Download PHYS1221 Physics 1B Solutions Tutorial 2 APotential(AV) = Work Q

PHYS1221 Physics 1B
Solutions Tutorial 2
Basic info:
Although the term “voltage” is used every day, in physics it is a measure of a fairly abstract
quantity called Electric Potential. It’s important to distinguish electric potential from electric
potential energy (U). They are similar in that they are both scalars but they are not the same. You
can think of electric potential as the potential for moving a positive charge at some point in space.
So, points closer to a positive charge have greater electric potential. However, more often we talk
about the potential difference between two points:
!Potential(!V) =
Work
Q
or
!V =
!U
Q
We can also define the electric potential difference between two points (A and B) in an electric
field E:
B
!V = " # E.ds
A
Where the integral is along the path from point A and point B.
Remember also that with electric potential the principle of superposition applies.
In all of the following solutions
k=
1
4! " 0
Questions:
1. (a) Electrons tend to be attracted towards regions of high potential. (b) Equipotential
surfaces do NOT intersect. (c) No. At a position equidistant between two equal positive point
charges the total electric field can equal zero because the electric fields will have equal
magnitudes but exactly opposite directions. However, the electric potential due to either of
these charges has no direction and, in fact, the electric potential from both charges will add
together. (d) No. You need to know the integral of E.ds along a path from a point of known
voltage. (e) From the mathematical relationship between V and E, -dV/ds = E, where V is
constant, therefore E = 0.
2.
(refer to the diagram above) For the first charge to be added to this configuration no work is
done because there is no pre-existing electric field. For the rest of the question we recognise
that work done = ΔU and that ΔU = QΔV. For the second (negative) charge to be brought
within “a” of the first we assume that the second charge originates at an infinite distance
away, where the electric potential = 0V (this assumption is repeated with all the subsequent
additions). Then, for the second point charge, the work done is:
q!V =
"kq 2
a
For the third we consider its proximity to both the first and second charges and add this to the
above:
"kq 2 "kq 2 kq 2
q!V =
+
+
a
a
a 2
and for the fourth
#% "kq 2 "kq 2 kq2 & "kq 2 "kq 2 kq2
q!V =
+
+
+
+
+
=
$ a
a
a 2'
a
a
a 2
(
q2
kq 2
= -0.21
2"4
"0 a
a
)
2. (a) The electric potential at a distance “r” from a point charge is given by
V = kQ/r. Where
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V = 30V, k = 8.99 X 10 Vm/C and Q = 1.5 X10 C, r must = 4.495m or, to two significant
figures r = 4.5m. (b) Potential is a hyperbolic function
of r (ie. V α !
1/r). For equipotential
!
surfaces to be evenly separated you would have to have V α r.
3.
(a) If q1 and q2 are point charges, VA = kq1/(0.15) + kq2/(0.05). Substituting in the values for
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5
both charges and k = 8.99 X10 Vm/C, VA = 0.6 X10 V.
(b) First we need to determine the potential at B, VB which can be done in similar fashion to
5
part (a). VB = kq1/(0.05) + kq2/(0.15) = -7.8 X10 V. Then the potential difference
5
VA - VB = 8.4 X10 V, and from:
!V =
Work
Q
then
Work = Q!V .
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Using the value of Q = 3.0 X10 C, the work done in moving that charge from B to
A = 2.5J. External work is done to move the positive charge to a higher electric potential
so the system of charges has a higher potential energy.
4. (a) VA = kq/(2.06) = 5062V and VB = kq/(1.17) = 8913V. Then the potential difference
VA – VB = -3.85 kV. (b) The electric potentials calculated in part (a) depended on the
distances to A and B not the displacements. Direction plays no part in the calculation, so the
potential difference remains the same.
CAPACITORS AND DIELECTRICS
Basic info:
A capacitor is an electronic device for storing charge (Q) given a supplied voltage (V). As shown
symbolically in circuit diagrams, and often in reality, a capacitor can be considered as two
charged metal plates separated by a non-conducting medium (eg. Air), The definition of
capacitance, C:
C=
Q
V
or in terms of the area (A), the separation (d) of the plates and the dielectric constant of the
intervening medium εm.
C=
A !0 !m
d
Total capacitance, of several capacitors connected in series:
1
1 1
1
= +
+
+ ...
CT C1 C 2 C 3
…connected in parallel:
The energy stored in a capacitor U:
CT =C1 +C 2 + C 3 + ...
1
U = CV 2
2
Questions:
6.
Using the following:
C=
2
2
A !0 !m
d
-12
where A = π(0.08) m , d = 0.001m and ε0 = 8.85 X 10 F/m and assuming a vacuum
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between the plates εm = 1, then C = 1.78 X 10 F. Then CV = Q and Q = 1.8 X 10 C.
-1
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7. First combine C1 and C2 in series, so C12 = (1/C1 + 1/C2) = 3.27 X10 F. Then combine
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C12 and C3 in parallel: 3.27 X10 + 3.9 X10 F =7.17 X10 F
8.
(a) The over-riding rule here will be the conservation of charge. First we need to find exactly
how much charge is stored and where. With 96.6 V across C1 there must be Q = CV = 1.12
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X 10 C on C1. By the same formula the charge on C2 is 3.11 X 10 C. When the switches
are closed the effective capacitance will be = C1 + C2 = 4.38 µF and the net charge across
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this will be = 3.11 X 10 C – 1.12 X 10 C = 1.99 X 10 C. Therefore the voltage can be found
by V = Q/C = 45.4V (b) Now, given the voltage across both capacitors, we can find their
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respective charges by Q = CV. For C1, Q1 = 52.7 X 10 C and (c) for C2, Q2 = 146 X 10 C.
9.
(a) With capacitors arranged in series the charge across each capacitor is the same even if
the voltage isn’t. Here we begin by calculating the net total capacitance =[1/(2µF) +
-1
1/(8µF)] = 1.6 µF and, given the voltage across this equivalent capacitor is 300V, the
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charge each of the capacitor components will be Q = CV = 4.8 X 10 C. Now, knowing the
charge across each capacitor – the voltage across the 2µF, V2 = Q/2µF = 240 V and
across the 8µF, V8 = Q/8µF = 60V. (b) Now the capacitors are effectively connected in
parallel with a combined capacitance of 10µF and across capacitors in parallel the voltage
is the same. Hence we can say, through conservation of charge, that the charge on this
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equivalent capacitor is Q = 9.6 10 C and from V=Q/10µF, the potential difference across
each of the component capacitors is = 96V. However the charge on each capacitor will be
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redistributed: C2 has Q2 = 96V X 2µF = 1.92 X 10 C; C8 has Q8 = 96V X 8µF =
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7.68 X 10 C. (c) With equal charge on each capacitor, reconnecting them in reverse will
result in a net charge = 0 everywhere. With zero charge there is also zero potential
difference.
2
10. Total capacitance = 6µF and the stored energy = 1/2 X CV . Then substituting in V = 300V
the stored energy = 0.27J