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PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
Name:
SAMPLE SOLUTION
Student No.: _______________ Section: _______
1. Two point charges, Q1 and Q2, are located on a plane. Q1 is +0.15 μC, located at coordinates
(x,y) = (-6 cm, 0 cm). Q2 is -400 nanocoulombs, located at coordinates (x,y) = (6 cm, 0 cm).
Furthermore, point A is located at (x,y) = (0 cm, 0 cm), and point B has the coordinates
(x,y) = (0 cm, 7 cm).
Compare Fig. 1. for a graphical representation (not to scale).
Fig. 1: Location of point charges Q1 and Q2 and points A and B.
a)
r
What is the force (expressed as F1 = F1 x iˆ + F1 y ˆj ) due to charge Q1
b)
on a test charge of +1.0 C located at point A?
r
What is the force (expressed as F2 = F2 x iˆ + F2 y ˆj ) due to charge Q2
c)
on a test charge of +1.0 C located at point A?
r
What is the total force (expressed as F = Fx iˆ + Fy ˆj ) due to charges Q1 and Q2
on a test charge of +1.0 C located at point A?
d)
What is the electric potential of point A with respect to point B (i.e. VAB) due to Q1?
e)
How much work is done to overcome the electric force due to charge Q1
when moving +1 C of charge from point A to point B?
f)
What is the electric potential of point A with respect to point B (i.e. VAB) due to Q2?
g)
How much work is done to overcome the electric force due to charge Q2
when moving +1 C of charge from point A to point B?
h)
What is VAB with both charges, Q1 and Q2, present?
i)
How much work is done when moving +1 C of charge from point B to point A
with both charges, Q1 and Q2, present?
p. 1 of 14
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
2. Two vertically oriented parallel planes intersect the x-y-plane of a coordinate system at right
angles.
One plane, referred to as plane “0”, is located at x = 0 m (i.e. it contains the y-axis) and is
uniformly charged with a charge density of σ1 = 2⋅10-9 C/m2.
The other plane, referred to as plane “2”, is located at x = 2 m and is uniformly charged with
a charge density of σ2 = -1⋅10-9 C/m2.
The planes are therefore separated by 2 m. Besides the planes, four more points are defined
along the x-axis:
Point M is located half way between the planes
(x, y, z) = (1 m, 0, 0)
Point O is located at the origin of the coordinate system (x, y, z) = (0, 0, 0)
Point A is located 10 cm away from the origin
(x, y, z) = (10 cm, 0, 0)
Point R is located far to the right
(x, y, z) = (3 m, 0, 0)
Compare Fig. 2. for a graphical representation (not to scale).
Fig. 2: Two uniformly charged planes “0” and “2”, and four points O, A, M, R.
left: 3D visualization. Right: 2D top view.
a)
b)
What is the amount of excess charge in a rectangular area of 40 cm by 60 cm
in the charged plane “0”?
r
What is the el. field at point M (x- and y-components, expressed as E0 = E0 x iˆ + E0 y ˆj )
due to the excess charge in the plane “0” (in the absences of plane “2”)?
c)
What are the x- and y-components of the electric field at point M in the presence of
both charged planes?
d)
What are the x- and y-components of the electric field at point R in the presence of
both charged planes?
e)
What is the electric potential of point A with respect to point O
in the absence of plane “0” (i.e. find VAO with only plane “2” present, or V2 AO).
f)
What is the electric potential of point A with respect to point O in the presence of both
charged planes?
p. 2 of 14
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
3. Each plate of a parallel plate capacitor has an area of 0.2 m2. The plates are 2⋅10-5 m apart,
and the voltage across the plates is 10 V.
a)
What is the capacitance of the capacitor if it is filled with air (assume air has a dielectric
constant of εr=1.00)?
b)
What is the amount of excess charge on each plate of the capacitor?
c)
What is the charge density in each plate?
d)
What is the electric field strength between the plates of the capacitor?
e)
How much work is required to move 10-22 C of charge from one plate to the other?
Note that this is such a small amount of charge that it can be assumed
that the removal from one plate and the movement to the other
does not change the voltage between the plates.
The air gap now gets filled with polystyrene (εr, polystyrene=2.60). V = 10 V is maintained.
f)
What is the capacitance now?
g)
What is the amount of excess charge on each plate of the capacitor?
h)
What is the effective charge Qeff on each plate of the capacitor?
I.e., what is the difference in magnitudes between the actual amount of charge on a plate
of the capacitor and the charge inside the dielectric very near to the surface that
interfaces with the plates of the capacitor?
i)
What is the electric field strength in the dielectric between the plates of the capacitor?
4. A thin but infinitely large sheet of dielectric with a dielectric constant κ is placed
perpendicular to a uniform electric field of strength Efield, compare Fig. 3.
Fig. 3: Sheet of dielectric placed perpendicular to a uniform electric field.
a)
What is the equation for the electric field strength in the dielectric? Explain all
components.
b)
The electric field induces ‘sheets of excess charge’ inside, but near the left and right
surfaces, of the dielectric.
What is the polarity of the excess charge on the left side, what on the right side?
c)
Let the excess charge density in these two ‘sheets’ be ±σdiel.
i)
Does σdiel account for the el. field strength being weaker inside the dielectric?
ii) Which equation relates the el. field strength between two parallel, uniformly and
oppositely charged, infinite planes (e.g. such as the two ‘sheets’)
to the charge density σ on the planes?
iii) Briefly derive an equation that relates σdiel to Efield and κ .
p. 3 of 14
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
Question 1:
a)
r κ Q1 Qt
magnitude of the force: Coulomb’s Law: F1 =
,
r2
with κ = 8.99⋅109 Nm2/C2, Q1 = 0.15 μC = 0.15⋅10-6 C, Qt = 1 C, r = 6 cm = 0.06 m
2
⇒ F1 = 8.99 ⋅ 10 9 Nm
C
2
( 0.15 ⋅ 10
C) ( 1 C)
= 374.6 kN
(0.06 m )2
−6
sign: Both charges are positive, and will therefore repel each other.
The repulsive force on Qt at point A must point away from point (x,y) = (-6 cm, 0 cm)
⇒ in positive x-direction
r
⇒ F1 = 3.75 ⋅ 105 N iˆ
b)
magnitude: as above, but Q2 = -400 nC = -4⋅10-7 C (instead of Q1)
2
⇒ F2 = 8.99 ⋅ 10 Nm
9
C
2
( − 0.4 ⋅ 10
C) ( 1 C)
= 9.99⋅105 N
2
(0.06 m )
−6
sign: The charges have opposite signs, and will therefore attract each other.
The force of attraction on Qt at point A must point towards point (x,y) = (6 cm, 0 cm)
⇒ in positive x-direction
r
⇒ F2 = 9.99 ⋅ 105 N iˆ
c)
r r
r
F = F1 + F2 = 3.75 ⋅ 105 N iˆ + 9.99 ⋅ 105 N iˆ
r
F = 1.37 ⋅ 106 N iˆ
p. 4 of 14
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
d)
magnitude of electric potential between points B and A in a field set up by point charge Q1
(as derived in class): V1 AB = κ Q1
with
1
1
−
r1 B r1 A
Q1 = 0.15 μC = 0.15⋅10-6 C
r1 A = distance from charge Q1 to point A. r1 A = 0.06 m
r1 B = distance from charge Q1 to point B. r1 B =
V1 AB = 8.99 ⋅ 10 9
(0.06 m )2 + (0.07 m )2 ≅ 0.092 m
Nm 2
1
1
≅ 7848 Nm/C = 7848 J/C = 7848 V
⋅ (0.15 ⋅ 10 −6 C )
−
2
C
0.092 m 0.06 m
sign: positive, if positive work required to move a positive test charge from point B to point A.
Here:
1st part: moving a positive charge against field line
1st part: ⇒ pos. work done by mover
2nd part: movement along equipotential line requires no work
⇒ pos. work overall ⇒ V1 AB has pos. sign
V1 AB = 7.8 kV
e)
V1 BA = W/Q ⇒ W = V1 BA ⋅ Q
with work W required to move a particle with excess charge Q from point A to point B
and with V1 BA = -V1 AB (see part d) ) = -7848 V; Q = 1 C
⇒ W = (-7848 V)⋅ (1 C) = -7848 J
f)
with details as in d),
and Q2 = -400 nC = -4⋅10-7 C, r2 A = 0.06 m, r2 B ≅ 0.092 m
V2 AB = κ Q2
1
1
−
r2 B r2 A
p. 5 of 14
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
V2 AB = 8.99 ⋅ 10 9
Nm 2
1
1
= 20.9 kV
⋅ (4 ⋅ 10 −7 C )
−
2
C
0.092 m 0.06 m
sign: positive, if positive work required to move a positive test charge from point B to point A.
Here:
1st part: moving a positive charge down the field line
1st part: ⇒ neg. work done by mover
2nd part: movement along equipotential line requires no work
⇒ neg. work overall ⇒ V2 AB has neg. sign
V2 AB = -20.9 kV
g)
V2 BA = W/Q ⇒ W = V2 BA ⋅ Q
with work W required to move a particle with excess charge Q from point A to point B
and with V2 BA = -V2 AB (see part f) ) = 20.9 kV; Q = 1 C
⇒ W = (20.9 kV)⋅ (1 C) = 20.9 kJ
h)
VAB = V1 AB + V2 AB = 7.8 kV – 20.9 kV = -13.1 kV (with results from parts d) and g) )
i)
with VAB = -13.1 kV (see part h) ): W = VAB ⋅ Q = (-13.1 kJ/C) ⋅ (1 C) = -13.1 kJ
or
with results from parts e) and g), but inversed sign (now moving from B-to-A instead of A-to-B):
total work = work to overcome the electric force due to charge Q1 + work to overcome the
electric force due to charge Q2
W = 7.8 kJ - 20.9 kJ = -13.1 kJ
p. 6 of 14
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
Question 2:
a)
charge density σ = charge Q / area A ⇒ Q = σ ⋅ A = 2⋅10-9 C/m2 ⋅ (0.4 m ⋅ 0.6 m) = 4.8⋅10-10 C
b)
The direction of the electric field must be perpendicular to
the charged plane (i.e. along the pos. or neg. x-axis).
⇒ to the right of the plane
(for positive x-values, where point M is located),
the field is in the direction of the positive x-axis
The electric field strength of a field set up by a uniformly charged plane does not depend on the
distance from the charged plane and is given by E =
with the permittivity of free space
−9
ε 0 = 8 . 85 ⋅ 10
C
2 ⋅ 10
σ
m2
⇒ E0 = 0 =
2 ε 0 2 ⋅ 8.85 ⋅ 10 −12 C 2
= 113
σ
2 ε0
−12
C2
.
N m2
N
C
N m2
r
⇒ with direction as derived above: E0 = 113 N
C
(
iˆ + 0 ˆj
)
c)
r
We already calculated the field E0 due to the excess charge on plane “0” in part b).
r
We now need to calculate the field E 2 due to the excess charge on plane “2”:
The direction of the electric field must be perpendicular to the
charged plane (i.e. along the pos. or neg. x-axis).
⇒ to the left of the plane (where point M is located),
the field is in the direction of the positive x-axis,
since the plane is negatively charged
p. 7 of 14
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
−9
C
− 1 ⋅ 10
σ
m2
E2 = 2 =
2 ε 0 2 ⋅ 8.85 ⋅ 10 −12 C 2
= 56.5
N m2
N
.
C
r
The field of this strength is oriented in pos. x-direction ⇒ E 2 = 56.5 N
C
iˆ
r r
r
Finally, both fields get superposed: E = E0 + E 2
r
r
⇒ E = (E0 x + E2 x ) iˆ = 113 N + 56.5 N iˆ ⇒ E = 169.5 iˆ N
C
C
C
(
)
d)
The electric field due to the excess charge in plane “0” has a direction away from the plane as the
plane is positively charged.
r
At point R, to the right side of the plane, the direction is along the positive x-axis: E0 = E0 x iˆ ,
with a positive value of E0x = 113 N/C from part b).
The electric field due to the excess charge in plane “2” has a direction towards the plane as the
plane is negatively charged.
r
At point R, to the right side of the plane, the direction is against the x-axis: E 2 = E 2 x − iˆ , with a
positive value of E2x = 56.5 N/C from part c).
( )
r r
r
Superposition of both fields: E = E0 + E 2
r
r
⇒ E = E0 x iˆ − E 2 x iˆ = 113 N − 56.5 N iˆ ⇒ E = 56.5 iˆ N
C
C
C
(
)
e)
Points O and A are located on an electric field line.
The strength along the field line is constant in this geometry of a
uniformly charged plane.
r
⇒ V2 AO = E 2 Δx
with distance Δx between points O and A, Δx = 0.1 m
N
and E2 = 56.5
from part c).
C
The field lines point from higher to lower potential
⇒ potential drops from points O to A ⇒ V2 AO will be negative
⇒ V2 AO = - (56.5 N/C) ⋅ (0.1 m) = -5.65 Nm/C = -5.65 J/C =-5.65 V
p. 8 of 14
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
f)
Scalar superposition of V2 AO (result from part e) ) and V0 AO (to be calculated now):
For the plane “0”, the field lines point in positive x-direction
(the same direction as the E2-field lines)
⇒ again, as in part e), this voltage (V0 AO) will be negative
⇒ with field strength E0 = 113 N/C from part b): V0 AO = - (113 N/C) ⋅ (0.1 m) = -11.3 V
VAO = V0 AO + V2 AO = -5.65 V – 11.3 V = -16.95 V
p. 9 of 14
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
Question 3:
a)
capacitance of a parallel plate capacitor is C = ε r ε 0
with
relative permittivity of air εr = 1.00
permittivity of free space
A
d
ε 0 = 8.85 ⋅ 10−12 C
2
N m2 ,
area of one plate A = 0.2 m2
distance between the plates d = 2⋅10-5 m
2
⎛
C 2 ⎞ 0.2 m2
−12
−8 C
−8 C
−8 C
(
)
⎟
⎜
C
1
8
.
85
10
=
8
.
85
⋅
10
=
8
.
85
⋅
10
=
8
.
85
⋅
10
=
⋅
⇒
2
−
5
⎜
J
N m ⎟⎠ 2 ⋅ 10 m
Nm
V
⎝
C
C = 8.85⋅10-8 F = 0.0885 μF
b)
Q=C⋅V
with excess charge on one (each) plate Q
capacitance C = 8.85⋅10-8 F = 8.85⋅10-8 C/V (as in part a) )
electric potential of one plate with respect to the other V = 10 V
Q = (8.85⋅10-8 C/V) ⋅ (10 V) = 8.85⋅10-7 C = 0.885 μC
c)
charge density σ = Q / A
with excess charge on one (each) plate Q = 8.85⋅10-7 C (as in part b) ), area of a plate A = 0.2 m2
⇒ σ = (8.85⋅10-7 C) / (0.2 m2) = 4.425⋅10-6 C/m2
d)
two ways to answer this question:
method 1:
The electric field between parallel plates must be perpendicular to the plates and constant
r r
⇒ V = E ⋅d
with vectors for the electric field and the distance between the plates in the same direction
r
⇒ V = E ⋅ d ⇒ E = V / d = (10 V) / (2⋅10-5 m) = 5⋅105 V/m
p. 10 of 14
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
method 2:
Treat the plates of the capacitor as uniformly charged, infinite planes
⇒ each plate creates an electric field with a strength of
Eone plate = σ / (2 ε0)
If plate 1 is the positively charged plate, and plate 2 the negatively
charged one, the field lines created at each plate, in the area between the
plates, all point towards the right
⇒ electric fields superpose so that field strengths gets added:
E = Eplate 1 + Eplate 2 = 2 Eone plate = 2 [ σ / (2 ε0)] = σ / ε0
With result from part c): σ =4.425⋅10-6 C/m2
⇒ E=
4.425 ⋅ 10 -6 C
8.85 ⋅ 10
−12
C
2
m2
⇒ E = 5⋅105 N/C
N m2
e)
In class we derived that W = V ⋅ Q
with voltage between plates V = const. = 10 V through which charge Q = 10-22 C is moved
⇒ W = (10 V) ⋅ (10-22 C) = 10-21 J
Note: to calculate this quantity exactly, without the assumption of unaffected fields, the
difference between the initial energy stored in the field between the two plates of the capacitor
and the final energy stored after the charge Q (now better called ΔQ) has been moved (and
therefore some excess charge getting neutralized), must be determined:
work to move ΔQ:
W = energyinitially stored in the field – energyfinally stored in the field
2
2
1 Qini
1 Q fin
W =
−
2 C
2 C
2
1
1
2
2
(
Qini
=
Qini
− Q 2fin ) =
− (Qini − ΔQ )
2C
2C
1
1
2
2
(
(
Qini
=
− (Qini
− 2 Qini ΔQ + ΔQ 2 )) =
2 Qini ΔQ − ΔQ 2 )
2C
2C
neglecting the extremely small quantity of ΔQ2
Q ΔQ
W ≈ ini
.
C
With Qini /C = Vini ⇒ W ≅ ΔQ ⋅ Vini
(exactly the equation used above, now that ΔQ2 has been neglected)
(
p. 11 of 14
)
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
f)
Solution as in part a), but with relative permittivity of polystyrene εr = 2.60
A
and the capacitance of a parallel plate capacitor C = ε r ε 0
d
⎛
−12
⇒ C = (2.6) ⎜⎜ 8.85 ⋅ 10
⎝
C 2 ⎞ 0.2 m2
⎟
= 2.3 ⋅ 10−7 F = 0.23 μF
2 ⎟
−5
N m ⎠ 2 ⋅ 10 m
C = 2.3⋅10-7 F = 0.23 μF
g)
Q=C⋅V
with excess charge on one (each) plate Q
capacitance C = 2.3⋅10-7 F = 2.3⋅10-7 C/V (as in part f) )
electric potential of one plate with respect to the other V = 10 V
Q = (2.3⋅10-7 C/V) ⋅ (10 V) = 2.3⋅10-6 C = 2.3 μC
Note: This solution is based on the assumption of constant voltage V = 10 V compared to the airfilled case.
Also acceptable would be calculations and results based on the assumption of constant actual
charge Q = QP on the capacitor plates:
QP = Qpart b) = 0.885 μC
h)
capacitance ∝
ε0 A
d
is the constant of proportionality between charge and voltage:
⇒ effective charge Qeff =
ε0 A
d
V
⇒ actual charge on a plate Q P =
⇒ QP = ε r Qeff ⇒ Qeff =
QP
and
εr ε0 A
d
V
= ε r Qeff , QP = 2.3 μC as in part g)
ε r ⇒ ⎪Qeff⎪ = 2.3 μC / 2.60 = 0.88 μC
Note: This solution is based on the assumption of constant voltage V = 10 V compared to the airfilled case.
Also acceptable would be calculations and results based on the assumption of constant actual
Qeff = QP / εr = 0.885 μC / 2.6 = 0.34 μC
charge Q = QP on the capacitor plates:
p. 12 of 14
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
i)
two ways to answer this question:
method 1:
The electric field between parallel plates must be perpendicular to the plates and constant
r r
⇒ V = E ⋅d
with vectors for the electric field and the distance between the plates in the same direction
r
⇒ V = E ⋅ d ⇒ E = V / d = (10 V) / (2⋅10-5 m) = 5⋅105 V/m
method 2:
The electric field strength can also be calculated using the formula derived in part d, method 2):
σ
σ
, here specifically E capacitor = eff
ε0
ε0
Q
σ
σ A
0.88 ⋅ 10 −6 C
= eff = eff
= eff =
ε0
ε0 A
ε0 A
⎛
C2 ⎞
−12
E capacitor =
E capacitor
⎜⎜ 8.85 ⋅ 10
⎝
⎟⎟ (0.2 m 2 )
Nm ⎠
= 5⋅105 N/C
2
Note: This solution is based on the assumption of constant voltage V = 10 V compared to the airfilled case.
(In order to maintain the same electric field strength between the capacitor plates when
introducing the dielectric instead of air (compare parts d) and i) ),
the voltage across the plates had to be maintained (at V = 10 V), which means that the actual
charge on the plates had to be increased by a factor εr ; compare parts b) and g) .)
Also acceptable would be calculations and results based on the assumption of constant actual
Ecapcitor = Qeff / (ε0 A) = 0.34 μC / … = 1.9⋅105 N/C
charge Q = QP on the capacitor plates:
(i.e. field strength decreased by factor εr
compared to air-filled case
when introducing a dielectric
and keeping the actual charge on the plates const.)
p. 13 of 14
PHYS 155 Assignment #3: Coulomb’s Law, Capacitance & Dielectrics
Question 4:
a)
E dielectric =
with
E field
κ
strength of the uniform el. field (into which the dielectric was eventually placed) Efield
dielectric constant κ
b)
The arrows on the electric field lines indicate that positive charges will get ‘pushed’ to the right
side.
⇒ The right surface of the dielectric will have a net positive charge.
⇒ Correspondingly, the left surface of the dielectric will have a net negative charge.
c)
i)
yes
ii) E = σ / ε0
iii) The strength of the electric field created by the ‘sheets’ of charge Esheets is
Esheets = Efield – Edielectric.
E
κ −1
With Edielectric = Efield /κ (as in part a) ), E sheets = E field − field = E field
κ
Also (compare part c ii) ), Esheets = σdiel / ε0
⇒ E sheets = E field
κ − 1 σ diel
=
,
κ
ε0
and therefore σ diel = ε 0
κ −1
E field
κ
with charge density on two surfaces of the dielectric σdiel.
p. 14 of 14
κ