Download School of Physics and Astronomy, University of Minnesota Physics

School of Physics and Astronomy, University of Minnesota
Physics 1102, Introductory Physics II, Midterm 2 Practise
1. Consider a charge distribution on three thin metal plates separated from each
other by distances d as shown. The charge on the top plate is +Q, the charge on
the next plate down is -2Q/3 and the charge on the bottom plate is -Q/3. The area
p
of each plate is A and (A) is >> d so that the plates can be treated as infinite
in area for the purpose of calculation the fields between the plates (but not for
computing the charge per unit area of the plates).
a. Write the electric fields (magnitude and direction) at points 1, 4, 7 and 10 in
terms of d, Q, A and the universal constant K.
at 1: E=0
at 4: E= 2π K [-(Q/A)-(2/3)(Q/A)-(1/3)(Q/A)] =-4π K (Q/A) (down)
at 7: E= 2π K [-(Q/A)+(2/3)(Q/A)-(1/3)(Q/A)] =-(4π K/3) (Q/A) (down)
at 10: E=0
b. Find the electrostatic potential differences V1 − V2 , V2 − V3 , V3 − V5 ,V5 − V6 , V6 − V8 ,
V8 − V9 and V9 − V10 in terms of the same variables d, Q, A and the universal constant
K.
V1 − V2 = 0
V2 − V3 = 0
V3 − V5 = 4πKd(Q/A)
V5 − V6 = 0
V6 − V8 = (4πKd/3)(Q/A)
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V8 − V9 = 0
V9 − V10 = 0
c. Suppose that you can control the voltage drop V1 − V9 = V and that, as you
increase or decrease it, the total charge remains zero, the positive charge all stays on
the top plate while 2/3 of the negative charge remains on the middle plate and 1/3
of the negative charge is stays the bottom plate, but the total amount Q of charge
on top changes as you change V . What is the capacitance of that arrangement of
plates?
C = Q/V = Q/(4πKd((Q/A)(1 + 1/3))) = (1/4πK)(A/d)(3/4) = (Aǫ0 /d)(3/4)
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2. The capacitor plate shown contains two slabs of material of different dielectric
constants κ1 and κ2 . Their thicknesses are respectively d1 and d2 as shown. A voltage
difference V between the top and bottom plate is imposed by attaching a battery
as shown and a positive charge Q appears on the top plate and a negative charge
-Q appears on the bottom plate. The areas of the plates are denoted A.
a. What are the electric fields at points 1 and 2 in terms of the variables Q, A,
d1 , d2 , κ1 and κ2 and ǫ0 ?
in 1: E1 = (Q/(Aǫ0 ))(1/κ1 )
in 2: E2 = (Q/(Aǫ0 ))(1/κ2 )
b. Use the answer to a. to write an equation relating V to Q, A, d1 , d2 , κ1 and
κ2 and ǫ0 . Use the equation to find the capacitance of the capacitor in terms of A,
d1 , d2 , κ1 and κ2 and ǫ0 .
V = E1 d1 + E2 d2 = (Q/(Aǫ0 ))(d1 /κ1 + d2 /κ2 )
so
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C = Q/V = (Q/(Aǫ0 ))
d1 /κ1 + d2 /κ2
c. When the voltage is applied, charges Q′1 , −Q′1 , Q′2 , −Q′2 appear at the surfaces of
the dielectric slabs as sketched below. Write equations relating each if those charges
to Q, the magnitude of the charge on the top plate, and the dielectric constants κ1
and κ2
Q′1 = Q(1 − 1κ1 )
Q′2 = Q(1 − 1/κ2 )
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d. What happens if the first dielectric slab is replaced by a piece of metal (which
is very close to , but does not touch the top plate or the bottom dielectric?
Q′1 = −Q, Q′2 = +Q(1 − 1/κ2 ), C = κ2 Aǫ0 /d2
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3. Atoms have been modeled as tiny solar systems in which the negatively
charged electrons orbit the positively charge nucleus. In such a model, the motion
of the nucleus can often be ignored, because it is much more massive than the
electron. Consider just one electron orbiting a nucleus and make the assumption
that the nucleus is stationary and has positive charge Q while the electron has
negative charge −e and mass m.
a. First suppose that the orbit is circular as sketched. What is the magnitude
and direction of the acceleration of the electron? Give the magnitude in terms of e,
the mass m of the electron, the coulomb constant K and the radius of the orbit r.
acceleration is pointed toward the nucleus, perpendicular to the direction of
motion. magnitude is K|e|2 /(r2 m)
b. What happens to the magnitude v of the velocity of the electron as it orbits
in a circular orbit? Give a quantitative answer (in terms of the symbols given, if
required). What happens to the direction of the velocity? Magnitude does not
channge. direction changes continuously.
c. What is the ratio of the kinetic energy of the electron to its electrostatic
potential energy in this circular orbit? (Hint: You can simplify the answer if you
remember that the magnitude of the acceleration of a body moving in a circular
orbit with radius r and speed v is v 2 /r.)
potential energy =−K|e|2 /r
kinetic energy = (1/2)mv 2
F=ma says mv 2 /r = K|e|2 /r so mv 2 = K|e|2 /r
and
(1/2)mv 2
kinieticenergy
=
= −1/2
potentialenergy
−K|e|2 /r
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Does the proportion change as the electron goes around? No
d. As with planets, elliptical orbits, which are not circular, are also possible in
this model of an atom. Such an orbit is sketched below. Consider points A and B
of such an orbit. Where is the electron moving faster (ie where does it have the
greatest speed), at A or at B? Explain your answer carefully.
greatest at A because the potential energy is most negative there so the kinetic
energy has be largest in order to conserve energy.
e. Find the speed vA in terms of vB and the variables rA , rB , K, Q, m and e.
Conservation of energy:
2
2
(1/2)mvA
− K|e|2 /rA = (1/2)mvB
− K|e|2 /rB
Solve for vA .
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4. Here is a schematic picture of the circuit which delivers electrical power to a
city. The wires, which are the big ones which you see on towers in the countryside
leading into the cities, have resistance R for the wire coming in and R for the wire
going out. The city absorbs energy at a fixed rate P and the voltage between the
input and output wires at the city end is a fixed amount V. ( Regard the city as a
dissipative circuit element in which the charges entering have higher electrostatic
potential energy than the charges leaving.)
a. Under those conditions, what is the current I in the circuit in terms of some
or all of the variables P, V and R given?
Poweri P = VI, so I =P/V
b. In terms of the same variables, what is the rate at which thermal energy is
dissipated in the wires?
2I 2 R = 2(P/V )2 R
c. Now suppose that P is fixed but you want to design the system so that thermal
energy losses are as low as possible. Use your previous results from a. and b. to
write an equation for the ratio of the rate of thermal energy loss to the rate of
power delivery in terms of P,V and R. If you can change R and V by redesigning
the system, which of those two variables (R or V) will most effectively reduce that
ratio and which way should you increase or decrease that variable?
(P/V )2 R
thermal energy loss
=2
= 2P R/V 2
power delivered
P
You can decrease R or increase V but increasing V is better because the losses go
down as the square of V but only go down linearly as you decrease R. d. In such a
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line from Oregon to Los Angeles, the resistance in the wires is about 2 ×10−2 ohms
per kilometer of wire, the line is about 1300 km long, the voltage is 8 ×105 volts
and the power delivered to Los Angeles is about 109 watts. Using these numbers,
what is the ratio of thermal energy losses in the wires to power delivered to Los
Angeles?
R = 1300kmx2 × 10−2 ohms/km = 26ohms
ratio =2 × 26 × 109 /((8 × 105 )2 ) = .08
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