FIELDS
[62 marks]
1. Two positive and two negative charges are located at the corners of a
[1 mark]
square as shown. Point X is the centre of the square. What is the value of
the electric field E and the electric potential V at X due to the four charges?
Markscheme
A
2. The graph shows the variation with distance r of the electric potential V
from a charge Q .
[1 mark]
What is the electric field strength at distance s?
A. The area under the graph between s and infinity
B. The area under the graph between 0 and s
C. The gradient of the tangent at s
D. The negative of the gradient of the tangent at s
Markscheme
D
A metal sphere is charged positively and placed far away from other charged
objects. The electric potential at a point on the surface of the sphere is 53.9 kV.
3a. Outline what is meant by electric potential at a point.
[2 marks]
Markscheme
the work done per unit charge ✓
In bringing a small/point/positive/test «charge» from infinity to the point ✓
Allow use of energy per unit charge for MP1
3b. The electric potential at a point a distance 2.8 m from the centre of the
sphere is 7.71 kV. Determine the radius of the sphere.
[2 marks]
Markscheme
use of Vr = constant ✓
0.40 m ✓
Allow [1] max if r + 2.8 used to get 0.47 m.
Allow [2] marks if they calculate Q at one potential and use it to get the
distance at the other potential.
A small positively charged object moves towards the centre of the metal sphere.
When the object is 2.8 m from the centre of the sphere, its speed is 3.1 m s−1. The
mass of the object is 0.14 g and its charge is 2.4 × 10−8 C.
3c. Comment on the angle at which the object meets equipotential surfaces
around the sphere.
[1 mark]
Markscheme
90° / perpendicular ✓
3d. Show that the kinetic energy of the object is about 0.7 mJ.
[1 mark]
Markscheme
1
2
× 0. 14 × 10−3 × 3. 12 OR 0.67 «mJ» seen ✓
3e. Determine whether the object will reach the surface of the sphere.
[3 marks]
Markscheme
«p.d. between point and sphere surface = » (53.9 kV –
7.71) «kV» OR 46.2 «kV» seen ✓
«energy required =» VQ « = 46 200 × 2.4 × 10-8» = 1.11 mJ ✓
this is greater than kinetic energy so will not reach sphere ✓
MP3 is for a conclusion consistent with the calculations shown.
Allow ECF from MP1
4. An object of mass m released from rest near the surface of a planet has
an initial acceleration z. What is the gravitational field strength near the
surface of the planet?
A.
z
z
B. m
C. mz
D. m
z
Markscheme
A
[1 mark]
5. The points X and Y are in a uniform electric field of strength
distance OX is x and the distance OY is y.
E. The
[1 mark]
What is the magnitude of the change in electric potential between X and Y?
A.
Ex
B. Ey
C. E(x + y)
D. E√x2
+ y2
Markscheme
A
6. A satellite orbits planet X with a speed vX at a distance r from the centre [1 mark]
of planet X. Another satellite orbits planet Y at a speed of vy at a distance r from
the centre of planet Y. The mass of planet X is M and the mass of planet Y is
4M . What is the ratio of vvX ?
y
A. 0.25
B. 0.5
C. 2.0
D. 4.0
Markscheme
B
The table gives data for Jupiter and three of its moons, including the radius r of
each object.
7a. Calculate, for the surface of Io, the gravitational field strength gIo due to [2 marks]
the mass of Io. State an appropriate unit for your answer.
Markscheme
« GrM
=
2
6.67×10−11×8.9×1022
( 1.8×106 )
2
= »1. 8 ✓
N kg−1 OR m s−2 ✓
A spacecraft is to be sent from
7b. Show that the
Io to infinity.
gravitational potential due to Jupiter at the orbit of Io
gravitational potential due to Io at the surface of Io
is about 80.
[2 marks]
Markscheme
1.9×1027
8.9×1022
AND
seen ✓
4.9×108
1.8×106
27
6
« 1.9×108 ×1.8×1022 = »78 ✓
4.9×10 ×8.9×10
For MP1, potentials can be seen individually or as a ratio.
7c. Outline, using (b)(i), why it is not correct to use the equation
of Io to calculate the speed required for the spacecraft to reach
√ 2G×mass
radius of Io
infinity from the surface of Io.
[1 mark]
Markscheme
«this is the escape speed for Io alone but» gravitational potential / field of
Jupiter must be taken into account ✓
OWTTE
7d. An engineer needs to move a space probe of mass 3600 kg from
[2 marks]
Ganymede to Callisto. Calculate the energy required to move the probe
from the orbital radius of Ganymede to the orbital radius of Callisto. Ignore the
mass of the moons in your calculation.
Markscheme
−GMJupiter (
1
1.88×109
−
1
)=
1.06×109
«5. 21 × 107 Jkg−1 » ✓
« multiplies by 3600 kg to get » 1.9 × 1011 «J» ✓
Award [2] marks if factor of ½ used, taking into account orbital kinetic
energies, leading to a final answer of 9.4 x 1010 «J».
Allow ECF from MP1
Award [2] marks for a bald correct answer.
A vertical wall carries a uniform positive charge on its surface. This produces a
uniform horizontal electric field perpendicular to the wall. A small, positivelycharged ball is suspended in equilibrium from the vertical wall by a thread of
negligible mass.
8a. The charge per unit area on the surface of the wall is σ. It can be shown [2 marks]
that the electric field strength E due to the charge on the wall is given by
the equation
E=
σ
.
2ε 0
Demonstrate that the units of the quantities in this equation are consistent.
Markscheme
identifies units of σ as
C
m2
×
Cm−2 ✓
Nm2
seen and reduced to
C2
N C−1 ✓
Accept any analysis (eg dimensional) that yields answer correctly
8b. The thread makes an angle of 30° with the vertical wall. The ball has a
mass of 0.025 kg.
Determine the horizontal force that acts on the ball.
[3 marks]
Markscheme
horizontal force
T=
F on ball = T sin 30 ✓
mg
✓
cos30
F « = mg tan 30 = 0. 025 × 9. 8 × tan 30» = 0. 14«N» ✓
Allow g = 10 N kg−1
Award [3] marks for a bald correct answer.
Award [1max] for an answer of zero, interpreting that the horizontal force
refers to the horizontal component of the net force.
8c. The charge on the ball is 1.2 × 10−6 C. Determine σ.
[2 marks]
Markscheme
E=
0.14
«
1.2×10−6
σ = « 2×8.85×10
= 1. 2 × 105 » ✓
−12
1.2×10
×0.14
−6
» = 2. 1 × 10−6 «Cm−2 » ✓
Allow ECF from the calculated F in (b)(i)
Award [2] for a bald correct answer.
8d. The thread breaks. Explain the initial subsequent motion of the ball.
[3 marks]
Markscheme
horizontal/repulsive force and vertical force/pull of gravity act on the ball ✓
so ball has constant acceleration/constant net force ✓
motion is in a straight line ✓
at 30° to vertical away from wall/along original line of thread ✓
−6
The centre of the ball, still carrying a charge of 1.2 × 10−6 C, is now placed 0.40 m
from a point charge Q. The charge on the ball acts as a point charge at the centre
of the ball.
P is the point on the line joining the charges where the electric field strength is
zero. The distance PQ is 0.22 m.
8e. Calculate the charge on Q. State your answer to an appropriate number [3 marks]
of significant figures.
Markscheme
Q
0.22
2
=
1.2×10−6
✓
0.182
−6
« + »1. 8 × 10
«C»✓
2sf ✓
Do not award MP2 if charge is negative
Any answer given to 2 sig figs scores MP3
8f. Outline, without calculation, whether or not the electric potential at P is [2 marks]
zero.
Markscheme
work must be done to move a «positive» charge from infinity to P «as both
charges are positive»
OR
reference to both potentials positive and added
OR
identifies field as gradient of potential and with zero value ✓
therefore, point P is at a positive / non-zero potential ✓
Award [0] for bald answer that P has non-zero potential
9. Satellite X is in orbit around the Earth. An identical satellite Y is in a
[1 mark]
higher orbit. What is correct for the total energy and the kinetic energy of
the satellite Y compared with satellite X?
Markscheme
B
10. The escape speed from a planet of radius R is vesc. A satellite orbits the
planet at a distance R from the surface of the planet. What is the orbital
speed of the satellite?
[1 mark]
A. 12 vesc
√2
B. 2
vesc
C. √2vesc
D. 2vesc
Markscheme
A
11. An electron is fixed in position in a uniform electric field. What is the
position for which the electrical potential energy of the electron is
greatest?
[1 mark]
Markscheme
D
A planet of mass m is in a circular orbit around a star. The gravitational potential
due to the star at the position of the planet is V.
12a. Show that the total energy of the planet is given by the equation
shown.
[2 marks]
E = 12 mV
Markscheme
E = 12 m GrM −
GM m
r
comparison with V
= − 12
= − GrM
GM m
✔
r
✔
«to give answer»
12b. Suppose the star could contract to half its original radius without any
loss of mass. Discuss the effect, if any, this has on the total energy of
the planet.
[2 marks]
Markscheme
ALTERNATIVE 1
«at the position of the planet» the potential depends only on the mass of the
star /does not depend on the radius of the star ✔
the potential will not change and so the energy will not change ✔
ALTERNATIVE 2
r / distance between the centres of the objects / orbital radius remains
unchanged ✔
since
ETotal = − 12
GM m
, energy will not change
r
✔
12c. The diagram shows some of the electric field lines for two fixed,
charged particles X and Y.
[2 marks]
The magnitude of the charge on X is Q and that on Y is q. The distance between X
and Y is 0.600 m. The distance between P and Y is 0.820 m.
Q
At P the electric field is zero. Determine, to one significant figure, the ratio q .
Markscheme
kQ
2
(0.600+0.820)
Q
q
=«
=
kq
0.8202
(0.600+0.820)2
0.8202
✔
= 2.9988 ≈ »3 ✔
13. A moon of mass M orbits a planet of mass 100 M. The radius of the planet [1 mark]
is R and the distance between the centres of the planet and moon is 22R.
What is the distance from the centre of the planet at which the total gravitational
potential has a maximum value?
A.
2R
B.
11R
C.
20R
D.
2R and 20R
Markscheme
C
14. The diagram shows the electric field and the electric equipotential
surfaces between two charged parallel plates. The potential difference
between the plates is 200 V.
[1 mark]
What is the work done, in nJ, by the electric field in moving a negative charge of
magnitude 1 nC from the position shown to X and to Y?
Markscheme
A
15. A positive point charge is placed above a metal plate at zero electric
[1 mark]
potential. Which diagram shows the pattern of electric field lines between
the charge and the plate?
Markscheme
C
16. A satellite orbiting a planet moves from orbit X to orbit Y.
[1 mark]
What is the change in the kinetic energy and the change in the gravitational
potential energy as a result?
Markscheme
C
17. The mass of the Earth is ME and the mass of the Moon is MM. Their
respective radii are RE and RM.
Which is the ratio
A.
MR M
√M
M ER E
B.
M ER E
√M
MR M
C. √
M ER M
M MR E
D. √
M MR E
M ER M
escape speed from the Earth
?
escape speed from the Moon
[1 mark]
Markscheme
C
A planet has radius R. At a distance h above the surface of the planet the
gravitational field strength is g and the gravitational potential is V.
18a. State what is meant by gravitational field strength.
[1 mark]
Markscheme
the «gravitational» force per unit mass exerted on a point/small/test mass
[1 mark]
18b. Show that V = –g(R + h).
[2 marks]
Markscheme
at height h potential is V = – GM
(R+h)
field is g =
GM
(R+h)2
«dividing gives answer»
Do not allow an answer that starts with g = – ΔV
and then cancels the deltas
Δr
and substitutes R + h
[2 marks]
18c. Draw a graph, on the axes, to show the variation of the gravitational
[2 marks]
potential V of the planet with height h above the surface of the planet.
Markscheme
correct shape and sign
non-zero negative vertical intercept
[2 marks]
18d. A planet has a radius of 3.1 × 106 m. At a point P a distance 2.4 × 107 m [1 mark]
above the surface of the planet the gravitational field strength is 2.2 N
kg–1. Calculate the gravitational potential at point P, include an appropriate unit
for your answer.
Markscheme
V = «–2.2 × (3.1 × 106 + 2.4 × 107) =» «–» 6.0 × 107 J kg–1
Unit is essential
Allow eg MJ kg–1 if power of 10 is correct
Allow other correct SI units eg m 2s–2, N m kg–1
[1 mark]
18e. The diagram shows the path of an asteroid as it moves past the planet. [3 marks]
When the asteroid was far away from the planet it had negligible speed. Estimate
the speed of the asteroid at point P as defined in (b).
Markscheme
total energy at P = 0 / KE gained = GPE lost
« 12 mv2 + mV = 0 ⇒» v = √−2V
v = «√2 × 6.0 × 107 =» 1.1 × 104 «ms–1»
Award [3] for a bald correct answer
Ignore negative sign errors in the workings
Allow ECF from 6(b)
[3 marks]
18f. The mass of the asteroid is 6.2 × 1012 kg. Calculate the gravitational
force experienced by the planet when the asteroid is at point P.
Markscheme
ALTERNATIVE 1
force on asteroid is «6.2 × 1012 × 2.2 =» 1.4 × 1013 «N»
«by Newton’s third law» this is also the force on the planet
ALTERNATIVE 2
mass of planet = 2.4 x 1025 «kg» «from V = – GM »
(R+h)
force on planet « GM m2 » = 1.4 × 1013 «N»
(R+h)
MP2 must be explicit
[2 marks]
[2 marks]
19. A positive charge Q is deposited on the surface of a small sphere. The
dotted lines represent equipotentials.
[1 mark]
A small positive point charge is moved from point P closer to the sphere along
three different paths X, Y and Z. The work done along each path is WX, WY and
WZ. What is a correct comparison of WX, WY and WZ?
A. WZ > WY > WX
B. WX > WY = WZ
C. WX = WY = WZ
D. WZ = WY > WX
Markscheme
B
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