PHYSICS 120 : ELECTRICITY AND MAGNETISM
TUTORIAL QUESTIONS
ELECTRIC FIELD AND ELECTRIC POTENTIAL
Question 10
Consider a point 2.00 m away from a -3.00 µC point charge. Calculate
(a) the electric field E and
(b) the electric potential V at this point.
(a) A positive test charge at P will experience a force in the -x̂ direction, attracted to the
-3 µC charge. The electric field, thus, acts in the -x̂ direction. The magnitude of the
electric field is
(9 × 109 ) × (3 × 10−6 )
1 q
=
= 6.75 × 103 V m−1
|E| =
2
2
4πǫ0 r
2
E = −6.75 × 103 x̂ V m−1 ( N C−1 )
(b) For the potential V , the sign of the charge must be used:
V =
1 q
(9 × 109 ) × (−3 × 10−6 )
=
= −1.35 × 104 V
4πǫ0 r
2
Question 11
An electron in a uniform electric field experiences a force of 8.00 × 10−16 N. What is the
magnitude and direction of E at this point?
Place a positive test charge q0 at P . The force on the test charge will act away from P ,
towards the electron. Thus the electric field must necessaarily be towards the electron, in
the -x̂ direction.
(9 × 109 ) × (1.6 × 10−19 )
1 q
=
|E| =
4πǫ0 r2
= 5.00 × 103 N C−1
∴ E = −5.00 × 103 x̂ N C−1
page 1 of 10
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 12
A point charge of -36.0 µC is located at the origin. Find the electric field (a) on the x-axis
at x = 2.00 m, (b) on the y-axis at y = −3.00 m and (c) at the point with coordinates
x = 1.00 m and y = 1.00 m.
(a) If a positive test charge is placed at (x = 2, y = 0), it will experience a force towards
the origin due to the -36.0 µC charge. The electric field thus points in the -x̂ direction.
1 q
(9 × 109 ) × (36 × 10−6 )
|E| =
=
4πǫ0 r2
22
−1
4
= 8.10 × 10 N C
∴ E = −8.10 × 104 x̂ N C−1
(b) As before, put a positive test charge at (x = 0, y = −3). It will experience a force
in the positive y-axis, towards the origin. The electric field then also points in this
direction.
1 q
(9 × 109 ) × (36 × 10−6 )
=
4πǫ0 r2
32
= 3.60 × 104 N C−1
∴ E = 3.60 × 104 ŷ N C−1
|E| =
(c) The electric field (directed towards the origin) is
(9 × 109 ) × (36 × 10−19 )
1 q
√
=
|E| =
4πǫ0 r2
( 2)2
= 1.62 × 105 N C−1
∴ E = 1.62 × 105 x̂ N C−1 at 45°to the x-axis
page 2 of 10
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 13
A charge of 16.00 × 10−9 C is fixed at the origin of coordinates, a second charge of unknown
magnitude is at x = 3.000 m and y = 0 and a third charge of 12.00×10−9 C is at x = 6.000 m,
y = 0. What is the magnitude of the unknown charge if the resultant field at x = 8.000 m,
y = 0 is 20.25 N C−1 ?
Use the principle of superposition and the definition of the electric field for a stationary
point charge.
Given that the electric field at P is
EP = 20.25x̂
The electric field at P due to q1 and q2 is
1
q1 q2 q3
E =
+
+
x̂
4πǫ0 r12 r22 r32
16 × 10−9 q2 12 × 10−9
9
x̂
+ 2+
20.25 x̂ = (9 × 10 )
82
5
22
h
i
q2
9
−10
9
= (9 × 10 ) 2.5 × 10
+
+ 3 × 10 x̂
25
q2
∴
= 2.25 × 10−9 − 2.5 × 10−10 − 3 × 10−9
25
= −1 × 10−9
∴ q2 = 25 × (−1 × 10−9 )
= −2.500 × 10−8 C
Question 14
A small object carrying a charge of −5.00 × 10−9 C experiences a downward force of 20.0 ×
10−9 N when placed at a certain point in an electric field.
page 3 of 10
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
(a) What is the electric field at the point?
(b) What would be the magnitude and direction of the force acting on an electron placed
at the point?
(c) What is the acceleration of the electron?
(a) The direction of the field is in the direction of the force a positive test charge experiences. Hence the field is upwards
E=
F
20 × 10−9
=
= 4.00 N C−1
q
5 × 10−9
(b) The force on an electron (a negative charge) would be downwards. The magnitude of
the force is
F = qE = eE = (1.6 × 10−19 ) × (4) = 6.40 × 10−19 N
(c) The acceleration is
a=
6.40 × 10−19
F
=
= 7.03 × 1011 ms−2
m
9.11 × 10−31
Question 15
A uniform electric field exists in the regien between two oppositely charged, plane, parallel
plates. An electron is released from rest at the surface of the negatively charged plate and
strikes the surface of the opposite plate, 2.00 cm distant from the first, in a time inverval of
1.50 × 10−8 ms−1 .
(a) Find the electric field.
(b) Find the velocity of the electron when it strikes the second plate.
page 4 of 10
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
(a) By definition
F = qE ⇒ E =
F
q
By Newton’s second law, the resultant force is
FNET = ma ⇒ E =
ma
q
From kinematics, the distance is given by s = ut + 12 at2 . Since the electron starts at
rest, then u = 0. The acceleration is then
2s
2 × 0.02
=
t2
(1.5 × 10−8 )2
= 1.78 × 1014 ms−2
(9.11 × 10−31 )(1.78 × 1014 )
∴E =
1.6 × 10−19
= 1.01 × 103 N C−1
a =
Remember that since the electron has negative charge, the electric field always acts
opposite the force on it, as the electron will move from a region of low potential to a
region of high potential.
(b) The velocity is give by
v 2 = u2 + 2as
√
∴v =
2as
= 2.67 × 106 ms−1
Question 16
What is the direction of the electric field at the centre of each of the charge distributions
shown below? The charges are located at the corners of a square.
page 5 of 10
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 17
Three charges are located at the corners of an equilibrium triangle as shown. (i) Determine
the magnitude and direction of E at the centre of the triangle. (ii) Find the magnitude and
direction of the resultant force on the charge −q.
page 6 of 10
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
(i) Place a positive test charge at O. The resulting electric fields are as shown:
(ii) Clearly |EA | = |EB |, since qA = qB . It can be shown that |E(A)x | = |E(B)x |, so therefore
there is no net field in the x direction, and that E(A)y = E(B)y . The y-component of
EB is
1 qB
E(B)y =
4πǫ0 r2
where
a
a
cos 30° =
⇒r=
2r
2 cos 30°
and E(A)y = E(B)y = EB cos 60°. The resulting electric field is then
ENET = EC + 2E(B)y
q
q cos 60°
1
+2
=
4πǫ0 r2
r2
1
q(4 cos2 30°)
q(4 cos2 30°)(cos 60°)
=
+2
4πǫ0
a2
a2
1 4q cos2 30°
=
[1 + 2 cos 60°]
4πǫ0
a2
1 4q
=
(0.75)(2)
4πǫ0 a2
1 6q
=
4πǫ0 a2
Question 18
Two charges q1 and q2 are placed 10.0 cm apart in a vacuum. At what point on the line
joining the two charges is the electric field zero if
(a) q1 = +1.00 µC and q2 = +2.00 µC?
(b) q1 = +1.00 µC and q2 = −2.00 µC?
(a) Only at a point like P can the test charge be repelled with equal and opposite forces.
At P
page 7 of 10
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
1 q1
q2
1
=
4πǫ0 x2
4πǫ0 (0.1 − x)2
2
0.1 − x
q2
= 2=
∴
q1
x
√
0.1 − x
∴ 2 =
x
√
2x = 0.1 − x
√
( 2 + 1)x = 0.1
x = 4.14 × 10−2 m
= 4.14 cm
E=
(b) The neutral point can only occur at a point like Q (since the force on a positive
test charge between q1 and q2 would point in the same direction due to each charge).
Therefore at Q:
|q2 |
1
1 |q1 |
=
2
4πǫ0 y
4πǫ0 (0.1 + y)2
2
q2
0.1 + y
∴
= 2=
q1
y
√
0.1 + y
∴ 2 =
y
√
2y − y = 0.1
√
( 2 − 1)y = 0.1
y = 0.241 m
= 24.1 cm
E=
Question 19
A negative charge −q is fixed to one corner of a rectangle as shown. What positive charge
must be fixed at corner A and what positive charge must be fixed at corner B so that the
total electric field at the remaining corner is zero? Express your answer in terms of q.
page 8 of 10
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Let the unknown charges be qA and qB , placed at A and B respectively. Consider point C,
by placing a positive test charge there. The electric field due to the other charges is shown:
By definition, the electric field at C is given by
1
EA =
4πǫ0
1
=
4πǫ0
1
EB =
4πǫ0
Resolve ED into its components:
qA
(2d)2
qA
4d2
qB
d2
1 qD
cos θ
4πǫ0 5d2
1 qD
sin θ
E(D)y = ED sin θ =
4πǫ0 5d2
Since the electric field at C must be zero, EA must equal E(D)x and likewise EB must equal
E(D)y . Therefore
E(D)x = ED cos θ =
EA = E(D)x
1 qA
1 qD
=
cos θ
2
4πǫ0 4d
4πǫ0 5d2
4
∴ qA =
qD cos 26.6°
5
= 0.715q
since
2d
cos θ = √ ⇒ θ = 26.6°
5d
Then
EB = E(D)y
1 qD
1 qB
=
sin θ
2
4πǫ0 d
4πǫ0 5d2
1
qD sin 26.6°
∴ qB =
5
= 0.0896q
page 9 of 10
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 20
Calculate the field and potential of a point charge of -4.00 nC at a distance of 20.0 cm.
A positive test charge at a point P , a distance of 20.0 cm away, will experience an attractive
force towards the -4.00 nC charge. The electric field will also point towards the -4.00 nC
charge.
1 4 × 10−9
= 900 N C−1 towards the -4.00 nC charge
E=
4πǫ0 (0.2)2
By definition, the electric potential of the point charge
V =
1 q
(9 × 109 )(−4 × 10−9 )
=
= −180 V
4πǫ0 r
0.2
page 10 of 10