Download TOPIC 5 , !0 (New) Part 1 electric_currents_and_fields

SOLUTIONS TOPIC 5, 10 (New) Part 1– Electric Currents, Fields
1. Electric Field and Electric Potential Tsokos pp.289 - 298
Examples pp. 6 of packet
540 - 541)
( taken from Wilson Buffa 41,42,45,46,48-54 p.
1. The electric field due to a positive point charge :
a) varies as 1/r b) points toward the charge c) points away from the charge d)
has a finite range
C-
2. The units of electric field are : a) C
B-
b) NC-1 c) N
d) J
E = Force / charge = F / q = NC-1
3. How can the relative magnitudes of the field in different regions be determined
from an electric field line diagram?
The relative strength of an electric field is determined by the spacing or density of
the filed lines :
1
4. a) If the distance from a charge is doubled the magnitude of the electric field will
decrease by one fourth
 1
(2)2
E=k Q
r2
=1
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b) If the original electric field is 1.0 x 10-4 NC -1and the distance is reduced to onethird the original distance, what is the magnitude of the new electric field?
E=
1 =
r2
1
. =
(1/3)2
so it is 9 times original field = 9.0 x 10-4 NC-1
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5. An isolated electron is acted on by an electric force of 3.2 x 10-14 N. What is the
magnitude of the electric field at the electron’s location?
no distance given so use E = F
q
electron = 1.6 x 10-19 C
E = 3.2 x 10-14 N
1.6 x 10-19 C
= - 2.0 x 105 NC-1
( 2 sig. dig.)
6. What is the magnitude of the electric field at a point 0.25 cm away from a point
charge of + 2.0 μC ?
k = 9.0 x 109 Nm2 C-2
Distance given so use E = k Q
r2
E=
9 .0 x 109 Nm2 C-2 x
=
+ 2.9 x 109 NC-1
( 2.0 x 10-6 C)
(0.0025 m)2
( 2 sig. dig.)
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7. At what distance from a proton is the magnitude of the electric field 1.0 x 105
NC-1?
E=k Q
r2
proton = 1.6 x 10-19 C
r = √ kQ
E
Note : everything inside √
= √(9.0 x 109 Nm2 C-2)( 1.6 x 10-19 C)
1.0 x 105 N/C
= 1.2 x 10-7 ( 2 sig. dig)
8. The diagram below shows electric field lines in an isolated region of space
containing two small charged spheres, Y and Z. Which of the following
statements is true?
(A) The charge on Y is negative and the charge on Z is positive.
Wrong Y = + Line arrows gor from + charge to – charge ( follow a positive point
charge)
(B) The strength of the electric field is the same everywhere.
Wrong dense line spacing changes
(C) The electric field is strongest midway between Y and Z.
Wrong strongest where spacing of line is more dense
(D) A small negatively charged object placed at point X would tend to
move toward the right.
Correct since Z is negative
(E) Both charged spheres Y and Z carry charge of the same sign.
NO way!
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2. Parallel Plates and Spheres
IB Exam May 2009 # 22 Paper 1 p. 8 of packet
Answer – B
Since sphere is NEGATIVE, field lines go TOWARD sphere.
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Example p. 8 of packet :
The hollow metal sphere shown above is positively charged. Point C is the center of the
sphere and point P is any other point within the sphere. Which of the following is
true of the electric field at these points?
(A) It is zero at both points.
(B) It is zero at C, but at P it is not zero and is directed inward.
(C) It is zero at C, but at P it is not zero and is directed outward.
(D) It is zero at P, but at C it is not zero.
(E) It is not zero at either point.
Answer - A
F = Eq and F = ma
Tsokos Q1 p. 291
F = Eq
F = 100.0 NC-1 x ( 2.0 x 10-6 C) = 2.0 x 10-4 N
F = ma 
a = F/m = (2.0 x 10-4 N) / 10-3 kg
= 0.2 ms-2 ( 1 sig. dig. )
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3. Electric Potential Energy ( Ue) , Electric Potential Difference –
Voltage ( V)
Q2 , Q3 p. 292
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Q4 p. 293
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Tsokos p. 296 : 4, 5, 6
NOTE: mass of electron = 9.11 x 10-31 kg
Final Answer : F = - 8.0 x 10-16 N “-“ : force
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Tsokos Q 8 p. 295
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Miscellaneous Questions
1. How much work is done moving 10.0 nC through a potential difference of 1.50 x
102 V?
W = qV = (10.0 x 10-9 C) x ( 1.50 x 102 V? ) = 1.50 x 10-6 J
2. The gap between two parallel plates is 1.00 x 10-3 m, and there is a potential
difference of 1.00 x 104 V between the plates. Calculate:
(a) the potential energy gained by an electron (q = -1.60 x 10-19 C) when it is moved
from the positive plate to the negative plate;
W = ∆ Ee = q V = (1.60 x 10-19 C ) x (1.00 x 104 V) = 1. 6 x 10-15 J
(b) the speed of the electron when it reaches the second plate if it is released from
rest (me = 9.11 x 10-31kg); skip for now
1.87 x 107 m/s
(c) the strength of the electric field between the plates.
E=V
d
= 1.00 x 104 V
= 1.00 x 107 V m-1
1.00 x 10-3 m
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Tsokos p. 296 : 9
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4. Electric Current and Resistance Tsokos pp. 310-316
Example 1 : The photo electric effect is where light falling on a metal surface causes
the metal to emit electrons. If 2.2 x 1015 electrons are leaving the metal surface every
second calculate :
a) the number of coulombs per second.
b) the current leaving the surface in amps
a) 1 e- = 1.6 x 10-19 C
2.2 x 1015 electrons x 1.6 x 10-19 C = 3. 5 x 10-4 coulombs per sec.
b) same as a) :
I = amps = Cs-1 =
∆Q
t
= 3. 5 x 10-4 coulombs per sec.
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Example 2.
A current of 1.30 A flows in a wire. How many electrons are flowing past
any point in the wire per second?
I = ∆Q
∆t
I = 1.3 A = 1.3 Cs-1
need to convert C to e-s: 1 e- = 1.6 x 10-19 C
1.3 C x 1 e= 8. 13 x 1018 e1.6 x 10-19 C
Example 3.
per second
A service station charges a battery using a current of 6.7 A for 5.0 h. How
much charge passes through the battery?
6.7 A = 6.7 Cs-1
Total charge = ( 6.7 Cs-1 ) ( 5.0 H) ( 3600s) = 1.2 x 105 C
Tsokos p. 316 : 8
8. a) If a current of 10.0A flows through a heater, how much charge passes through in
one hour?
I = q  q = I x t = (10. 0 A ) ( 1.0 h) ( 3600s/h) = 36000 C ( 2 sig. dig.)
t
b) How many electron does this charge correspond to?
need to convert C to e-s: 1 e- = 1.6 x 10-19 C
36000C x 1 e= 2.3 x 1018 e1.6 x 10-19 C
per second
( 2 sig. dig.)
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5. Resistance and Ohm’s Law
Tsokos p. 316: 10 – 13
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Tsokos p. 316 : 3- 5, 14
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6. Power
Examples p. 34 - 35 of packet
Example1 : If the current through a resistor is halved by what factor will the power
change?
P = RI2  P = ( ½) 2 = ¼
so power reduced by factor of 4
Example 2 : If the voltage across a resistor is doubled by what factor will the power
change?
P = V2  P = ( 2) 2 = 4 so Power increased by factor of 4
R
Filaments and Light bulbs
Light bulbs have power ratings of 40 watts , 60 W , 100W etc. Higher rating light bulbs
will have thicker filaments ( made of tungsten) in order to withstand the heat produced
by the power output when turned on. This is why lights burn out.
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Example: Which heater consumes more power from a 12 V battery, a heater with an
overall resistance of 5 Ω or a resistor or a heater with an overall resistance of 10 Ω.
Can Use all three but if you use one that involves current (I) must calculate current first
from I = V/R
P = V2
R
= ( 12) 2 = 28.8 W
5
P = V2
R
= ( 12) 2 = 14.4W
10
P = I2 R ( 5 ohm resistor)
power is inversely proportional to R
calculate current first :
I = V/R = 12V/ 5 = 2.4 A
P = ( 2.4)2 x 5 = 28.8 W
P = I2 R ( 10 ohm resistor)
calculate current first :
I = V/R = 12V/ 10 = 1.2 A
P = ( 1.2)2 x 5 = 14.4 W
P=VI ( 5 ohm resistor)
calculate current first
I = V/R = 12V/ 5 = 2.4 A
P = 12V x 2.4A = 28.8 W
P=VI ( 10 ohm resistor)
calculate current first
I = V/R = 12V/ 10 = 1.2 A
P = 12V x 1.2A = 14.4 W
Example 3 : What would be different about the filament in a 60 –W bulb compared to a
40-W bulb?
THICKER FILAMENT
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Examples p. 36 of packet
Problems taken from Wilson / Buffa pp. 598-599: 55-61,63,67,68,70,72
1. List 3 different formulas and units for power.
P=W
t
P = qV
t
J sJ s-1
P = IV
AV
P = V2
R
V2 /Ω
P = I2 R
Ω A2
2. A hair dryer is designed for 120 V and does not have dual voltage. What
would happen to it if you plugged it into a 220 V outlet? Use the formula that
explains this in terms of power and voltage.
P = V2
R
V almost doubled so P increased by ( 2) 2 = 4 x
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3. OMIT (2015)
4. A DVD player is rated at 100W at 120V. What is its resistance?
 R = V2
P
P = V2
R
= 144Ω
5. The current through a refrigerator with a resistance of 12Ω is 13 A. What is
the power delivered to the refrigerator?
P = I2 R
= 2028W
6. An ohmic resistor in a circuit is designed to operate at 120 V and dissipate
excess heat in the circuit by a certain power rating.
a) If you connect the resistor to a 60-V voltage source , the resistor will
dissipate heat by what factor of the power rating it was originally designed
for?
P = V2
R
V reduced by ½ 
P = (1/2)2 = ¼ reduced by factor of 4
b) If the designed power is 90 W at 120 V, but the resistor is connected to 40
V, what is the power delivered to the resistor at the lower voltage?
P = V2
R
V reduced by 1/3 
P = (1/3)2 = 1/9 reduced by factor of 9
90 W  10 W
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Tsokos p. 317: 17
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7. Electric Circuits – Series Circuits
Examples p. 41 of packet ( taken from Wilson/ Buffa p. 627)
1. Are the voltage drops across resistors in series generally the same? If not,
under what circumstances could they be the same?
No. If the resistors are all equal value.
2. Which of the following is always the same for resistors in series?
a) voltage b) current c) power d) energy
Current
3. Three resistors have values of 5Ω , 2Ω , and 1Ω ,respectively, are
connected in series to a battery. Which resistor gets the most power and
why?
Larger resistor will receive more voltage and therefore more power. P = I 2 R
4. If a large resistor and a small resistor are connected in series, will the
effective resistance be closer in value to that of the large resistance or the
small one?
Rtot = R1 + R2 + R3
Larger one
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5. Three resistors with values of 5.0Ω , 10 Ω and 15Ω ,respectively, are
connected in series with a 9.0 – V battery.
a) What is the total equivalent resistance?
Rtot = 30.0 Ω
b) What is the current in each resistor?
I = V/R = 9.0 V / 30.0 Ω
= 0.3A
c) At what rate is energy delivered to the 15 Ω resistor? ( rate = power)
P = I2 R = ( 0.3)2 x 15 Ω = 1.35 W
d) Calculate voltage drop across each resistor
V = RI
V5 = ( 5) ( 0.3 ) = 1.5 V
V10 = ( 10 ) ( 0.3) = 3.0 V
V15 = ( 15 ) ( 0.3) = 4.5 V
_____________________________
total = 9 V
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Examples p. 45 of packet ( taken from Wilson/ Buffa p. 627)
1. Are the currents across resistors in parallel generally the same? If not,
under what circumstances could they be the same?
NO. When all resistors have equal value
2. Which of the following is always the same for resistors in parallel?
a) voltage b) current c) power d) energy
Voltage
3. Three resistors have values of 5Ω , 2Ω , and 1Ω ,respectively, are
connected in parallel to a battery. Which resistor gets the most power
and why?
Smaller resistor gets more current and therefore more power : P = V2
R
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4. Three resistors with values of 5.0Ω , 10 Ω and 15Ω ,respectively, are
connected in parallel with a 9.0 – V battery.
a) What is the total equivalent resistance?
1 =
Rtot
1
5
+
1 +
10
1
15
1 =
Rtot
6 +
30
3 +
30
2
30
= 11
30
Rtot = 30 = 2.7 Ω
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b) What is the current in each resistor? I = V/R
I5 = 9/ 5 = 1.8 A
I10 = 9/10 = 0.9A
I15 = 9/15 = 0.6A
____________________________
total = 3.3 A
check : Total V = 9.0 V Total R = 2.7
Total I = 9V/ 2.7 = 3.3A
c) At what rate is energy delivered to the 15 Ω resistor? ( rate = power)
P = V2
R
= 9.02/ 15 = 5.4W
P = I2 R = ( 0.6A)2 (15 ) = 5.4W
d) What is the voltage drop across each resistor? same = 9V
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Tsokos pp. 328: 1- 5
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The II - Series Combo. – Determining V and I in each resistor
Example Try – Determine V and I p. 63 of packet
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FINAL ANSWER
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Example ( taken form Wilson Buffa p. 628 # 27)
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Tsokos p. 329 : IB Example 1,2 , 9, Q5, IB example3
IB Example 1
Six light bulbs , each of constant resistance 3.0 Ω, are connected in parallel to a battery
of emf = 9.0V and negligible internal resistance. The brightness of a light bulb is
proportional to the power dissipated in it. Compare the overall resistance, current and
power of the circuit when all six are on, to that when only five are on, the sixth having
been burned out.
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IB Example Now try this:
Six light bulbs , each of constant resistance 3.0 Ω, are connected in parallel to a battery
of emf = 9.0V and negligible internal resistance. The brightness of a light bulb is
proportional to the power dissipated in it. One light bulb burns out. What happens to the
brightness of each light bulb when 5 bulbs are on compared to when 6 light bulbs are
on. Prove your answer by calculating the power , resistance and current of only one
light bulb in each circuit.
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IB Example 2
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IB Example 3
Four light bulbs each of conctant resistance 60 Ω are connected as show in the figure
below.
a) Find the power in each light bulb.
b) If light bulb A burns out, find the power in each light bulb.
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INTERNAL RESISTANCE
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IB EXAM EXAMPLE – Internal Resistance
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Tsoksos: MISCELLANEOUS EXAMPLES Q6,Q7,Q8
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