Exam EE1P21
Electricity and magnetism
This exam consists of 3 pages with 4 exercises.
A list of formulas is appended after the exercises, starting with page 4.
The maximum attainable number of points is 90.
The number of points assigned to each exercise is stated explicitly.
Start each exercise on a new exam sheet and fill in on each sheet
your name, your study number and the number of the current exercise.
The EE1P21 team wishes you a lot of success!
20
points
Exercise 1
Determine the following quantities by making use of Coulomb’s constant k = 9·109 Nm2 /C2 .
Let Q = 3q be a point charge located at (x, y, 0).
a) Determine the force F exerted by the charge Q on another point charge Q0 located at
the origin (0, 0, 0). (7 points)
generated by the point charge Q at the origin
b) Give the expression of the electric field E
(0, 0, 0). (3 points)
Consider now three point charges: Q1 = 3q located at (a, a, 0), Q2 = q located at (−a, a, 0) and
Q3 = q located at (−a, −a, 0).
generated by the three point charges at the
c) Give the expression of the electric field E
origin (0, 0, 0). (5 points)
d) Determine the value of a point charge Q4 that must be added to this configuration at
1 = E1 ŷ. Derive the expres(a, −a, 0) such that the electric field at the origin becomes E
sion of E1 . (5 points)
2
25
points
Exercise 2
Let the case of a long coaxial cable with a perfectly conducting, massive core of radius r = R0
(with r being measured orthogonally to the core’s axis), a perfectly conductive, thin mantel M1
of radius r = R1 > R0 and a second mantel M2 of radius r = R2 > R1 . The thickness of both
mantels is negligible. The space between the massive core and the mantel M1 is filled with a
homogeneous medium with permittivity 1 = 20 . The space between the mantel M1 and the
mantel M2 is filled with a homogeneous medium with permittivity 2 = 30 . The massive core
carries on its outer surface an electric charge of Q per meter along the cable. The mantel M1
carries on its outer surface an electric charge of 2Q per meter along the cable. The mantel M2
carries on its outer surface no charge. The concentric media are free of charge, ‘fringing effects’
at the end of the cable are neglected and we assume that the potential at infinity is zero.
a) Make a clear sketch of the examine situation. (1 point)
b) Determine the electric field E(r)
in the region r < R0 and in the region r > R2 . (4 points)
c) Determine the electric field E(r)
in the region R0 < r < R1 and in the region R1 < r < R2 .
(5 points)
d) Determine the surface charge density σ in C/m2 on the inner surface of the mantel M1 .
(5 points)
e) Determine the electric potential ΔV (R1 ) on the mantel M1 . (5 points)
f) Determine the capacitance C between the mantel M1 and the mantel M2 . (5 points)
20
points
Exercise 3
Let the configuration in Fig. 1 consisting of two parallelplate capacitors and an ideal DC voltage source with
voltage V . The capacitor C1 has plates of area A, the
distance between its plates is d1 (d1 A) and relative
permittivity εr,1 . The capacitor C2 has plates of identical
area A, the distance between its plates is d2 (d2 A)
and relative permittivity εr,2.
V
C1 , εr,1 , d1
+
a) Determine the capacitance C of the two capacitances in series. (6 points)
b) Derive the charges Q1 and Q2 stored on the positive
plate of C1 and the positive plate of C2 , respectively.
(5 points)
c) Determine the magnitude of the electric fields E1
and E2 inside the capacitors C1 and C2 . (5 points)
d) Assume now that the dielectric strength of medium
1 allows a maximum electric field of E1,b (V/m) and
that the dielectric strength of medium 2 is infinite.
V1
−
C2 , εr,2 , d2
+
V2
−
Figure 1: Explicative for Exercise 3.
3
Determine the minimum distance d1,min at which
breakdown occurs in the capacitor for the given
source voltage V . (4 points)
25
points
Exercise 4
Let the configuration in Fig. 2 consisting of non-uniform
electric charge distribution across an infinitely thin circular ring C of radius R and centre point (0, 0, 0), located
in the plane z = 0. The non-uniform charge distribution
has the expression
λ(ϕ) = λ0 sin(ϕ) (C/m)
y
C
R
O z
ϕ
x
with λ0 being a constant and ϕ being defined as in the
figure. The circular charge distribution is located in free
space, the permittivity being ε0 .
a) Determine the elementary charge dq along C and
Figure 2: Explicative for Exercise 4.
calculate the total charge Qtot of circular charge
distribution. (5 points)
b) Determine the potential V (0, 0, z), with z 0, assuming that V∞ = 0 V. (8 points)
c) Determine the magnitude of the electric field com for z 0.
ponent Ez (0, 0, z) of the electric field E
(6 points)
0, 0) at the ring’s
d) Determine the electric field E(0,
centre. (6 points)
– End of exam –
4
List of formulas
Universal constants
electrons charge: e = 1.6·10−19 C
electrons mass: me = 9.11·10−31 kg
protons mass: mp = 1.66·10−27 kg
mp ≈ 1836 × me
Coulombs constant: k = 9·109 (N·m2 /C2 )
Avogadros number: NA = 6.022·1023 (1/mol)
atomic mass unit: u = mp = 1.66·10−27 kg
electric permittivity of vacuum: ε0 = 8.854·10−12 F/m
k=
1
4πε0
Mathematical instruments
1) Polar coordinates: {ρ, ϕ}
elementary surface: dS = ρ dρ dϕ
elementary arc along a circle of radius R: dL = R dϕ;
elementary surface of a ring: dS = 2πρ dρ.
2) Cylindrical coordinates: {ρ, ϕ, z}
elementary volume: dV = ρ dρ dϕ dz
elementary surface on a cylindrical surface of radius R: dS = R dϕ dz;
elementary surface of a band on a cylindrical surface of radius R: dS = 2πR dz.
3) Spherical coordinates: {ρ, ϑ, ϕ}
elementary volume: dV = ρ2 sin(ϑ) dρ dϑ dϕ
elementary surface on a spherical surface of radius R: dS = R2 sin(ϑ) dϑ dϕ;
elementary volume of a thin shell: dV = 4πρ2 dρ.
Electrostatics
Colomb’s force exerted by a point charge qa at a over a point charge qb at b:
F = k
qa qb
qa qb
(rb − ra ) = k 2 r̂ab
3
|rb − ra |
rab
with rab = rb − ra , rab = |rab | and r̂ab = rab /rab
5
Electric field (strength) at a point:
r ) = F (r)
E(
q
with F being the force exerted on a test point charge q.
Electric field at b due to a point charge qa at a:
=k
E
qa
qa
(rb − ra ) = k 2 r̂ab
3
|rb − ra |
rab
Electric field at a due to a collection of point charges qi at ri , i = 1, . . . , N:
ra ) = k
E(
N
i=1
qi
qi
(
r
−
r
)
=
k
r̂
a
i
2 a,i
|ra − ri |3
r
a,i
i=1
N
Electric field at a due to a volume charge distribution:
ρ(
r
)
ρ(r ) ra − r
E(ra ) = k
(ra − r ) dV (r ) = k
dV (r )
3
2
ra − r |
ra − r | |ra − r |
V |
V |
Electric field at a due to a surface charge distribution:
σ(r ) ra − r
E(ra ) = k
dS(r )
2
ra − r | |ra − r |
S |
Electric field at a due to a line charge distribution:
λ(r ) ra − r
dL(r )
E(ra ) = k
|2 |
|
|
r
−
r
r
−
r
a
a
L
Electric dipole moment – two point charges +q and −q at a small distance d:
p = qd ˆl = p ˆl
with ˆl oriented from the −q point charge towards the +q one.
Electric field at r due to an electric dipole p centred at ro (the centre of the d ˆl line
segment); the observation point is located such that rp d, with rp = r − ro :
– on the perpendicular bisector (rp · ˆl = 0):
r) = −k p ˆl
E(
rp3
– along ˆl (rp · ˆl = rp ):
r) = 2k p ˆl
E(
rp3
6
Torque experienced by an electric dipole p in a (uniform) electric field E:
τ = p × E
Electric flux through an arbitrary surface S:
· n̂ dA
E · dA =
E · dA =
E
Φ=
S
S
S
with n̂ being the normal to the surface S
Gauss’s law for a closed surface S:
· n̂ dA = qenclosed
E · dA =
E
ε0
S
S
with n̂ being the outward oriented normal to the surface S
Gauss’s law in the case of charge distributions inside of a volume VS enclosed by a closed
surface S:
1
E · dA =
E · n̂ dA =
ρ(r )dV
ε0
S
S
VS
Electric field at the surface of a charged, perfectly conducting surface:
· n̂ = σ
En = E
ε0
Electric potential between the points a and b:
ΔVab
ΔUab
=−
= Vb − Va =
q
a
b
· dl
E
with dl along an arbitrary curve between a and b
Electric potential between the points a and b in a uniform field:
· Δl
ΔVab = Vb − Va = −E
Electric potential at b due to a point charge qa at a:
V =k
qa
qa
=k
|rb − ra |
rab
Electric potential at a due to a collection of point charges qi at ri , i = 1, . . . , N:
V (ra ) = k
N
i=1
qi
qi
=k
|ra − ri |
r
i=1 a,i
N
7
Electric potential at a due to a volume charge distribution:
ρ(r )
V (ra ) = k
dV (r )
|
|
r
−
r
a
V
Electric potential at a due to a surface charge distribution:
σ(r )
dS(r )
V (ra ) = k
ra − r |
S |
Electric potential at a due to a line charge distribution:
λ(r )
V (ra ) = k
dL(r )
|
|
r
−
r
a
L
Electric potential at r due to an electric dipole p centred at ro (the centre of the d ˆl line
segment); the observation point is located such that rp d, with rp = r − ro :
– in general:
V (r) = k
p · r̂p
rp2
– on the perpendicular bisector (rp · ˆl = 0):
V (r) = 0
– along ˆl (rp · ˆl = rp ):
V (r) = k
p
rp2
Calculating the electric field from the electric potential:
= −∇V = ∂x x̂ + ∂x ŷ + ∂x ẑ V
E
Capacitance:
C=
Q
V
Capacitance of a parallel-plate capacitor (fringing effects neglected)
C=
εr ε0 A
d
with A the aria and d the distance between the plates
Energy stored in a capacitor:
Q2
CV2
=
U=
2
2C
8
Energy in the electric field electric energy density:
1
u e = εr ε0 E 2
2
Integral electric energy:
1
εr ε0 E 2 dV
Ue =
2 V
Instantaneous electric current:
I=
dQ
dt
Electric current density:
J = nqvd
with n being the number of charges per unit volume, q the charge value and vd the drift
velocity
Ohm’s law (local form):
J = σ E
= ρJ
E
with σ being the conductivity and ρ the resistivity
Ohm’s law (integral form):
I=
V
R
Resistance of a cylindrical wire of length L and cross section A:
R=
ρL
A
Electric power:
P = I V = I 2R =
V2
R