Le Fevre High School
SACE Stage 2 Physics
Electric Fields 1 Solution
1. (a) State Coulomb's Law.
Any two point charges have acting on them equal sized oppositely directed forces acting
along the line joining their centres. The magnitude of these equal sized forces is directly
proportional to the product of the charges and inversely proportional to the square of their
distance apart. The forces are attractive for unlike charges and repulsive for like charges.
1 q1q 2
.
F
4o d 2
e.g.
f
+1 C
F
+2 C
f
-6 C
-10 C
F
(b) Find the electrostatic force between charges of +2.0 C and +5.0 C separated by a distance
of 75 m in a vacuum.
q1 = 2.0 C q2 = 5.0 C r = 75 m
F

1 q1q 2
4o d 2
9 x 10 9 x 2 x 5
75 2
1.6 x 10 7 N repulsion
2
Two charges of +8.0 mC and -6.0 mC attract each other with a force of 3.0 x 103 N in a
vacuum. What is the distance between the charges?
-3
q1 = 8 x 10-3 C
q2 = -6 x 10 C
3
F = 3 x 10 N
1 q1q 2
F
4o d 2
d 
1 q1q 2
4o F
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9 x 10 9 x 8 x 10 3 x 6 x 10 3
3 x 10  3
d
 12 m
3
A metal sphere of mass 6.0 x 10-3 kg is found to just float in air above a similar metal sphere
when both have a charge of 4.0 mC. Assuming that the only upwards force is electrostatic
repulsion, find the distance between the spheres.
M = 6.x 10-3 kg g = 9.8 ms-2 q1 = 4 mC
q2 = 4 mC (Could both be + or - charges)
Forces acting have equal magnitudes, opposite directions
F coulombic
+
F gravity
Fcoulombic = Fgravitational
Since the gravitaional force is equal and opposite to the coulombic
1 q1q 2
= mg
4o d 2
d 

d
+
1 q1q 2
4o mg
9 x 10 9 x (4 x 10 3 ) 2
6 x 10  3 x 9.8
 1.56 k m
(This indicates how small the gravitational force is compared to the electrostatic!)
4.
A charge of 12 C when placed in an electric field experiences a force of 648 N. What is the
magnitude of the electric field strength?
q = 12 mC
F = 648 N
E

F
q
648
12 x 10 6
E = 5.4 x 10 -7 N C -1
5.
Compare the magnitude of the electrostatic repulsion with the magnitude of the force of
gravitational attraction between two electrons 0.01 cm apart.
m m
Fgravitational  G e 2 e
d
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
6.67 x 10 11 x (9.11 x 10 31 ) 2
(0.01 x 10  2 ) 2
 5.54 x 10  63 N attraction
Fcoulombic 

6.
1 q1q 2
4o d 2
Fgravitational is extremely small and attractive
Fcoulombic is much larger and repulsive.
The point of this question is that in any
dealings with electrons we can ignore
gravitational forces as they are insignificant
compared to electric forces.
9 x 10 9 x (1.6 x 10 19 ) 2
(0.01 x 10  2 ) 2
 2.3 x 10  20 N repulsion
Two small charged spheres have their centres 0.05 m apart and repel each other with a force of
5 x 10-5 N.
(a) What would be the force of repulsion if the spheres were 0.10 m apart?
r = 0.05 m F = 5 x 10-5 N repulsion.
distance of separation increased to 0.10 m, i.e. doubled.
1
Since F  2 ; q1 and q2 constant
r
then force of repulsion quartered
5 x 10 5
F
4
F = 1.25 x 10-5 N repulsion.
(b) What would be the force of repulsion if the spheres were 0.025 m apart?
Separation now 0.025 m
i.e. halved,
 F quadrupled q1 , q2 the same.
 F' = 20 x 10 -5 N repulsion
(c) What would be the force at 0.05 m separation if the charge on one of the
spheres was reduced by a half?
charge on of spheres halved
q q
from F  1 2 2 , q2 and r same
r
 F is halved
F = 2.5 x 10 -5 N repulsion.
7.
What happens to the force between two charged metal spheres in a vacuum if the charge on
each is doubled and the distance between them is multiplied by three?
F
1 q1q 2
4o d 2
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F
q 1q 2
r2
F2 

each charge doubled, r is tripled
2q1 x 2q 2
(3r ) 2
4
F2  F
9
ie. Force is decreased to
4
of original value
9
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8.
Find the magnitude and direction of the force on a charge of +10 C placed at X in each of the
following:
(a)
+5C
+5C
X
1.5 m
1.5 m
You must calculate the force on the +10 C charge due to each of the end charges (here called A and
B ) separately. Then add the two together to get the resultant force.
1 q1q 2
4o d 2
9 x 109 x 5 x 10 6 x 10 x 10 6

(1.5 ) 2
FA = FB  F 
 0.2 N
FB = 2 N
x
FA = 2 N
 Fresult = 0
(b)
+5C
X
2m
+5C
1m
9 x 109 x 5 x 10 6 x 10 x 10 6
FA =
(2) 2
 0.1125 N to the right

FB = 4 x 0.1125 N = 0.45 N to the left. [F 
FB = 0.45
N
x
FA = 0.1125
N
 F res = 0.3375 N to the left
1
d2
]
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(c)
X
1.0 m
+5  C
+5  C
1.0 m
1.0 m
similarly FA = 0.2 N to the right
FB = 0.2 N to the right
 Fresult = 0.4 N to the right
FA = 0.45 N to the right
FB = 0.1125 N to the right
 Fresult = 0.5625 N to the right
distance between points =
1 q1q 2
F
4o d 2
FA = FB

2m
9 x 10 9 x 5 x 10 6 x 10 x 10 6
( 2 )2
 0.225 N
Fresult
0.225 N
Fresult = 0.225 x 2 =0.318 N up along the
perpendicular bisector of line joining charges .
+5  C
0.225 N
+5  C