PHYSICS 120 : ELECTRICITY AND MAGNETISM
TUTORIAL QUESTIONS
COULOMB’S LAW
Question 1
A small sphere A carries a charge of +120 µC. Calculate the magnitude and direction of the
force which it exerts on a second small sphere B carrying a charge of −300 µC if the distance
between them is 30.0 cm.
Coulomb’s Law:
1 q1 q2
r̂
4πǫ0 r2
The magnitude of the force is therefore given by
F=
1 |qA ||qB |
4πǫ0 r2
(120 × 10−6 )(300 × 10−6 )
= (9 × 109 )
(0.3)2
= 3.60 × 103 N
F =
The direction is given by the rule: like charges repel and unlike charges attract. The two
charges are unlike thus sphere A will attract sphere B with a force of 3.60 × 103 N (the force
acts along the line joining the charges).
Question 2
A small metal sphere is given a charge of 20.0 µC and a second identical sphere is located
25.0 cm away is given a charge of -5.00 µC.
(a) What is the force of attraction between the charges?
(b) Calculate the number of excess electrons on the negative sphere.
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
(a) By Coulomb’s Law:
F=
1 q1 q2
r̂
4πǫ0 r2
the magnitude is
1 |qA ||qB |
4πǫ0 r2
(20 × 10−6 )(5 × 10−6 )
= (9 × 109 )
(0.25)2
= 14.4 N
F =
Since the charges are unlike (one positive and one negative) the force will be attractive.
Depending which sphere is considered, the other sphere will move towards it (e.g. if
one considers sphere A, then sphere B will move towards it).
(b) Charge is quantized, therefore the charge Q on sphere B is made up of N electrons
each with a charge e of −1.60 × 10−19 C:
Q = Ne
Q
∴N =
e
−5.00 × 10−6
=
−1.60 × 10−19
= 3.13 × 1013
Question 3
The two spheres of Question 2 are allowed to touch and are again placed 25.0 cm apart.
(a) What charge is transferred between the spheres?
(b) What force exists between them now?
(a) When spheres A and B touch, the total charge of the combination is the sum of the
two charges:
qTOTAL = qA + qB = 20 + (−5) = 15.0 µC
The charge will then be divided equally between the two spheres since they are identical. Thus, each sphere will have a charge of 7.50 µC.
To determine the amount of charge transferred, the transferred charge is given by the
original minus the final:
qTRANSFERRED = qORIGINAL − qFINAL
= 20.0 − 7.50
= 12.5 µC
Therefore sphere A loses 12.5 µC of charge while sphere B gained 12.5 µC of charge.
(b) Calculate the number of excess electrons on the negative sphere.
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 4
Two non-conducting spheres carry a total charge of 190 µC. When placed 1.00 m apart, the
force each exerts on the other is 54.0 N and is repulsive.
(a) What is the charge on each sphere?
(b) What would the charge on each sphere be if the force was attractive?
(a) Let the unknown charges be q1 and q2 . Since F is repulsive, both q1 and q2 are positive
(the total charge is positive).
q1 + q2 = 190 µC
By Coulomb’s Law
F = 54.0 =
1 q1 q2
4πǫ0 r2
= (9.00 × 109 )
q1 q2
1.002
∴ q1 q2 = 6.00 × 10−9
Multiplying q1 with the expression for the total charge gives
q1 (q1 + q2 ) = (190 × 10−9 )q1
q12 + q1 q2 − (190 × 10−9 )q1 = 0
q12 − (190 × 10−9 )q1 + 6 × 10−9 = 0
where q1 can be solved using the quadratic formula:
p
(190 × 10−9 ) ± (190 × 10−9 )2 − 4(1)(6 × 10−9 )
q1 =
2
⇒ q1 = 1.50 × 10−4 or 4.00 × 10−4
Therefore the charges are 150 µC and 40 µC.
(b) If the force is attractive, assume q1 to be postive and q2 to be negative. The total is
still 190 µC, therefore
q1 + q2 = 190 µC
Once again by Coulomb’s Law
F = −54.0 =
1 q1 q2
4πǫ0 r2
= (9.00 × 109 )
∴ q1 q2 = −6.00 × 10−9
Solving as before leads to
q1 = 218 µC
q2 = −27.6 µC
page 3 of 8
q1 q2
1.002
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 5
A 4.5 µC and a -0.50 µC charge are placed 18 cm apart. Where can a third charge be placed
so that it experiencs no net force?
If a third charge, q3 , is placed in between the two original charges, it would not be possible
for it to experience a net force of zero since the force due to both charges would point in the
same direction.
For a positive q3 , the repulsive force due to the 4.5 µC charge would point in the positive x
direction. The attractive force due to the -0.5 µC charge would also point in the positive x
direction.
Similarly, a negative q3 would experience an attractive force due to the 4.5 µC charge, pointing in the negative x direction. The repulsive force due to the -0.5 µC charge would also
point in the negative x direction.
The only possibility is for the third charge q3 to be placed beyond the -0.5 µC. Let this
charge be a distance x away from B. The force FB due to the -0.5 µC charge on q3 is
FB =
1 (0.5 × 10−6 )q3
4πǫ0
x2
and FA due to the 4.5 µC charge on q3 is
FA =
1 (4.5 × 10−6 )q3
4πǫ0 (0.18 + x)2
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
At the neutral point (with a net force of zero), FA must be equal to FB :
FA
1 (4.5 × 10 )q3
∴
4πǫ0 (0.18 + x)2
(4.5 × 10−6 )
(0.18 + x)2
x2
(0.18 + x)2
−6
x
(0.18 + x)
⇒x
⇒ 0.67x
x
= FB
1 (0.5 × 10−6 )q3
=
4πǫ0
x2
−6
(0.5 × 10 )
=
x2
0.5 × 10−6
=
4.5 × 10−6
r
0.5 × 10−6
=
4.5 × 10−6
= 0.33
= 0.059 + 0.33x
= 0.059
= 0.089 m
So the charge q3 must be 8.9 cm from q2 .
Question 6
Two equally charged balls, each of mass 0.10 g, are suspended from the same point by threads
13 cm long. The balls come to rest 10 cm apart due to repulsion. Determine the charge q on
each ball.
The balls are at rest, hence they are in equilibrium and the forces acting on them must
balance.
The force causing the repulsion is the Coulombic force
F12 =
q2
1
1 q1 q2
=
4πǫ0 r2
4πǫ0 (0.10)2
since q1 = q2 .
Now, consider one of the balls. The forces acting on it are:
First, resolve the tenstion T into its components:
Tx = T cos θ
Ty = T sin θ
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
where θ = arcsin(5/13) = 22.6°. Since the forces are balanced,
T cos θ = W
T sin θ = F21
W
∴T =
cos θ
m1 g
=
cos 22.6°
(0.10) × 9.8
=
cos 22.6°
= 1.06 × 10−3 N
⇒ F21 = T sin 22.6°
1
q2
= (1.06 × 10−3 )(sin 22.6°)
4πǫ0 (0.10)2
1
∴ q2 =
(0.1)2 (1.06 × 10−3 )(sin 22.6°)
9.0 × 10−9
= 4.53 × 10−16
∴ q = 2.13 × 10−8 C
Question 7
Three point charges +2.00, +3.00 and +4.00 µC are at the vertices of the equilateral triangle
ABP having sides 10.0 cm. What is the magnitude of the resultant force FR acting on the
+4.00 µC charge?
page 6 of 8
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
The force on the +4.00 µC due to each of the other two is repulsive since they all carry the
same sign. The forces act along the line separating the charges. By Coulomb’s Law
−6
−6
1 q2 q4
9 (2.00 × 10 )(4.00 × 10 )
=
(9.00
×
10
)
4πǫ0 r2
(0.100)2
= 7.2 N
−6
−6
1 q3 q4
9 (3.00 × 10 )(4.00 × 10 )
=
=
(9.00
×
10
)
4πǫ0 r2
(0.100)2
= 10.8 N
F24 =
F34
The cosine rule can be used to get the resultant force FR :
FR2 =
=
=
∴ FR = =
2
2
F24
+ F34
− 2F24 F34 cos θ
2
(10.8) + (7.2)2 − (2)(10.8)(7.2) cos 120°
246.2
15.7 N
The angle with respect to the x-axis can be found using the sine rule
sin θ
sin δ
=
F34
FR
10.8 sin 120°
⇒ sin δ =
15.7
∴ δ = 36.9°
Question 8
Two small spheres are mounted on identical horizontal springs and rest on a frictionless table
as shown. When the springs are uncharged, the spacing between them is 5.00 cm and the
springs are unstressed. When each sphere carries a charge of 1.60 µC, the spacing doubles.
Determine a value for the spring constant.
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Using Coulomb’s Law, the force on one sphere due to the other when the spheres are charged
is
1 q1 q2
4πǫ0 r2
1 (1.6 × 10−6 )2
=
4πǫ0
(0.1)2
= 2.304 N
F21 =
But there is also a force on the spring given by Hooke’s Law:
F = −k∆x
The force F in this case is the Coulombic force F21 , since the Coulombic force on the sphere
is causing the spring to compress
∴ F = −k∆x = F21
−k(−0.025) = 2.304
2.304
k =
0.025
= 92.2 N m−1
(If each spring has an initial length x0 and the separation between the spheres increases from
5.00 cm to 10.0 cm, then x0 must decrease by (10.0 − 5.00)/2 = 2.50 cm.)
page 8 of 8