Le Fevre High School
SACE Stage 2 Physics
Motion of Charged Particles in Electric Fields 1 Solutions
1.
(a)
Define electric field strength, electric potential energy and electric
potential difference.
The electric field strength at a point is the electric force per unit positive charge
placed at that point,
i.e. E 
F
q
(b)
The diagram below shows a set of points in a region between two
charged conductors.
+ 30 V
A
C
- 10 V
B
(i)
Copy the diagram and sketch the electric field in the region shown.
A
Electric field
lines
+ 30 V
C
-10 V
B
(ii)
Calculate the electric field strength at C if AB = 10 cm and AC = 3
cm. Explain your working.
V 40

d 0.1
 E = 400 V m -1 downwards
F = Eq (or F = Ee for an electron )
= 400 x 1.6 x 10 -19
F = 6.4 x 10 -17 N upwards
E
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(iii) An alpha particle (Helium nucleus, ie doubly ionised) is moved from A
to B. Calculate the change in its electric potential energy.
W = q V
= 3.2 x 10-19 x 40
= 1.28 x 10-17 J
2.
Two insulated metal plates, P and Q stand horizontally 10 cm apart, and are
connected to a 1000 volt battery.
(a) From your knowledge of the field associated with charged parallel plates to
draw a clearly labelled diagram of the electric field in the same region.
+
+
+
+
+
+
-
-
-
+
+
+
+
+
+
+
- - - - Electric
field
lines
+
+
+
+
+
1000 volt
-
-
-
-
0 volt
- -
(b) If point A is 2 cm from plate P, and B is 8 cm from plate P, as shown, find
the work done in moving a -1 C charge from A to B .
W = qV =q(VB - VA) = -1 x 10-6(200 - 800)
= -1 x 10-6 x -600
= +6 x 10-4 J (+’ve indicates work was done on the
object)
(c) What happens to this energy?
Converted to electric potential energy.
(d) If point C is also 8 cm from plate P as shown, find the work done in moving
the -1C charge from C to B. Explain your answer.
P
A
10 cm
C
B
1000 volt
Q
In moving from B to C no work is done as the displacement is at right angles to the
direction of the force acting on it. Under these conditions no work is done by
or against the electric field.
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3.
(a)
Define an electron Volt.
One electron volt is the energy lost or gained by an electron when it moves
through a potential difference of one volt.
(b)
Calculate the speed of an electron, which has been accelerated through
a potential difference of 30 V from rest.
Gain in K = qV
1

mv2 = qV
2
2qV
v2 =
m
2 x 1.6 x 10 19 x 30
=
9.11 x 10 31
v = 3.25 x 10 6 m s -1
4.
An electron starting from rest is accelerated by a uniform electric field of 1000
Volt/metre for a distance of 5 cm. Calculate:
(a)
the acceleration of the electron
Electric force = resultant force

Eq = ma
Eq
a=
m
1000 x 1.6 x10 19
=
9.11 x 10 31
a = 1.76 x 1014 m s-2 towards the positive plate.
(b)
the velocity of the electron after 5 cm of travel
v22 = v12 + 2a s
= 0 + 2 x 1.76 x 10 14 x .05
v2 = 4.2 x 10 6 m s -1 in the direction of 'a'
(c)
the time it takes to travel the 5 cm
v
a=
t
v
 t =
a
4.2 x 10 6
=
1.76 x 1014
t = 2.38 x 10 -8 s
(d)
the kinetic energy of the electron at the 5 cm position
1
K= mv2
2
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1
x 9.1 x 10 -31 x (4.2 x 10 6) 2
2
K = 8.03 x 10 -18 J
=
(e)
5.
the average speed of the electron.
v  v2
s
vave = 1
OR vave =
t
2
vave = 2.1 x 10 6 m s -1
Two parallel metal plates are separated by a distance of 5.00 mm. If the voltage
between the plates is 2.00 x 102 V and the top plate is positive find:
(a)
the force on a charge of +1.50 C placed midway between the plates
d = 5.00 mm = 5.10-3 m
V = 200 Volt
V
E=
(parallel plates)
d
200
=
5 x 10 3
= 40000 V m-1 downwards
It doesn't matter where the charge is placed because the field
between the plates is uniform
 F = qE
= 40 000 x 1.5 x 10 -6
F = 0.06 N downwards
(b)
plate.
the kinetic energy gained by the charge when it moves to the negative
Assume it moves from the mid point to the negative plate
gain in K = qV
= 200 x 1.5 x 10-6
= 3.0 x 10-4 J
OR W by field = Fs
= 0.06 x 2.5 x 10-3
= 1.5 x 10-4 J
= gain in K
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6.
(a)
A test charge of 1.6 x 10-19 C experiences a force of 3.2 x 10-19 N.
What is the electric field strength?
F
E = = 2 N C-1
q
(b)
A charged body shows an acceleration of 10 m s-2. What is the
electric field strength, assuming that it alone is responsible for the
acceleration, if the body has a mass of 1.0 x 10-15 kg and carries a
charge of 1.6 x 10-19 C?
a = 10 m s-2
m = 1.0 x 10-15 kg
q = 1.6 x 10-19 C
F = ma
= 1.0 10-15 x 10
= 1.0 x 10-14 N
F
E=
q
=
1.0 x 10 14
1.6 x 10 19
E = 62.5 x 103 N C-1 parallel to force on + charge
(c)
In the absence of an electric field a charged sphere falls freely under
gravity. The weight of the sphere is 10-14 N. When the electric field is
operational the sphere remains stationary. What is the electric field
strength if the charge on the sphere is 1.6 x 10-19 C?
Fg = 1.0 x 10-14 N
q = 1.6 x 10-19 C
As there is no net force Fg = FE
Fg = Eq
Fg

E=
q
=
1.0 x 10 14
1.6 x 10 19
E = 6.25 x 104 N C-1
direction of field will depend on sign of charge.