Le Fevre High School
SACE Stage 2 Physics
Electric Fields 2 Solution
1.
(a) Define electric field strength. Derive an expression for the electric field strength in the
vicinity of a single point charge.
Electric field strength is defined as the force per unit positive test charge
F
q
To determine the electric field strength in the vicinity of a point charge, consider a small
positive test charge ,qt at a distance r from a source charge q.
ie. E 
E 
F
 qt
r
q
source
charge
1 qq t
4 o r 2

qt

qt
test
charge
1 q
4 o r 2
as by Coulomb's Law F 
1 qq t
4 o d 2
(b) Find the electric field intensity at a distance of 10.0 cm from a point charge of 250 pC in
a vacuum.
r = 10 cm = 0.1 m,
for a point charge,
1 q
E
4o r 2

q = 250 pC = 250 x 10-12 C
9 x 10 9 x 250 x 10 12
(0.1) 2
 2.25 x 10 2 N C 1 directed away
or radially outwards
(c) Find the electric field intensity at a point midway between two charges of +400 pC and 500 pC situated 10 cm apart in a vacuum. (NB 1 pC = 10-12 C.)
EA
A
400 pC
5 cm
5 cm
P
B
-500 pC
EB
electric field strength at P due to A →
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EA 
9 x 10 9 x 400 x 10 12
(0.05) 2
 1.44 x 10 3 N C 1 directed to the right
E B = 1.8 x 10 3 N C -1 to the right
similarly,

 E total = 3.24 x 10 3 N C -1 to the right
2.
What is the electric field intensity at a point 2.4 m from a point charge of 5.7 m C in air?
q = 5.7 m C =5.7 x 10-3 C
. r = 2.4 m
E

1 q
(point charge)
4o r 2
9 x 10 9 x 5.7 x 10 3
(2.4) 2
 8.9 x 10 6 N C 1 directed away
3.
A
If in the fig below A, B and C are
-12
each positive charges of 8.0 x 10
Coulomb,
(a) Find the electric field intensity at a
point C 2.0 cm from A and 2.0 cm
from B.
Electric field strength at C due to A
1 q
E
4o r 2
EA 
23 cm
B
1 cm
2 cm
2 cm
C
30o
30o
9 x 10 9 x 8 x 10 12
(0.02) 2
180 N C 1 directed away from A
E B = 180 N C -1 directed away from B
 E total = E A + E B
E total
(b)
(vector addition )
120o
Hence, find the force acting on the charge C.
180 N C-1
EA
EB
= 180 N C -1 vertically down
F  Eq
= 180 x 8 x 10-12 = 1.44 x 10-9 N vertically down.
4.
180 N C-1
60o EA
Etotal
60o
60o EB
Find the point between charges of +8.00 C and +6.00 C, which are situated 4.00 m apart in
a vacuum, where the electric field strength is zero.
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A
r cm
(4 - r) cm
B
let P be the point where there is no Pnet field at P →
+8 C
+6 C
EA
EA = EB EB
1 q
EA 
but
4o r 2

9 x 10 9 x 8 x 10 6
r2
9 x 10 9 x 6 x 10 6
EB = E A 
(4  r ) 2
Hence,
9 x 10 9 x 8 x 10 6
r2
4
r
2


9 x 10 9 x 6 x 10 6
(4  r ) 2
3
(4  r ) 2
Hence,
2
3

r 4r
2(4  r )  1.732 r
8  2r  1.732 r
3.732 r  8
r  2.145 m
Hence answer is 2.1 m from 8 C charge . (or 1.9 m from the 6 C charge)
5.
A Uranium nucleus can be regarded as a spherical object of radius 2 x 10-14m containing 92
protons as well as neutrons. The electric field around the nucleus is the same as a point charge
concentrated at the centre of the nucleus.
(a)
Estimate the electric field strength at the surface of the nucleus
E

1 q
assuming point charge
4o r 2
9 x 10 9 x 92 x 1.6 x 10 19
(2 x 10 14 ) 2
 3.31 x 10 20 N C 1 radially outwards
(b)
If an alpha particle (this is an helium nucleus containing two protons) is close to the
uranium nucleus estimate the size of the electric force acting on it.
For alpha particle q = 2 x 1.6 x 10 -19 C (remember the charge on a proton is
the same size as that on electron)
assume close distance = 10 x radius
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1 th
as large.
→
 F = Eq
100
1
= 3.31 x 10 20 x
x 2 x 1.6 x 10 -19
100
F = 1.06 N radially outwards. This is a huge force on such a small particle.
Hence estimate the initial acceleration of the helium nucleus away from the uranium
nucleus.
1.06
F
a=
=
m
4 x 1.67 x 10 27
 E will be
(c)
= 1.58 x 10 29 ms -2
6
huge!
(a) How are lines used to represent the electric field strength in an electric field?
The spacing of the lines is used to represent the magnitude of the field strength and the direction of
the lines is used to represent the direction of the field. The larger the spacing the weaker is
the field.
(b) Draw diagrams and sketch on each the electric field lines that correspond to each situation.
(i)
Single Point Positive Charge
(ii)
Single Point Negative Charge
+q
-q
(iii) Two equal magnitude positive charges (iv) Two equal magnitude negative charges
+q
+q
-q
(v) Two equal magnitude but opposite sign charges
+q
-q
-q
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(vi) Oppositely charged parallel plates
(vii) A positively charged hollow sphere
+
+
+
+
+
+
+
+
+
+
+
-
-
-
-
-
-
-
-
-
-
(viii) Negatively charged hollow
sphere
-
-
-
-
-
-
-
-
-
-
-
(ix) Equally charge co-axial cylinders (similar to parallel plate). Application is to a coaxial cable
-
-
+
-
-
+
+
+
+
+
+
-
-
-
+
-
(x) Point charge and a parallel plate. Similar to the electrode system
gun.
+
-
+
+
+
+
of an electron