NORTHERN CAPE DEPARTMENT OF EDUCATION
PHISICAL SCIENCES
GRADE 11
PHYSICS
ELECTROSTATICS
THEORY & EXERCISES
COMPILED BY:
G. IZQUIERDO RODRIGUEZ
2020
1
Copyright reserved
ELECTROSTATICS
Electrostatics is the study of static electricity (charges at rest). We know:
 an object acquires a positive or negative charge through the loss of electrons from
the object (+ charged) or the gain of electrons by the object (- charged).
 insulators (e.g. plastic, rubber, wool) can be charged through friction and are able
to retain the charge (e.g. a ruler acquires a static charge when rubbed).
 conductors (metals) can be charged through contact or induction, but the charge
can pass through the conductor and flow away.
INDUCTION: An already charged object is held close to another object, which is then
earthed (touched with a finger). It acquires an opposite charge to the already charged
object.
The smallest charge an object could have is when it gains or losses one electron.
This is known as the elementary charge, i.e. e = 1,6 x 10.19 C.
Law of conservation of charge:
Charge cannot be created or destroyed, but can only be transferred from one object to
another.
Or
In a closed system, the total amount of charge remains constant.
COULOMB'S LAW
The scientist Charles de Coulomb investigated the force that charges/charged objects
exert on each other. He found:
F21
F21
F12
-Q2
Q1
Unlike charges
attract each other
F12
Q1
Q2
Like charges
repel each other
The electrostatic force experienced by the objects, is:
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1.
directly
proportional
to
the
2. inversely
proportional
to
the
product of their charges
square of the distance between
FQ1  Q2
them
F
Therefore: F α
1
r2
KqQ
r2
Coulomb's Law in symbols: F 
kQ1Q 2
r2
Coulomb's Law:
The magnitude of the electrostatic force exerted by one point charge (Q 1) on
another point charge (Q2) is directly proportional to the product of the magnitudes
of the charges and inversely proportional to the square of the distance (r) between
them.
Remember

Force is a vector quantity, so the direction of the force has to be indicated, i.e.:
attracting if the charges are unlike and repelling if the charges are like.

Newton third law applies: F12 =- F21

The net/resultant force (Fres) on a single charge can be obtained by the vector
addition of all the forces acting on the charge (take direction into account).
F net  F1  F2  ...  F n
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PROBLEM
SOLVING
STRATEGY
FOR
ELECTROSTATIC
FORCES
AND
COULOMB’S LAW
 MODEL: Identify point charges or objects that can be modelled as point charges.
 VISUALISE: Use a pictorial representation to establish a coordinate system, show
the position of the charges, show the force vectors on the charge, define distance
and angles and identify what the problem is trying to find. This is the process of
translating words to symbols.
 SOLVE: The mathematical representation is based on Coulomb’s law:
F
KqQ
r2

Show the direction of the forces- repulsive for like charges, attractive for
opposite charges- on the pictorial representation.

When possible, do graphical vector addition on the pictorial representation.
While not exact, it tells you the type of answer you should expect.

Write each force vector in terms of its x- and y- components, then add the
components to fined the net force. Use the pictorial representation to
determine which components are positive and which are negative.
 ASSESS: Check that your result has the correct units, is reasonable and that is
answers the question.( Night R. 2008)
EXAMPLE 1
Three point charges Q1, Q2 and Q3 are placed in vacuum as shown in the diagram below.
Charge Q1 has a charge of +6 x 10-6 C, Q2 has a charge of +3 x 10-6 C, and Q3 has a
charge of +6 x 10-6 C.
1.1
State Coulombs’s Law in words.
1.2
Draw a free body diagram to show the forces acting on the point charge Q2 and
find the resultant force that acts on Q2 graphically. Ignore gravitational force.
1.3
Calculate the magnitude of the resultant force that acts on point charge Q 2.
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SOLUTION
1.1
Coulomb’s law states that the magnitude of the electrostatic force between two
point charges is directly proportional to the product of the magnitudes of the charges
and inversely proportional to the square of the distance between them.
OR
The attractive or repulsive force exerted by one point charge on another is directly
proportional to the product of the charges and inversely proportional to the square of the
distance between them.
OR
Coulomb’s law states that the force of attraction or repulsion between two charges is
directly proportional to the product of the charges and inversely proportional to the square
of the distance between their centres.
1.2
1.3
Fnet  ( F12 ) 2  ( F32 ) 2
2
Fnet
 kq q   kq q 
  1 22    3 22 
 r12    r32  
2
2
 9  109  6  10 6  3  10 6   9  109  6  10 6  3  10 6 
 

Fnet  
 

0.062
0.062

 

2
 162  10 3   162  10 3 
  

Fnet  
3 
3 
 3,6  10   3,6  10 
2
2
Fnet  45  45
Fnet  63,64 N
2
2
EXAMPLE 2
Three +100 μC point charges, A, B and C, are equally spaced on a straight line in a
vacuum. The charges are a distance of 3 cm from each other as shown in the sketch
below.
2.1
Calculate the net electrostatic force experienced by point charge C due to
charges A and B.
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SOLUTION
2.1
force on C due to A :
force on C due to B:
FA = kQ1Q2
d2
= (9 x 109)(100 x 10-6)2
(0,06)2
FB = kQ1Q2
d2
= (9 x 109)(100 x 10-6)2
(0,03)2
=
25 000 N right
=
100 000 N right
OR
Distance is halved
 FB = 4 FA
= 4 (25 000) = 100 000 N
⃗ net = F
⃗A + F
⃗B
F
Positive to the right
= 25 000 + 100 000
= 125 000 N; right
ELECTRIC FIELD
An electric field is a region of space in which an electric charge experiences a force. The direction
of the electric field at a point is the direction that a positive test charge would move if placed at
that point.
q
q = point charge
= direction of field
Q
Examples of electric field lines around:
1)
Point charges
2)
Unlike charges
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3)
Like charges
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Field lines:







occur in 3 dimensions around a charge.
start or end perpendicular to the surface of the charged object.
never touch or cross each other.
are closer together (denser) where the E-field is stronger
are further apart (less dense) where the E-field is weaker
are parallel (evenly spaced) where the field is of uniform
strength throughout (e.g. between 2 charged parallel plates) their direction, by
convention, is taken from +  -
Electric field at a point
Electric field at a point is the electrostatic force experienced per unit positive charge
at that point.
E
F
q
SI unit
E [ N·C-1]
or
E [V·m-1]
Electric field at a point is a vector quantity and the direction has to be indicated, i.e. the
direction in which a + charged particle will move, from +
 -.
The electrostatic force exerted on another charge in the field can be calculated as follows:
F  qE
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Electric field strength
The field is the strongest at the point charge, but the further from it, the weaker the field gets.
The strength (magnitude) of the field is indicated by the density of the field lines.
Electric field strength at distance r from point charge Q
If a test charge q is placed at a distance r from a charge Q, the magnitude of the electrostatic
force F that Q and q are exerting on each other is:
F
KqQ
r2
But the magnitude of the force on q as a result of the field around Q is:
F=qE
Therefore
qE 
KqQ
r2
Hence the electric filed strength at a point can be calculated as follows.
E
KQ
r2
The electric field strength at a point is inversely proportional to the squared of the distance
between the point and the charge.
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Electric field at a point due to a number of point charges
If electric fields around charges overlap, the net field strength at any point is determined by
the vector addition of all the electric fields at that point. For example, consider the three
charges +QA,+QB and +Qc respectively at a distance rA, rs and rc from a point P.
At point P the net electric field is:
E res  E A  E B  E C
OR
Eres= EA + EB + EC
PROBLEM-SOLVING STRATEGY: THE ELECTRIC FIELD OF MULTIPLE POINT
CHARGES.
STEP 1: MODEL: Model charged objects as point charges.
STEP 2: VISUALIZE: Make a pictorial representation (drawing/sketch).
For the pictorial representation:
 Stablish a coordinate system
 Identify the point P at which you want to calculate the electric field.
 Draw the electric field of each charge at point P.
 Use symmetry to identify if any components of 𝐸⃗ net are zero.
STEP3: SOLVE: Apply the principle of superposition of fields. The mathematical
representation is 𝐸⃗𝑛𝑒𝑡 = 𝐸⃗1 + 𝐸⃗2 + ⋯ 𝐸⃗𝑛
OR
𝐸⃗𝑛𝑒𝑡 = ∑ 𝐸⃗𝑖
 For each charge, determine its distance from point P.
 Calculate the electric field strength (magnitude) of each charge at point P.
 Sum all the electric fields at point P to determine 𝐸⃗𝑛𝑒𝑡
 If needed, determine the magnitude and direction of 𝐸⃗𝑛𝑒𝑡 .
STEP 4: ASSES: Check that your result has the correct unit, is reasonable, and agrees
with any known limiting cases.
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EXAMPLE 3
The sketch below shows two point charges that form a right-angled triangle with point A.
Their charges are Q1 = -3 x 10-9 C and Q2 = -6 x10-9 C. The distance between A and Q1
is 3 x 10-2 m and the distance between A and Q2 is 5 x 10-2 m.
Q1 = -3 x10-9 C
3 x10-2 m
A
Q2 = -6 x10-9C
5 x10-2 m
3.1
3.2
Define the term electric field at a point in words.
Calculate the net electric field (magnitude and direction) at point A due to the
two point charges.
Solution
3.1
Electric field at a point is the force per unit charge experienced by a charge
placed at that point.
3.2
Right as positive
9 x10 9 x3x10 9
kQ
=
E (Q1 )  2
(3x10 2 ) 2
r
E (Q2 ) 
kQ
r2
=
9 x10 9 x6 x10 9
(5 x10  2 ) 2
= 3X104 N∙C-1
= 2,16 X104 N∙C-1
Enet = (2,16 X 104 ) 2  (3 X 104 ) 2
=3,7x104 N∙C-1
ycomponent
tan θ=
xcomponent
3  10 4
2.16  10 4
θ= tan-11,39= 54,260 OR 54,240
OR
E  3,7x104 N∙C-1 acting at 54,240 to the negative x-axis
OR
E  3,7x104 N∙C-1 acting at 125,760 to the positive x-axis.
tan θ=
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EXAMPLE 4
Two charges, Q1 = 6 nC and Q2 = -3nC, are placed 6 cm apart in a horizontal line, as
shown below.
4.1. Calculate the magnitude of the electric field at x as a result of charge Q1 only.
4.2. Calculate the magnitude of the electric field at x as a result of charge Q2 only.
4.3. Calculate the net electric field at X as a result of the presence of both Q1, and Q2.
4.4. Calculate the magnitude and direction of the force that will experience a charge of
μC at point x.
4
Solution
4.1. E  KQ
2
r
E
9  109  6  10 9
 33750 N  C 1
2
(0,04)
4.2. E  KQ
2
r
E
9  109  3  10 9
 67500 N  C 1
2
(0,02)
4.3. E net  E 1  E 2
Enet at x= E1 + E2
Enet at x= 33750 + 67500=101 250 N∙C-1
4.4. F=qE
F= 4 x 10-6 x 101 250 = 405 000 x10-6 = 0,405 N to the right/in the direction of Q2.
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EXAMPLE 5
The diagram below shows a metal sphere X of negligible mass on an insulated stand in
a vacuum. 3,125 x 1010 electrons have been removed from the sphere.
0,8 m
0,4 m
X
M
P
5.1
Draw the electric field pattern associated with sphere X.
(2)
51.2
Describe an electric field.
(2)
5.3
Calculate the net charge on the sphere.
(3)
5.4
Calculate the electric field at point P.
(3)
5.5
How does the magnitude of the electric field at point M compare with the value
calculated in QUESTION 5.4? Write down only GREATER THAN, EQUAL TO
or SMALLER THAN. Give a reason for the answer.
(2)
5.6
A metal sphere Y, on an insulated stand carrying a charge of -4 nC, is now
placed at point M. Show by calculations where a positive point charge Q should
be placed so that it is in equilibrium.
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(4)
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SOLUTION
5.1
Shape (radial)
Correct direction 
(2)
5.2
5.3
5.4
An electric field is a region of space in which an electric charge experiences a
force. The direction of the electric field at a point is the direction that a
positive test charge would move if placed at that point.
Q=ne
Q=(3,125 ×1010 ) (1,6 ×10-19 )
Q=+5×10-9 C
POSITIVE MARKING FROM 5.3
kQ
E= r2 
E=
(2)
(3)
(9,0×109 )(5×10-9 )

(0,8)2
-1
E=70,31 N∙C away from the charge
5.5
(3)
GREATER THAN 
The electric field at a point due to a point charge is inversely proportional to the
1
square of the distance between the point and the charge/(𝐸 ∝ 𝑟 2 ) 
OR
The distance from the charge to point M is smaller than the distance from the
charge to point P. 
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(2)
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5.6
d
x
+
⃗FxonQ Q ⃗FyonQ Q
y
+
Qx
x
OPTION 1
⃗ x on Q +F
⃗ y on Q =0
⃗ OR -Fx on Q +Fy on Q =0 OR Fx on Q =Fy on Q
F
k
Qx Q
(d+x)2
=k
√(5×10-9 )
0,4+x
Qy Q
x2
=
OR √
Qx
(d+x)2
Qy
=√ x2 OR
√Qx
d+x
=
√Qy
Any one
x
√(4×10-9 )
x

x=3,39 m
OPTION 2
⃗Fx on Q +F
⃗ y on Q =0
⃗  OR -Fx on Q +Fy on Q =0 OR Fx on Q =Fyon Q
k
Qx Q
(d+x)
2
=k
Qy Q
x2
OR Qx x2 =Qy (d+x)2
OR
Qx x2 =Qy d2 +2Qy dx+Qy x2
OR
Qx x2 -Qy d2 -2Qy dx-Qy x2 =0
OR
(Qx -Qy )x2 -2Qy dx-Qy d2 =0
Any one/
(5×10-9 -4×10-9 )x2  -2(4×10-9 )(0,4)x-4×10-9 (0,4)2 =0
OR
(10-9 )x2  -(3,2×10-9 )x+0,64×10-9 =0
OR
x2 -3,2x-0,64=0
OR
3,2±√(3,2)2 -4(1)(-0,64)
x=
2
x = 3,39 m
 OR x=
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3,2±√(3,2)2 +4(1)(0,64)
2
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OPTION 3
⃗ net =0
⃗
F
⃗ x on Q +F
⃗ y on Q =0
⃗ OR -Fx on Q +Fy on Q =0 OR Fx on Q =Fy on Q
F
⃗⃗ x +QE
⃗⃗ y =0
⃗ OR
QE
k
(d+x)
-QEx +QEy =0 OR QEx =QEy
Q
Qx
2
Q
Qx
=k x2y OR √
(d+x)
√(5×10-9 )
0,4+x
=
Any one/
2
=√ x2y OR
√Qx
d+x
=
√𝑄𝑦
𝑥
√(4×10-9 )

x
x=3,39 m
OPTION 4
d
x
⃗⃗ y
E
⃗⃗Ex
Qy
+
Qx
⃗Fnet =0
⃗
⃗Fnet =qE
⃗⃗ net =0
Therefore
⃗⃗ net =0
⃗
E
⃗⃗Ex +E
⃗⃗ y =0
⃗ OR -Ex +Ey =0 OR
k
Qy
Qx
(d+x)
2
√(5×10−9 )
0,4+𝑥
(d+x)
=
x=3,39 m
Qy
Qx
=k x2 OR√
2
x
Ex =Ey
=√ x2 OR
√Qx
d +x
=
√ Qy
x
√(4×10−9 )
𝑥

(4)
[16]
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SUMMARY
ELECTROSTATICS
Electrostatics Force
Electric field
Coulomb`s law
The magnitude of the
electrostatic force exerted
by one point charge (Q1)
Electric field is an
area of space in
which an electric
charge
experiences a
force. The
direction of the
electric field at a
point is the
direction in which
a positive test
charge would
move if placed at
that point.
on another point charge
(Q2) is directly proportional
to the product of the
magnitudes of the charges
and inversely proportional
to the square of the
distance (r) between them:
F
KqQ
r2
Electric field at a
point
The electric field at
a point is the
electrostatic force
experienced per
unit positive
charge placed at
that point.
E=
F
q
For a point charge:
E=
KQ
r2
Lines
Electric field lines are
IMAGINARY LINES along
which a small POSITIVE
test charge would move.
The force experienced by
the positive test charge
is always in the direction
of the tangent to the field
line
They start on a positive
charge and end on a
negative charge.
Principle of
superposition of forces
The force that a
system of point
charges exerts on
another point charge
is equal to the vector
addition of all the
forces each one
exerts on it.
Principle of
superposition of fields
The electric field
strength at a point due
to a system of point
charges is equal to the
vector addition of all
the electric field
strengths of each one
at a specific point.
F net  F1  F2  ...  Fn
E net  E1  E2  ...  En
𝐸⃗
Electric field pattern
Positive like charges
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Unlike charges
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EXERCISES
QUESTION 1: MULTIPLE-CHOICE QUESTIONS
Four options are provided as possible answers to the following questions. Each question
has only ONE correct answer. Choose the answer and write only the letter (A–D) next
to the question number (1.1–1.10) in the ANSWER BOOK, for example 1.11 D.
1.1
Two charged spheres of magnitudes 2Q and Q respectively are placed a
distance r apart on insulating stands.
If the sphere of charge Q experiences a force F to the east, then the sphere of
charge 2Q will experience a force …
1.2
A
F to the west.
B
F to the east.
C
2F to the west.
D
2F to the east.
(2)
P, Q and R are three charged spheres. When P and Q are brought near each
other, they experience an attractive force. When Q and R are brought near
each other, they experience a repulsive force.
Which ONE of the following is TRUE?
A
P and R have charges with the same sign.
B
P and R have charges with opposite signs.
C
P, Q and R have charges with the same sign.
D
P, Q and R have equal charges.
16
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(2)
Two charges, + Q and – Q, are placed a distance d from a negative charge
– q. The charges, + Q and – Q, are located along lines that are perpendicular
to each other as shown in the diagram below.
1.3
+Q●
d
●– Q
●
–q
d
Which ONE of the following arrows CORRECTLY shows the direction of the
net force acting on charge – q due to the presence of charges + Q and – Q?
A
B
C
D
(2)
1.4
Two identical conducting spheres are charged such that sphere 1 has a charge of
4 C and sphere 2 has a charge of −4 C. A third identical sphere is initially uncharged.
If sphere 3 touches sphere 1 and separates, then touches sphere 2 and separates.
The final charge on sphere 3 is…
A
-1 C
B
+ 1C
C
-2C
D
+2C
(2)
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1.5
The sketch below shows a negative and a positive point charge. The magnitude of
the positive charge is greater than that of the negative charge.
Midpoint
Where on the line that passes through the charges is the total electric field zero?
1.6
A
To the right of the positive charge.
B
To the left of the negative charge.
C
Between the charges, to the left of the midpoint.
D
Between the charges, to the right of the midpoint.
(2)
Two spheres S and T on isolated stands carry charges Q and q respectively and
their centres are a distance r apart.
S
Q
A
2F
B
1/2 F
C
1/4 F
D
4F
r
q
The magnitude of the electrostatic force
T exerted by S on T is F. If the distance between
them is doubled the new electrostatic force is:
(2)
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1.7
A learner drew a diagram shown below to show the electric field pattern due to
two point charges and the electric field strength at different points.
Which ONE of the vectors that show the electric field strength at a point was
represented incorrectly?
1.8
(2)
Which ONE of the sketches below correctly represents the electric field lines
of a positive point charge?
B
A
C
D
(2)
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1.9 Two spheres S and T on isolated stands carry charges Q and q respectively and
their centres are a distance r apart. The magnitude of the electrostatic force
exerted by S on T is F.
S
Q
r
q
T
If the distance between them is tripled the new electrostatic force is …
1
A
9
1
B
3
F
F
C
F
D
9F
(2)
1.10
Which ONE of the following combinations is CORRECT regarding the
properties of electric field lines?
Direction
Strength of field
A
Positive to negative
Strongest where the lines are the most dense
B
Negative to positive
Weakest where the lines are the least dense
C
North to south
Strongest where the lines are the most dense
D
North to south
Weakest where the lines are the least dense
(2)
[20]
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QUESTION 2 (DBE/November 2015)
A very small graphite-coated sphere P is rubbed with a cloth. It is found that the sphere
acquires a charge of + 0,5 µC.
2.1
Calculate the number of electrons removed from sphere P during the charging
process.
(3)
Now the charged sphere P is suspended from a light, inextensible string. Another
sphere, R, with a charge of – 0,9 µC, on an insulated stand, is brought close to sphere P.
As a result sphere P moves to a position where it is 20 cm from sphere R, as shown
below. The system is in equilibrium and the angle between the string and the vertical is
7o.
7o
20 cm
R
P
2.2
Draw a labelled free-body diagram showing ALL the forces acting on
sphere P.
(3)
2.3
State Coulomb's law in words.
(2)
2.4
Calculate the magnitude of the tension in the string.
(5)
[13]
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QUESTION 3 (DBE/November 2015)
Two charged particles, Q1 and Q2, are placed 0,4 m apart along a straight line.
The charge on Q1 is + 2 x 10-5 C, and the charge on Q2 is – 8 x 10-6 C. Point X is
0,25
m east of Q1, as shown in the diagram below.
Q1
0,25 m
N
Q2
X
●
E
W
0,4 m
S
Calculate the:
3.1
Net electric field at point X due to the two charges
(6)
3.2
Electrostatic force that a – 2 x 10-9 C charge will experience at point X
(4)
The – 2 x 10-9 C charge is replaced with a charge of – 4 x 10-9 C at point X.
3.3
Without any further calculation, determine the magnitude of the force
that the – 4 x 10-9 C charge will experience at point X.
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(1)
[11]
QUESTION 4 DBE/Feb.–Mar. 2016)
Two identical spherical balls, P and Q, each of mass 100 g, are suspended at the same
point from a ceiling by means of identical light, inextensible insulating strings. Each ball
carries a charge of +250 nC. The balls come to rest in the positions shown in the diagram
below.
Ceiling
20o 20o
P
4.1
Q
In the diagram, the angles between each string and the vertical are the same.
Give a reason why the angles are the same.
(1)
4.2
State Coulomb's law in words.
(2)
4.3
The free-body diagram, not drawn to scale, of the forces acting on ball P is
shown below.
T
Fe
w/Fg
Calculate the:
4.3.1
Magnitude of the tension (T) in the string
(3)
4.3.2
Distance between balls P and Q
(5)
[11]
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QUESTION 5 (DBE/Feb.–Mar. 2016)
A sphere Q1, with a charge of -2,5 μC, is placed 1 m away from a second sphere Q2,
with a charge +6 μC. The spheres lie along a straight line, as shown in the diagram
below. Point P is located a distance of 0,3 m to the left of sphere Q1, while point X is
located between Q1 and Q2 . The diagram is not drawn to scale.
-2,5 μC
Q1
P
●
X
●
1m
0,3 m
5.1
5.2
+6 μC
Q2
Show, with the aid of a VECTOR DIAGRAM, why the net electric field at
point X cannot be zero.
Calculate the net electric field at point P, due to the two charged spheres
Q1 and Q2.
(4)
(6)
[10]
QUESTION 6
The diagram below shows a point charge A with a charge of - 4 × 10-6 C and two points
M and N.
1m
N
A
3m
QA= - 4 × 10-6 C
M
6.1
Define electric field at a point in words
(2)
6.2
Draw the electric field pattern due to point charge A.
(2)
6.3
At what point,M or N, is the magnitude of the electric field due to the point (3)
charge A greater? Explain the answer.
6.4
A positive point charge B with charge + 8 x 10-6 C is placed at point M. Point
charges A and B exerts forces on each other.
6.4.1 State Coulomb’s law in words.
(2)
6.4.2 Calculate the electrostatic force exerted by charge A on charge B.
(4)
6.4.3 Calculate the net electric field at point N.
(5)
[18]
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QUESTION 7 (DBE/November 2016))
7.1
In an experiment to verify the relationship between the electrostatic force, FE,
and distance, r, between two identical, positively charged spheres, the graph
below was obtained.
GRAPH OF FE VERSUS
•
0,030
•
FE (N)
0,025
0,020
•
0,015
•
0,010
•
•
0,005
0
0
1
2
3
4
5
6
-2
(m )
7.1.1
State Coulomb's law in words.
7.1.2
Write down the dependent variable of the experiment.
(1)
7.1.3
What relationship between the electrostatic force FE and the square
of the distance, r2, between the charged spheres can be deduced
from the graph?
(1)
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(2)
7.1.4
7.2
Use the information in the graph to calculate the charge on each
sphere.
(6)
A charged sphere, A, carries a charge of – 0,75 µC.
7.2.1
Draw a diagram showing the electric field lines surrounding
sphere A.
Sphere A is placed 12 cm away from another charged sphere, B, along
straight line in a vacuum, as shown below. Sphere B carries a charge
+0,8 μC. Point P is located 9 cm to the right of sphere A.
(2)
a
of
12 cm
– 0,75 µC
P
+ 0,8 µC
●
A
7.2.2
B
9 cm
Calculate the magnitude of the net electric field at point P.
(5)
[17]
QUESTION 8
Three point charges QA, QB and QC are placed in vacuum as shown in the sketch below.
The distance between point charges QA and QC is 20 cm while that between QB and QC
is 5 cm.
QB=-3.0 x10-6 C
QA=+ 5.0 x10-6C
rAB
QC
5 cm
20 cm
8.1
State Coulomb’s Law in words.
(2)
8.2
How does the magnitude of the electrostatic force exerted by point charge QA
on point charge QB compare with the magnitude of the electrostatic force
exerted by point charge QB on point charge QA ? Write down only GREATER
THAN, SMALLER THAN or EQUAL TO.
(1)
Determine the nature (positive or negative) and calculate the number of
protons or electrons in charge Qc so that the net electrostatic force on QB is
zero.
(8)
8.3
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[11]
QUESTION 9
Two point charges Q1= +5,0 μC and Q2= -3,0 μC 1 m apart, are placed in vacuum
as shown in the sketch below .
Q2
Q1
P
1,0 m
9.1
Define the term electric field at a point.
(2)
9.2
Calculate the net electric field at point P located at the middle of the distance
between the point charges Q1 and Q2.
(6)
[8]
QUESTION 10
The diagram below shows a point charge A with a charge +4,36 μC and two points M and
N.
A
23,2 cm
M
QA= +4,36 μC
10.1
N
Define electric field at a point in words
10.2
Draw the electric field pattern due to point charge A.
10.3
At what point, M or N, is the magnitude of the electric field due to the point
charge A greater? Explain the answer.
10.4
(2)
(2)
(3)
A negative point charge B with charge – 7 x 10-6 C is placed at point M. Point
charges A and B exerts forces on each other.
10.4.1 State Coulomb’s law in words.
(2)
10.4.2 Calculate the magnitude of the electrostatic force exerted by charge A
(4)
on charge B.
10.4.3 Calculate the distance from charge sphere A along the line that passes
through the point charges where the net electric field is zero.
(5)
[18]
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QUESTION 11
Two small objects A and B, equally positively charged, are placed in vacuum as shown in
figure below.
A
B
200 cm
Charge B repels charge A with an electrostatic force of 3,0 x 10-6 N.
11.1
What is the magnitude of the electrostatic force that A exerts on B?
11.2
Calculate the magnitude of charge A and B.
11.3
Draw the resultant electric field lines of the electric field around charges A and B.
11.4
Calculate the electric field strength at a point in the centre of charge A and B.
QUESTION 12
Two metal spheres, P and T, on insulated stands, carry charges of +3 x 10-9 C and
-6 x 10-9 C respectively.
The spheres are allowed to touch each other and are then placed 1,5 m apart as shown
below.
12.1. In which direction will electrons flow while spheres P and T are in contact?
Write down only FROM PTO T or FROM T TO P.
12.2. Calculate the net charge gained or lost by sphere P after the spheres have been in
contact.
12.3. Calculate the number of electrons transferred during the process in QUESTION 12.2.
12.4. A third sphere R, carrying a charge of -3 x 10-9 C, is NOW placed between P and T at
a distance of 1 m from T. Calculate the net force experienced by sphere R as a result
of its interaction with P and T.
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QUESTION 13
Two identical charged spheres A and B are 30 cm apart with charges of +3x10-4 C and – 2
x 10-4 C respectively. A small positive charge C of +10-8 C is placed 10 cm from sphere A as
shown in the diagram below.
A
B
C
10 cm
30 cm
13.1 Calculate the magnitude of the force exerted on the small positive charge C by the
charge A.
13.2 Calculate the magnitude of the force exerted on the small positive charge C by the
charge B.
13.3 Calculate the magnitude of the resultant force exerted on the small positive charge C.
13.4 Calculate the magnitude of the electric field at point C, 10 cm from A as a result of
charge A.
13.5 Calculate the resultant electric field strength at a point situated in the centre of A and
B if the small charge C is removed.
13.6 Charge C makes contact with charge B until the total charge between B and C is
distributed. Calculate the new charge on C.
QUESTION 14
Two small, identical metal spheres A and B carrying charges of +30nC and -14nC
respectively are mounted on insulated stands are placed at a certain distance r from each
other as shown below.
30 nC
A
r
B
- 14 nC
14.1. Sketch the field pattern surrounding the charges.
14.2. Sphere B is moved and makes contact with sphere A. It is then moved back to its
original position.. The magnitude of the electrostatic force between them is now 6,4 x 10 4 N. Calculate the magnitude of the charge on each sphere after contact.
14.3. Determine the original distance between the spheres.
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QUESTION 15
Two metal spheres, P and Q, on insulated stands, carrying charges of +5 x 10-9 C and
+5 x 10-9 C respectively, are placed with their centres 20 mm apart. X is a point at a
distance of 10 mm from sphere Q, as shown below.
15.1. Define the term electric field.
15.2. Sketch the net electric field pattern for the two charges.
15.3. Calculate the net electric field at point X due to the presence of P and Q.
15.4. Use your answer to QUESTION 15.3 to calculate the magnitude of the electrostatic
force that an electron will experience when placed at point X.
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