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
ELECTRIC FIELD

COE, DC

EM
Name
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20
Name : _________________________
1
Time: 25 mins
(a)
Define electric potential.
(b)
A helium nucleus with a charge of +3.20  1019 C and mass of 6.74 × 10–27 kg approaches a
stationary gold nucleus of charge +1.26  1017 C along the line joining the two nuclei, as shown
in Fig. 13. When the helium nucleus is at a distance of
1.77 × 10–12 m away from the gold
6
–1
nucleus, it is traveling at a speed of 7.0 × 10 m s . Assume that the helium nucleus and gold
nucleus are point charges.
Gold
nucleus
[1]
7.0  106 m s1
X
Helium
nucleus
1.77  1012 m
Fig. 13
(i) 1.
2.
Calculate the electric potential due to the gold nucleus at a distance of 1.77 ×
10–12 m.
[1]
Hence, determine the electric potential energy of the helium nucleus when the
separation of the helium nucleus and the gold nucleus is 1.77 × 10–12 m.
[1]
(ii) At point X, the helium nucleus comes to rest momentarily. Given that the gold
nucleus remains stationary, find the distance of point X from the gold nucleus.
[3]
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(iii) If the gold nucleus recoils away from the helium nucleus, explain whether your
answer to (b) (ii) will be larger, smaller or the same.
[1]
2
An electron is directed into a uniform electric field at an angle of 300 to the vertical as shown in Fig. 14
below. Its initial speed is 1.0 × 106 m s─1. The parallel plates are arranged 10 cm apart in vacuum. The
right plate is at a potential of +20 V relative to the left plate. The electron moves along the path as
shown by the dotted line and hits the positive plate. The mass of an electron is 9.11 x 10−31 kg.
e
1.0 × 106 m s ─1
30
0V
0
10 cm
+20 V
(a)
Fig. 14
State the shape of the path followed by the electron.
[1]
(b)
Find the magnitude and direction of the acceleration experienced by the electron.
[2]
(c)
Show that the time taken for the electron to reach the positive plate is 6.26  10 8 s .
(d)
[2]
Hence, find the minimum length of the plates required to stop the electron from leaving
the uniform electric field.
[2]
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3)
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18
Name : _________________________
1
(a)
Time: 25 mins
P and Q are two identical conducting spheres each carrying a charge +Q. They are placed in
a vacuum with their centres distance d apart as shown in Fig. 4.1.
d
P
Q
Fig. 4.1
Explain why the force F between them is not given by the expression
F
Q2
4 0d 2
……………………………………………………………………………………………
……………………………………………………………………………………………
………………………………………………………………………………………………
(b)
[2]
Electric fields and magnetic fields may be represented by lines of force. Fig. 4.2 shows some
lines of force.
B
A
Fig. 4.2
(i)
State whether the field strength at the vicinity of A and at the vicinity of B is constant,
increasing or decreasing when measured in the direction from A towards B.
at A: ………………………………………………………………………..
[1]
at B: ………………………………………………………………………..
[1]
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(ii)
Explain why the field lines can never touch or cross.
……………………………………………………………………………………
……………………………………………………………………………………
………………………………………………………………………………
2)
[2]
Two equally charged solid conducting spheres with small radii, of equal uniform mass m are hanged from the
top plate of two parallel charged plates, with insulated strings of negligible mass. Each sphere has a charge of
3.2 1012 C
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3)
(a)
Define electric potential energy at a point.
[1]
(b)
The figure below shows the path of a proton
nucleus.
1
1
p travelling head-on towards a uranium
235
92
U
1.5x10-14 m
path of proton
uranium nucleus
By considering only the electrostatic repulsion, the distance of closest approach of the proton
to the uranium nucleus is 1.5x10-14 m. Mass of proton is 1.67  1027 kg
(i)
Name the form of the energy to which the kinetic energy of the proton has been
converted when it is at the distance of closest approach.
[1]
(ii)
Calculate the initial speed of the proton, and state the assumption that you have made
in the calculation.
[4]
(iii)
In nuclear fission reactions, protons are not normally used to induce the reactions.
Suggest an alternative particle that you would use to penetrate the nucleus and explain
your choice.
[2]
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24
Name : _________________________
1)
Time: 25 mins
The two lamps on a bicycle are powered by a small dynamo. The dynamo’s internal resistance is 20
 and at one particular speed, the dynamo generates an e.m.f. of 6.0 V. Each lamp may be
considered to have a constant resistance of 5.0 .
(a)
(b)
2)
Calculate the current in each lamp if the lamps are connected
(i) in series
[1]
(ii) in parallel
[1]
Calculate the power dissipated in each light bulb for (i) and (ii)
[1]
Fig below shows a circuit for measuring a small e.m.f. produced by a solar cell.
10 V
0.5 
5.0 
Solar cell
V
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(a)
The galvanometer shows null deflection when the variable resistor is set to 300 . Determine
the value of the e.m.f., V of the solar cell. [1]
(b)
Fig 3.2 shows the 5.0  resistor being replaced with a 1.2 m uniform resistance wire PQ of
total resistance of 7.0 . The variable resistor remains at 300.
10 V
0.5 
I
Q
P
Solar cell
V
moveable
contact R
(i)
Calculate the current I, when the galvanometer shows null deflection. [2]
(ii)
Calculate the distance from P that contact R must be connected to wire PQ such that the
galvanometer shows null deflection. [3]
(iii)
Explain why, this circuit is not suitable for measuring the e.m.f. of the solar cell when the
value of the e.m.f. of the solar cell is of the order of millivolts. [1]
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4
A potentiometer has a uniform resistance wire AB of length 1.00 m and resistance R. It is powered by a battery
of e.m.f. 2.00 V and negligible internal resistance, as shown. The resistance of the connecting wires for the
potentiometer circuit is represented by RW.
RW
2.00 V
R
A
B
X
(a)
Y
With a standard cell of e.m.f. 1.018 V, connected between X and Y, the balance length is found to be
51.2 cm of the wire. Determine the ratio
(b)
RW
. [3]
R
When a dry cell is connected between X and Y, the balance length is 74.8 cm. Calculate the
e.m.f. of this dry cell. [3]
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(a)
Distinguish between resistance and resistivity of a conductor.
[2]
(b)
A cell of e.m.f. 2.50 V and internal resistance R is connected to two uniform resistive wires in
series as shown in Fig. 2.1. The wires are made of the same material but have different
lengths and diameters. Wire AB is 50.0 cm long and has a diameter d, whereas wire BC is
30.0 cm long and has a diameter 0.30 d. The ammeter and connecting wires are assumed to
have no resistance.
2.50 V
A
R
A
Show that
B
RAB
 0.150
RBC
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C
[1]
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(c)
A battery of e.m.f. 2.00 V and internal resistance r is connected across wire BC in
parallel with another resistor of resistance r as shown in Fig 2.2. The galvanometer
shows no deflection when the jockey J is at the midpoint of wire BC.
2.50 V
A
R
A
B
C
J
2.00 V
r
r
(i)
Show that VBC = 2.00 V
(ii)
Determine the internal resistance R of the 2.50 V cell if the ammeter shows a
[2]
reading of 0.400 A.
[3]
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18
Name : _________________________
1
Time: 25 mins
The circuit in Fig. 18.1 makes use of a thermistor immersed in a water bath to monitor the temperature of the
bath. The thermistor T is connected in series with a fixed resistor R of resistance 500  and a 3.0 V battery of
negligible internal resistance.
A
T
3.0 V
B
R
C
o
o
When the temperature of the water bath is 80 C and 20 C, the potential difference across BC is 2.5 V and 0.5
V respectively.
(a)
Determine the resistance of T when the temperature of the water bath is
(i) 80 oC
resistance =

[2]
resistance =

[2]
(ii) 20 oC
(b)
State the variation of the resistance of the thermistor with its temperature.
[1]
(c)
The circuit in Fig. 18.1 can be used to check that the temperature of the water bath is maintained below
80 oC. A student suggests connecting a 2.5 V light bulb across BC and if the temperature of the water
bath is at 80 oC, the bulb will light up. State why his suggestion is not feasible.
[2]
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2
r
E
Switch, S
A
RL
An ammeter A of negligible resistance is connected in series with a variable load resistor of resistance RL to
determine the electromotive force E and the internal resistance r of a cell (Fig. 20.1). The following values of
current I are obtained for different values of RL: The graph of R against
1
is plotted as shown, with gradient =
I
1.5085 and y-intercept = -0.573.
I/A
0.590
3.00
0.420
4.00
0.330
5.00
0.270
6.00
0.230
Graph of R L ag 1/I
7
R = 1.5085 / I - 0.573
6
5
4
L
RL / Ω
2.00
3
2
1
0
0
1
2
3
4
5
-1
(1/I) / A
(a)
Find the EMF of the battery and the internal resistance r.
[2]
(b)
The circuit in Fig. 20.1 can be modified to vary the brightness of showroom lights by adding bulbs in
parallel to the variable load resistance RL. Fig. 20.3 shows a simplified model of such a circuit.
r
E
Switch, S
A
RL
RB
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(i) Resistances RL and RB may be combined and treated as a single load R. The resistance of
the bulb RB remains constant at 4.00 Ω and maximum power delivered by the cell occurs
when R = r. When the cell delivers maximum power, calculate
1. the value of RL.
RL =
Ω
[2]
current =
A
[2]
power =
W
[2]
2. the current in the circuit.
3. the power dissipated in the cell.
(ii)
Although the cell is supplying maximum power to the combined external load, the light
bulb is not at its brightest. Explain why.
[1]
3 (a)
The figure shows a cell of e.m.f. 2.0 V and internal resistance 0.20  connected in parallel to two identical
lamps L1 and L2. The ammeters A1 and A2 in the circuit have negligible resistance and A2 reads 0.50 A.
A1
A2
2.0 V
0.20 
(i)
L1
L2
Calculate the potential difference across L1.
potential difference = …………V [2]
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(ii)
If another identical lamp L3 is connected in parallel with L1 and L2, explain whether the
current in ammeter A1 remains the same, increases or decreases.
[2]
(b)
Fig. 4.2 shows a circuit which is used to measure the emf of Cell Y.
Cell Y
A
J
B
8.0 
Cell X, 2.00 V
Fig. 4.2
Cell X has an e.m.f. of 2.00 V and negligible internal resistance. It is connected in series with a 8.0  resistor and
resistance wire AB. The resistance wire AB has a length 100.0 cm and a resistance, 2.0 .
(i)
Calculate the potential difference across AB.
potential difference = …………V [1]
(ii)
(iii)
The movable contact J is now moved along AB. When the galvanometer indicates a zero
reading, the length AJ is 30.0 cm. Calculate the e.m.f., in mV, of Cell Y.
e.m.f.= ……..….. V [2]
Without changing the length AB, suggest three modifications to the circuit of Fig. 4.2 that
would cause the contact J to be closer to A when the galvanometer gives a zero reading.
[2]
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18
Name : _________________________
1)
(a)
Time: 25 mins
State what is meant by a field of force.
[1]
(b)
Atoms of Neon-20 are ionised by the removal of one electron from each atom. These
ions are accelerated through a potential difference of 1400 V. They are then injected into
a region of space where there are uniform electric and magnetic fields acting in right
angles to the original direction of motion of the ions. The region of magnetic and electric
fields and polarities of the electric plates are shown in Fig 1.1.
+ + + + + + + +
Region of uniform
electric
and
magnetic fields
Path of Neon ions
- - - - - - - -
Fig 1.1
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The electric field has field strength E and the flux density of the magnetic field is B.
(i)
State the charge of a Neon ion.
Charge of a Neon ion: …………………………. [1]
(ii)
Calculate the speed of the Neon ions before they enter into the region of the electric
and magnetic fields. Mass of Neon 20 atom can be assumed to be 20 X mass of
27
proton. Mass of proton = 1.66 10 kg
Speed = …………………….. m s-1
(iii)
(1)
[3]
State the direction of the electric force acting on a Neon ion when it enters into
the shaded region.
Direction of electric force: ……………………. [1]
(2)
(iv)
Draw the direction of the magnetic field onto the shaded region in Fig 1.1 such
that the Neon ions pass through the shaded region undeflected.
The electric field strength E is 6.2 x 103 V m-1. Calculate the magnitude of the magnetic
flux density so that the ions are not deflected in the region of the fields.
Magnetic flux density = …………………… T
(c)
[1]
[2]
The mechanism by which the Neon atoms in (b) are ionised is changed so that each
atom loses two electrons instead of just one. State the changes that occur in
(i)
the speed of the ions entering the region of the electric and magnetic fields in (b).
[1]
(ii)
the path of the ions in the two fields.
[1]
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2)
A narrow beam of electrons at a speed of 3.2 × 107 m s─1 travels along a circular path in a uniform
magnetic field of flux density, B, as shown in Fig. 3.1 below.
B
incident beam of
electrons
Fig. 3.1
(a) (i) Explain why the electrons undergo uniform circular motion.
[3]
..………………………………………………………………………………...…
..…………………………………………………………………………………...
..………………………………………………………………………………...…
..…………………………………………………………………………………
(ii) Show that the speed, v, of the electrons in the field is given by
v
Ber
m
where r is the radius of the circular path of the beam in the field.
[2]
(iii)
The radius of the circular path of the beam in the field was found to be 25 mm. Determine the
magnetic flux density of the field.
Flux density = ….................... T [2]
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3)
(a) A charged particle may experience a force in an electric field and in a magnetic field.
State the difference between the forces experienced in the two types of field.
.....................................………………………………………………………………………………………… [1]
(b) A proton, travelling in a vacuum at a speed of 4.5 × 106 m s–1, enters a region of uniform magnetic field of
flux density 0.12 T. The path of the proton in the field is a circular arc, as illustrated in Fig. 3.1.
Fig 3.1
(i) State the direction of the magnetic field.
.…………………………………..……………………………………………………… [1]
(ii) Calculate the radius of the path of the proton, in terms of cm, in the magnetic field.
27
Mass of proton = 1.66 10 kg
radius = ………………………… cm [3]
(c) A uniform electric field is now created in the same region as the magnetic field in
proton passes undeviated through the region of the two fields.
(i) On Fig. 3.1, mark with an arrow labelled E, the direction of the electric field.
Fig. 3.1, so that the
[1]
(ii) Calculate the magnitude of the electric field strength.
electric field strength = ………………………… N C-1 [2]
(d) Suggest why gravitational force on the proton has not been considered in the calculations in (b) and (c).
.....................…………………………………..……………………………………………..…………… [1]
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28
Name : _________________________
1)
Time: 30 mins
(a)
Electrons from a filament source enter a region between the parallel plates after being
accelerated by an electric field. Fig. 7.1 below shows the electrons travelling horizontally at a speed
of 2.50 x 107 m s-1 entering the pair of parallel plates.
+V
40 mm
7
2.5 x 10 m s
-1
Fig. 7.1
0V
(i)
The electrons deviate by 30° on leaving the parallel plate of 80.0 mm long. The separation
between the plates is 40.0 mm. Calculate the time taken for the electrons to traveling through the
plates.
time = ………………….. s
(ii)
Calculate the vertical component of the velocity when the electrons exit the parallel plates.
vertical component of velocity = ……………… m s-1
(iii)
[1]
[2]
Hence calculate the acceleration of the electrons. Given the mass of electron is 9.11 10 31 kg
acceleration = ………………..m s-2
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(iv)
Calculate the p.d. V between the two plates.
p.d. = ………………… V
[2]
(b)
A magnetic field is applied to the region of the electric field in Fig. 7.1 so that the electrons
pass straight through undeflected.
(i)
Indicate in words or sketch the direction of the magnetic flux density for Fig. 7.1 in the space
below.
[1]
(ii)
Calculate the magnetic flux density required to produce this effect.
magnetic flux density = ………………… T
[2]
(c) The parallel plates in Fig. 7.1 are removed but the magnetic field is kept at the same value and
direction.
(i)
Draw the path taken by the electrons in the magnetic field in the space below.
(ii)
Calculate the radius of curvature of these electrons.
radius = ……………….. m
(iii)
[1]
[2]
Determine the speed of the electrons upon leaving the field. Explain your reasoning.
…………………………………………………………………………………
………………………………………………………………………………[2]
2)
(a) Define magnetic flux density.
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.......................................................................................................................................
.................................................................................................................................
[1]
(b) Describe, by means of a well-labelled diagram, the motion of electrons moving at right angles in a uniform
(i) electric field,
..................................................................................................................................
...........................................................................................................................
[2]
(ii) magnetic field.
..................................................................................................................................
...........................................................................................................................
[2]
(c) A rectangular strip of copper of dimensions 10 mm  5 mm  1 mm carries a conventional current I.
A magnetic field B of flux density 1.0 T is applied in a direction perpendicular to the strip as shown in
Fig. 4.1.
10 mm
I
P
Q
S
R
5 mm
1 mm
B  1.0 T
Fig. 4.1
(i) Explain how a voltage is set up across side PQ with respect to SR.
..................................................................................................................................
..................................................................................................................................
..................................................................................................................................
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..................................................................................................................................
..................................................................................................................................
...........................................................................................................................
[3]
(ii) Calculate the voltage, given that the velocity of the electrons is 2.52  10 5 ms–1.
voltage = ........................................ V [2]
3)
(a) Define the tesla
[1]
(b) Fig. 4.1 shows an electric motor which is made up of a rectangular coil of wire of 150 turns. The
coil is 0.20 m long and 0.12 m wide. The coil has a current of 0.32 A flowing through it and its
plane is parallel to a field of magnetic flux density 0.36 T.
magnetic field
0.12 m
Y
0.20 m
0.32A
X
(i) Draw arrows on Fig. 4.1 to represent the directions of the magnetic forces acting
on the coil. Label them F.
[1]
(ii) Calculate the magnitude of the magnetic force acting on one side of the coil.
force = ………………………… N [1]
(iii) Calculate the torque which is exerted on the coil.
torque = ………………………… N m [1]
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