˆ
(σ 1 − σ 2 ) = xˆ .
2ε O
εO
c. To the right of the two sheets
1
E = xˆ
(σ 1 + σ 2 ) = 0 .
2ε O
Question 5:
(b)
A solid conducting sphere (of radius a) is sited concentrically within a spherical
conducting shell (of inner and outer radii b and c respectively). Determine the
electric field (for r > 0) for the following cases:
(i)
The sphere has a charge of Q1 whereas the shell has no net charge.
(ii)
The shell has a charge of Q2 whereas the sphere has no charge.
(iii) The sphere has a charge of Q1 whereas the shell has a net charge of Q2.
Solution to Question 5:
There are two key points here:
1) The inner injected charge Q1 on the conducting inner sphere of radius a will produce an
induced charge –Q1 on the inner conducting shell surface at r = b and another induced
charge +Q1 on the outer conducting shell surface at r= c; and
2) An outer injected Q2 [as in the case (iii) of this example] will not produce anything
inside of the system (because of the shielding effects).
Therefore, we have Q1 on the surface at r = a; –Q1 on the surface at r = b; and Q1+ Q2 on the
surface at r = c. Please also note that the charges can distribute only on the conducting
surfaces, and there is no charge distributions in conductors.
Due to this reason, we will consider (iii) only and reduce the results for (i) and (ii) by
assuming Q2 = 0 and Q1 = 0, respectively. This is a symmetrical problem, so the electric
field should be radial in direction.
Considering a Gaussian surface S1 at r, as E is radial and
constant at a spherical surface, we thus have
Q
ε 0 E 4πr 2 = QEnclosed or E = Enclosed2 .
4πε 0 r
Therefore, we have
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EE2011 Engineering Electromagnetics
Qenclosed
Solutions to Part I: Tutorial-2
⎧0,
⎪Q
⎪
=⎨ 1
⎪Q1 + (-Q1 ) = 0,
⎪⎩Q1 + (-Q1 ) + (Q1 + Q2 ) = Q1 + Q2 ,
r<a
a<r <b
b<r<c
r>c
So, we have the solution for Part (iii) as follows:
⎧0,
r < a;
⎪ Q1
,
⎪
a < r < b;
Q
⎪ 4πε 0 r 2
EIII = enclosed2 = ⎨
0,
b < r < c;
4πε 0 r
⎪
Q
Q
+
2
⎪ 1
r > c.
,
⎪⎩ 4πε 0 r 2
Similarly, the solutions for Part (i) by letting Q2 vanish in the above solution and Part (ii) by
considering Q1 to be zero in the above solution will be
⎧0,
r < a;
⎪ Q1
,
⎪
2
a < r < b;
Q
⎪ 4πε 0 r
EI = enclosed2 = ⎨
0,
b < r < c;
4πε 0 r
⎪
Q
1
⎪
r > c;
,
⎪⎩ 4πε 0 r 2
and
⎧0,
r < a;
⎪0,
⎪
a < r < b;
Q
EII = enclosed2 = ⎨0,
b < r < c;
4πε 0 r
⎪ Q2
,
⎪
r > c.
2
⎩ 4πε 0 r
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