Transcript Electric Field

SPH4UI Questions
Coulomb’s Law
and Electric Fields
Electric
FE 
kq1q2
r2

FE kq1 V
 2 
q
r
d
Electric Field. The strength of field a distance r from a
charge.
kq1q2
r
Electric Potential Energy. The energy stored in a system
of two charges.
U E kq1

q
r
Electric Potential. The potential energy per unit charge.
UE 
V
Electric force (Coulombs Law). The magnitude of the force
between two charges. If more than two charges, then use
vector sum of forces.
W  U E  qV
 FE r   qr
Work done.
Nm2
k  9.0 10
,
2
C
9
k
1
4 0
e  1.602 1019 C
 0  8.85 10
12
C 2 Permittivity
Nm2 of free space
Recall Coulomb’s Law

Magnitude of the force between charges
q1 and q2 separated a distance r:
k q1q2
Fe 
r2

N  m2
k  9.0 10
C2
9
Force on nucleus of Hydrogen from eQp=1.6x10-19 C
F
+
Qe = -1.6x10-19 C
-
r = 1x10-10 m
F(1on 2)
To the right
2

9 N m 
19
19
9.0

10

1.6

10
C

1.6

10
C




2
C 


2
10
1.0 10 m 
 2.3 108 N
Field
What is the direction of the electric field at point C?
1) Left
Away from positive charge (right)
2) Right
Towards negative charge (right)
3) Zero
Net E field is to right.
y
C
x
Comparison:
Electric Force vs. Electric Field


Electric Force (F) - the actual force felt by a charge at
some location.
Electric Field (E) - found for a location only – tells what
the electric force would be if a charge were located
there:
F = qE

Both are vectors, with magnitude and direction. Don’t
forget to Add x and y components.
Question
What is the direction of the electric field at point A?
1) Up
2) Down
3) Left
4) Right
5) Zero
A
y
Question
What is the direction of the electric field at point A,
if the two positive charges have equal magnitude?
1) Up
2) Down
3) Right
4) Left
A
5) Zero
y
Question
A
Charge A is
B
Field lines start on positive charge, end on negative.
1) positive
2) negative
3) unknown
Question
X
A
Compare the ratio of charges QA/ QB
1) QA= 0.5QB
2) QA= QB
Y
B
# lines proportional to |Q|
3) QA= 2 QB
Question
X
A
Y
B
The electric field
is stronger when
the lines are
located closer to
one another.
The magnitude of the electric field at point X is greater than at point Y
1) True
2) False
Density of field lines gives E
Field Lines
B
A
Compare the magnitude of the electric field at point A and B
1) EA>EB
2) EA=EB
3) EA<EB
E inside of conductor

Conductor  electrons free to move
 Electrons feels electric force - will move until they
feel no more force (F=0)
 F=qE: if F=0 then E=0

E=0 inside a conductor (Always!)
Question
"Charge A" is actually a small, charged metal ball (a conductor).
The magnitude of the electric field inside the ball is:
(1) Negative
(2) Zero
(3) Positive
Question
C
F
Uniform E
A
B
In what direction does the force on a negative
charge at point A point?
1) left
2) right
3) up
Electric field points in the direction a
POSITIVE charge would feel force.
Question
F
F
F
F
F
- C
- A
B
Uniform E
When a negative charge is moved from A to C
the ELECTRIC force does
1) positive work.
2) zero work.
3) negative work.
The work is zero because the path is
perpendicular to the field
Question
C
F
-
F
A
-
F
-
F
-
F
-
B
motion
Uniform E
When a negative charge is moved from A to B
the ELECTRIC force does
1) positive work.
2) zero work.
3) negative work.
The work is negative, the electric
force opposes the direction of
motion
Question
C
A
B
Uniform E
When a negative charge is moved from A to B, the
electric potential energy of the charge
1) Increases
2) is constant
3) decreases
UE = -WE field
Like “climbing up hill” –
increases potential energy
Electric Potential Energy
+
AC: W=0
E
C
CB: W<0
B
A
-
-
-
-
-
When a negative charge is moved from A to B, the electric
potential energy of the charge
(1)
increases
(2)
is constant
(3)
decreases
1) The electric force is directed to bring the
electron closer to be proton.
2) Since the electron ends up further from the
proton the electric field did negative work.
3) So the electric potential energy increased
Work (by field) and  Potential Energy
W = F d cos(q)
Gravity
 Brick raised yi yf
Electric
• Charge moved ∞  rf
• FG = mg (down)
• WG = -mgh
• UG= +mgh
• FE = kq1q2/r2 (left)
• WE = -kq1q2/rf
•  UE= +kq1q2/rf
yf
h
yi
rf
Work done by YOU to assemble 3 charges

W1 = 0J
• W2 =
kq1q2
r12
• W3 =
kq1q3 kq2 q3

r13
r23
• Wtotal =
Remember, no work to
bring the first charge.
kq1q2 kq1q3 kq2 q3


r12
r13
r23
•
 kq1q2 kq1q3 kq2 q3 
WE =   r  r  r 
13
23 
 12

kq1q2 kq1q3 kq2 q3
UE = r  r  r
12
13
23
Watch signs!
3
5m
5m
2
1
5m
Work done by YOU
to assemble 3 negative charges
How much work would it take YOU to assemble 3 negative charges (4.0 uC)?
W
kq1q2 kq1q3 kq2 q3


r12
r13
r23
2
2

9 N m 
6
9.0

10

4.0

10
C



2
C

 3 
5m
 0.086 J
-
5m
1. W = +0.086 J
2. W = 0 J
3. W = -0.086 J
5m
-
-
5m
Electric Potential Energy
of 3 negative charges
Likes repel, so potential energy is still
positive!
-
5m
1. UE = +0.086 J
2. UE = 0 J
3. UE = -0.086 J
5m
-
-
5m
Question
1
5m
W
kq1q2 kq1q3 kq2 q3


r12
r13
r23
2

9 N m 
6
6

4.0

10
C
4.0

10
C


 9.0 10

2
C 

5m
 0.029 J
2
+
+
5m
5m
-
3
The total work required by you to assemble this
set of charges is:
(1)
positive
(2)
zero
(3)
negative
Bring in (1): zero
Bring in (2): positive
Bring in (3): negative x 2
Electric Potential (like height)*

Units Joules/Coulomb Volts
 Batteries
 Outlets
 EKG
Really Potential differences
 Equipotential lines at same height
 Field lines point down hill
 V = k q/r (distance r from charge q)

 V(∞)
=0
Question
C
A
Uniform E ->
B
The electric potential at point A is _______ at point B
1) greater than
2) equal to
3) less than
To go from B to A, a positive charge
must climb “up hill” – increases
potential energy. Hence A is at higher
potential than B
Question
C
Uniform E ->
B
A
Conductor
The electric potential at point A is _______ at point B
1) greater than
2) equal to
3) less than
The electric field is zero at any
point within a conducting material
Electric Potential
+
E
C
B
A
The electric potential at A is ___________ the electric
potential at B.
1) greater than
1) Electric field lines point “down hill”
2) AC is equipotential path (perpendicular to E)
2) equal to
3) less than
3) CB is down hill, so B is at a lower potential than
(“down hill from”) A
Two Charges

In the region II (between the two charges) the
electric potential is
1) always positive
2) positive at some points, negative at others.
3) always negative
I
II
Q=+7.0mC
III
Q=-3.5 mC
Very close to positive charge potential is
positive
Very close to negative charge potential is
negative
Question
Two identical +6 uC charges with mass 10-6 kg are placed as shown
below.
a) Determine the electric potential at P, with position (0, 3)
b) An identical third charge is brought slowly from far away to P. How
much work did this take?
c) The third charge is then released. Calculate the maximum speed it
will retain.
(0, 3)
P
(3, 0)
Question
Two identical +6 uC charges with mass 10-6 kg are placed as shown below.
a) Determine the electric potential at P, with position (0, 3)
b) An identical third charge is brought slowly from far away to P. How much work
did this take?
c) The third charge is then released. Calculate the maximum speed it will retain.
q q 
V k 1  2 
 r1 r2 
(0, 3)
P
(3, 0)
2
6

6.0 106 C 
9 N  m  6.0  10 C
  9.0 10



2
C
3
3
2



 3.1104 V
Question
Two identical +6 uC charges with mass 10-6 kg are placed as shown below.
a) Determine the electric potential at P, with position (0, 3)
b) An identical third charge is brought slowly from far away to P. How much work
did this take?
c) The third charge is then released. Calculate the maximum speed it will attain.
(0, 3)
P
W  qV
(3, 0)
  6.0  106 C  3.1 104 V 
 0.19 J
We could have used:
W
kq1q3 kq2 q3

r13
r23
To calculate the work done
Question
Two identical +6 uC charges with mass 10-6 kg are placed as shown below.
a) Determine the electric potential at P, with position (0, 3)
b) An identical third charge is brought slowly from far away to P. How much work
did this take?
c) The third charge is then released. Calculate the maximum speed it will retain.
(0, 3)
When the third charge is
brought to point P, it has
increased U by an amount
equal to the work done in
bringing it in, 0.19J. When it
is released, the maximum
speed attained will occur
when the U is converted to
Kinetic Energy.
P
(3, 0)
U K
1
0.19 J  mv 2
2
2  0.19 J 
v
m
2  0.19 J 

1106 kg
 6.2 102
m
s
Question: Direction of Electric Field
Two charges are 1.0 m apart. The force acting on charge number 2 is 1.8x1010 N [E],
If charge number 2 is removed, what is the resulting electric field produced by charge
number 1, given:
a) q1 = +1.0 C and q2 = +2.0 C
b) q1 = +1.0 C and q2 = -2.0 C
F
1m
The electric field is given by:
a:
b:
1.8 1010 N
E
2.0C
N
 9.0 109
C
E
or
Ffinal
qadded
or E 
kq1
r2
If point 1 is the only
charge generating the
Field , then you can
also use this equation.
2

9 N m 
 9.0 10
 1.0C 
2
C 
E
2
1.0
m


 9.0 109
N
C
Direction is
same as force,
to the right.
Same magnitude as above, but since the second charge is
negative and still moving away from charge 1, the field as 2
must be to the left.
Question: Electric Forces
Calculate the net force on particle 1:
q1
+
+
q1  5.0  10 6 C
q2
2.0 cm
300
q2  4.0  10 6 C
q3  3.5  10 6 C
1.0 cm
q3
We will need to calculate the horizontal and vertical force on particle 1 from both
particle 2 and particle 3. First for the horizontal forces.
F x  F 12  F 13x

kq1q2 kq1q3
 2  cos  30 
r122
r13
2
2


9 N m 
6
6
9 N m 
6
6
 9.0 10
  5.0 10 C  4.0 10 C   9.0 10
  5.0 10 C  3.5 10 C 
2
2
C 
C 



 cos  30 
2
2
 0.020m 
 0.010m 
 450 N  1364 N
 914 N
Question: Electric Forces
Calculate the net force on particle 1:
q1
q2
2.0 cm
300
+
+
q1  5.0  10 6 C
q2  4.0  10 6 C
q3  3.5  10 6 C
1.0 cm
q3
Now for the vertical forces:
Fx  914 N
F y  F 13y x

Now combine:
kq1q3
 sin  30 
r132
Fy  788 N
FNet 
 914 N    788 N 
2
2
2
3

9 N m 
6
6

1.2

10
N
9.0

10
5.0

10
C
3.5

10
C




2
C 

 sin  30 
2
F
 788 N 
1  y 
 0.010m 
  tan    tan 1 
  41
F
914
N


 x
 788 N
Therefore, the net force on charge 1 is 1200 N [E410S]
Question: Electric Field
Determine the electric field at the indicated point is space (x)
q1
+
q2  5.0C
0.15 m
+
x
q1  6.0C
q2
730
0.10 m
Let’s look at the electric field diagram for the point chosen. Remember that the
electric field lines are drawn away from positive charges and towards negative
charges. Since both q1 and q2 are both negative, the field lines are drawn towards
the charges.
E1
x
730
E2
Question: Electric Field
Determine the electric field at the indicated point in space (x)
q1
+
q1  6.0C
q2  5.0C
0.15 m
+
x
E1
q2
730
0.10 m
x
E2
730
Now let’s calculate the magnitudes of the electric fields using positive values of
charges (q), by using the horizontal and vertical components.
Enetx  E2  E1 cos  73 
Enet y  E1 sin  73 
kq2 kq1
 2 cos  73 
2
r
r
2
2


9 N m 
9 N m 
 9 10
  5.0C   9 10
  6.0C 
2
2
C 
C 



cos  73 
2
2
 0.10m 
 0.15m 

 5.2 1012
N
C


kq1
sin  73 
r2
2

9 N m 
 9 10
  6.0C 
2
C 

 0.15m 
 2.3 1012
N
C
2
sin  73 
Question: Electric Field
Determine the electric field at the indicated point is space (x)
q1
+
q2  5.0C
0.15 m
+
E1
q2
730
x
Enetx
q1  6.0C
0.10 m
N
 5.2 10
C
12
Enet y
x
N
 2.3 10
C
730
E2
12
Now we combine components to determine net electric field.
Enet 

Enetx
 
2
 Enet y
2

2
N 
N

  5.2 1012    2.3 1012 
C 
C

N
 5.7 1012
C
 Enet y
  tan 
 Enet
x

1
2





2.3 1012

 tan 1 
 5.2 1012

 24
N
C
N
C





Since both Ex and Ey
are positive, the
electric field is
5.7x1012 N/C [E240N]
Question: Electric Field
The net force on charge 2 is 18 N [E310N]. If charge 2 is removed, what is the net
electric field at its position?
q
1
+
q1  6.0  C
0.15 m
+
q2
q3
730
0.10 m
q2  4.0  C
q3  5.0  C
F
E
q
F
E
q2
18 N
4.0 106 C
N
 4.5 106
C

Since a negative charge is removed and the field
lines are calculated using positive particles, the
direction of the electric field is opposite to the
direction of the force. Therefore, the electric field
strength is 4.5x106 N/C [W310S]
Question: Conducting Plates
Two identical square conducting plates of side L are separated by a distance d. They
are connected to a battery that maintains a potential difference V between the plates.
A charge q, with mass m, moving parallel to the plates, enters the region midway
between the plates with speed v0 and is deflected so that it just misses the top plate
as it leaves the region. Express your answers to the following in terms of d, V, q, m,
and v0.
a) Determine the magnitude and direction of the nearly uniform electric field in the
region between the plates.
b) What is the sign of the charge?
c) Determine the vertical speed of the charge as it exits the region between the
plates.
d) Determine the length L of the plates.
q
d
v0
L
Question: Conducting Plates
Two identical square conducting plates of side L are separated by a distance d. They are
connected to a battery that maintains a potential difference V between the plates. A charge q,
with mass m, moving parallel to the plates, enters the region midway between the plates with
speed v0 and is deflected so that it just misses the top plate as it leaves the region. Express your
answers to the following in terms of d, V, q, m, and v0.
a) Determine the magnitude and direction of the nearly uniform electric field in the region
between the plates.
q
d
v0
L
By looking at the battery connections, the
top plate is at a higher potential than the
bottom plate (top is positive). Therefore
the field points down between the plates.
The magnitude of the uniform field is:
V

d
Question: Conducting Plates
Two identical square conducting plates of side L are separated by a distance d. They are
connected to a battery that maintains a potential difference V between the plates. A charge q,
with mass m, moving parallel to the plates, enters the region midway between the plates with
speed v0 and is deflected so that it just misses the top plate as it leaves the region. Express your
answers to the following in terms of d, V, q, m, and v0.
b)
What is the sign of the charge?
q
d
L
Since the charge was attracted to the top
while the field was pointing down, the
charge must be negative:
Question: Conducting Plates
Two identical square conducting plates of side L are separated by a distance d. They are
connected to a battery that maintains a potential difference V between the plates. A charge q,
with mass m, moving parallel to the plates, enters the region midway between the plates with
speed v0 and is deflected so that it just misses the top plate as it leaves the region. Express your
answers to the following in terms of d, V, q, m, and v0.
c) Determine the vertical speed of the charge as it exits the region between the plates.
q
d
L
We can use the Work Energy Theorem caring only about the work in moving the
partical in the vertical direction:
½ because
midway
between
voltage
extremes
Welectric  K
1
1
 1

q V   mv 2fy    mviy2 
2
2
 2

mv 2fy  q V  0
v 
2
fy
v fy 
qV
m
qV
m
Question: Conducting Plates
Two identical square conducting plates of side L are separated by a distance d. They are
connected to a battery that maintains a potential difference V between the plates. A charge q,
with mass m, moving parallel to the plates, enters the region midway between the plates with
speed v0 and is deflected so that it just misses the top plate as it leaves the region. Express your
answers to the following in terms of d, V, q, m, and v0.
d) Determine the length L of the plates.
d
L
We need the time it takes for the negative charge to travel the distance L: L=vt
Since the particle starts only with
horizontal velocity, and ends with a
combination of horizontal and vertical
velocity, We can solve for the time for
the particle when it travels vertically
towards the top plate, which is the
same time as the particle takes to travel
the distance L .
Question: Conducting Plates
Two identical square conducting plates of side L are separated by a distance d. They are
connected to a battery that maintains a potential difference V between the plates. A charge q,
with mass m, moving parallel to the plates, enters the region midway between the plates with
speed v0 and is deflected so that it just misses the top plate as it leaves the region. Express your
answers to the following in terms of d, V, q, m, and v0.
d) Determine the length L of the plates.
L  v0t
Therefore
 v0 d
m
qV
L
So,
vy 
d
Now,
qV
m
We know the initial velocity,
final velocity, if we find the
acceleration then we can use
kinematics to find the time, t
ay 
Fy
m
qE

m
qV

md
Applying Kinematics
v y  viy  a y t
 ayt
qV

t
md
qV
m

qV
t
md
td
m
qV
Question: Potential Energy
Calculate the closest distance an alpha particle with charge +3.2x10-19 C gets to a
gold nucleus with charge +1.26x10-17 C if the alpha particle has 7.2x10-13 J of kinetic
energy.
Alpha particle consists of two protons and two neutrons (helium nucleus)
This problems involves the law of conservation of energy. All of the alpha
particles’ kinetic energy is converted to potential energy by the time it is
stopped by the repulsive force between the two positive charges.
Ki  U i  K f  U f
kq1q2
r
kq1q2
r
7.2 1013 J
2

9 N m 
19
17
9.0

10
3.2

10
C
1.26

10
C




2
C 

7.2 1013 J
 5.0 1014 m
7.2 1013 J  0 J  0 J 
Question: Electric Potential
Find the electric potential at a point 4.0x10-3 m away from an  particle (+3.2x10-19 C )
Applying:
k q
V
r
2

9 N m 
19
 9.0 10
  3.2 10 C 
2
C 

V
4.0 103 m
 7.2 107 V
Question: Work
Find the speed of a mass of 2.5x10-6 kg if it moves through a potential difference of
30.0 V and has a charge of 4.1 C.
Applying W=qV gives you the energy the charge gains in the form of kinetic energy
W  qV
  4.1106 C   30.0V 
 1.23 104 J
Equate this to the Kinetic energy formula
1 2
mv  1.23 104 J
2
4
2
1.23

10
J

2
v 
m
2 1.23 104 J 

2.5 106 kg
m
v  9.9
s
Question: Millikan’s Experiment
Derive an equation in terms of m, g, d, and V for the charge on an oil droplet
that is balanced between charged plates.
Sum of the forces
must equal zero to
balance droplet
FE
F net  F E  F g
0   qE  mg
qE  mg
mg
q
E
mg

V
d
mgd

V
Free body
diagram
Fg
Question: Forces Everywhere
Given the angle θ, derive the steps needed to find the charge on an object
held up by a light string between two charged plates, such that it in balance
and not moving.

Using the component methods
y-direction
F net y  FT cos    mg
0  FT cos    mg
mg
FT 
cos  
x-direction
F net x  FT sin    FE
0  FT sin    FE
FE  FT sin  
Therefore
FE 
mg
sin  
cos  
d
 mg tan  
FT
But,

FE  qE
V
FE
d
qV
mg tan   
d
mdg tan  
q
V
q
FT sin  
Fg
FT cos  
Question
Which of the following statements is/are true:
I. If the electric field at a certain point is zero, then the electric potential at the
same point is also zero.
II. If the electric potential at a certain point is zero, then the electric field at the
same point is also zero.
III. The electric potential is inversely proportional to the strength of the electric
field.
A counter example for statement I is provided
A. I only
by two equal positive charges; at the point
B. II only
midway between the charges, the electric field
C. I and II only
is zero, but the potential is not.
D. I and III only
A counter example to statement II is provided
E. None are true
by a electric dipole (a pair of equal but
opposite charges); at the point midway
between the charges, the electric potential is
zero, but the electric field is not.
As for statement III, consider a single positive point charge +Q. Then at
a distance r from the charge the potential is : V=rE, so V is not inversely
proportional to E
Question
If the electric field does negative work on a negative charge as the charge
undergoes a displacement from position A to position B within an electric field,
then the electric potential energy:
A.
B.
C.
D.
E.
Is negative
Is positive
increases
Decreases
Cannot be determined from the information given
By definition UE=-WE so if work is negative then UE is positive, thus
the potential energy increases.
Question
In the figure shown, four equal charges (blue positive, red negative)are
situated at the corners of a square with sides s.
A.
B.
C.
D.
What is the total electrical potential energy of this array of fixed charges?
What is the electric field at the centre of the square?
What is the electric potential at the centre of the square?
Sketch on diagram the portion of the equipotential surface that lies in the
plane of the figure and passes through the centre of the square.
E. How much work would the electric field perform on a charge q as it is
moved from the midpoint of the right side of the square to the midpoint of
the top of the square?
Question
In the figure shown, four equal charges (blue positive, red negative)are
situated at the corners of a square with sides s.
A. What is the total electrical potential energy of this array of fixed charges?
Bring in charge 1:
1
4
2
3
0J
kQ 2
Bring in charge 2: 
J
s
2
2
Bring in charge 3:  kQ  kQ  J
2s 
 s
2
2
2


kQ
kQ
kQ
Bring in charge 4: 


J
s
s
2s 

Total Potential : add up the above

2kQ 2 
 
 J
s 

Question
In the figure shown, four equal charges (blue positive, red negative)are
situated at the corners of a square with sides s.
B. What is the electric field at the centre of the square?
From 2 and 4 we have E1
kQ kQ
 2
2
r
r
1
1
4
4
From 1 and 3 we have E2
kQ kQ
 2
2
r
r
kQ
kQ
1
cos
45


4


2
r2
 1  2
s

2



2
45 45
.5s
3
2
Combining :
4
.5s
4 2kQ
s2
1  1 
r   s  s
2  2 
1 2
s
2
1

s
2

3
2
2
ETotal
Question
In the figure shown, four equal charges (blue positive, red negative)are
situated at the corners of a square with sides s.
C. What is the electric potential at the centre of the square?
The distance, r to the
centre is the same
for all charges
kQ kQ kQ kQ
V 1 2 3 4
r1
r2
r3
r4

1
4
2
3
kQ kQ kQ kQ



r1
r2
r3
r4
0
The potential at the centre of the square is zero
Question
In the figure shown, four equal charges (blue positive, red negative)are
situated at the corners of a square with sides s.
D. Sketch on diagram the portion of the equipotential surface that lies in the
plane of the figure and passes through the centre of the square.
V

kQ1 kQ2 kQ3 kQ4



r1
r2
r3
r4
kQ kQ kQ kQ



r1
r2
r3
r4
0
At every point on the
centre horizontal line,
the corresponding
lines above and below
have the same length,
so V will equal zero.
Question
In the figure shown, four equal charges (blue positive, red negative)are
situated at the corners of a square with sides s.
E. How much work would the electric field perform on a charge q as it is
moved from the midpoint of the right side of the square to the midpoint of
the top of the square?
W  qV
E
The work done by
the electric force as q
is displaced from
point A to Point B is
given by:
B
WE  qVAB
 q VB  VA 
 qVB
Remember
VA=0
B
 kQ kQ
kQ
kQ 
 q  1  2  3  4 
 r1 B r2 B r3 B r4 B 




kQ
kQ
kQ 
 kQ
 q 



2
2
1
1 
1
1




s
A
 s
s2   s 
s2   s 
2
2

2 
2 




 4kQ 2kQ 

 q 

s
5 

s


2 
qQk  4 5  4 



s 
5 
Question
In the figure shown, a 5 uC point charge is located at the origin, and a second
point charge of -2 uC is located on the x-axis at the position (3, 0) m.
A. Determine the electric potential due to these charges at point P, whose
coordinates are (0, 4) m.
B. How much work is required to bring a third charge of 4 uC to point P?
C. What is the total potential energy of these three
P
4m
3m
Question
In the figure shown, a 5 uC point charge is located at the origin, and a second
point charge of -2 uC is located on the x-axis at the position (3, 0) m.
A. Determine the electric potential due to these charges at point P, whose
coordinates are (0, 4) m.
B. How much work is required to bring a third charge of 4 uC to point P?
C. What is the total potential energy of these three
For electric Potential
VP 
kq
V 
r
kq1 kq2

rp1 rp 2
 q1 q1 
k 

r
 p1 rp 2 
2
6

2 106 C 
9 N  m  5  10 C
  9 10



2
C
4
m
5
m



 7650V
P
4m
3m
A 3-4-5 triangle
Question
In the figure shown, a 5 uC point charge is located at the origin, and a second
point charge of -2 uC is located on the x-axis at the position (3, 0) m.
A. Determine the electric potential due to these charges at point P, whose
coordinates are (0, 4) m.
B. How much work is required to bring a third charge of 4 uC to point P?
C. What is the total potential energy of these three
P
W  U p  Vp q p
Work to bring in P
4m
W  Vp q p
  7650V   4 106 C 
3m
 0.0306 J
Work to bring
in P is also
equal to the
potential
energy
increase due
to q1 and q2
U
kq p q1
rp1

kq p q2
rp 2
2
2


9 N m 
6
6
9 N m 
6
6
 9 10
  4 10 C  5 10 C   9 10
  4 10 C  2 10 C 
2
2
C 
C 


4m
5m
 0.0306 J
Question
In the figure shown, a 5 uC point charge is located at the origin, and a second
point charge of -2 uC is located on the x-axis at the position (3, 0) m.
A. Determine the electric potential due to these charges at point P, whose
coordinates are (0, 4) m.
B. How much work is required to bring a third charge of 4 uC to point P?
C. What is the total potential energy of these three charges?
Potential energy is
equivalent to the
Work to bring in each
particle
P
UT 
kq1q2 kq1q3 kq2 q3


r12
r13
r23
4m
3m
6
6
6
6
6
2  4  10 6 C
5

10
C
4

10
C

2

10
C
5

10
C

2

10
C










N

m
9

U T   9 10



2
C
4
m
5
m
3
m


 
 6.0 104 J
We can also solve it by adding the work done to bring in the 4 uC charge
to the potential energy of the two particle system
Question
In the figure shown, a 5 uC point charge is located at the origin, and a second
point charge of -2 uC is located on the x-axis at the position (3, 0) m.
A. Determine the electric potential due to these charges at point P, whose
coordinates are (0, 4) m.
B. How much work is required to bring a third charge of 4 uC to point P?
C. What is the total potential energy of these three charges?
UT  W  U12
kq1q2
 Vp q p 
r12
P
4m
6
2  5  10 6 C

2

10
C




N

m
9

U T   0.0306 J    9 10

2
C
3
m


 
 0.0306 J  0.03 J
 6.0 104 J
We can also solve it by adding the work done to bring in the 4 uC charge
to the potential energy of the two particle system
3m
Question
You have 3 charges ( q1=+6 uC, q2=-3 uC and q3=+2uC) all at an infinite
distance away from the grid whose unit indicators are measured in metres.
y
6
4
2
x
2
4
6
8
10
a) How much work did it take you to place q1 at the origin on the grid?
b) q3 is placed at the location (3, 0) on the grid (q1 is still at the origin).
i) How much work does it take you to assemble this q3 at (3, 0)
ii) How much work did it take the field to assemble q3 at (3,0)?
iii) What is the change in potential energy when q3 is brought in?
c) q2 is now brought in and placed at location (0, 4) (other two charges are still
on grid).
i) How much work does it take you to assemble this q2 at (0,4)?
ii) How much work did it take the field to assemble q2 at (0,4)?
iii) What is the change in potential energy when q2 is brought in?
d) What is the total potential energy of these three assembled particles?
e) What is the electric field’s magnitude at the origin (0, 0) of the assembled
particles q1 and q2?
Question
You have 3 charges ( q1=+6 uC, q2=-3 uC and q3=+2uC) all at an infinite
distance away from the grid whose unit indicators are measured in metres.
y
6
4
2
x
2
4
6
8
10
a) How much work did it take you to place q1 at the origin on the grid?
0J
It costs you nothing to bring in
a charge when no other charge
is present.
Question
You have 3 charges ( q1=+6 uC, q2=-3 uC and q3=+2uC) all at an infinite
distance away from the grid whose unit indicators are measured in metres.
y
6
4
2
x
2
4
6
8
10
b) q3 is placed at the location (3, 0) on the grid (q1 is still at the origin).
i) How much work does it take you to assemble this q3 at (3, 0)
ii) How much work did it take the field to assemble q3 at (3,0)?
iii) What is the change in potential energy when q3 is brought in?
Wyou
2

9 N m 
6
6
6

10
C
2

10
C


 9 10

2
C 
kQ1Q3 


 0.036 J
r
3
W field  0.036 J
U  Wfield    0.036J   0.036J
Question
You have 3 charges ( q1=+6 uC, q2=-3 uC and q3=+2uC) all at an infinite
distance away from the grid whose unit indicators are measured in metres.
c) q2 is now brought in and placed at location (0, 4) (other two charges are
still on grid).
i) How much work does it take you to assemble this q2 at (0,4)?
ii) How much work did it take the field to assemble q2 at (0,4)?
iii) What is the change in potential energy when q2 is brought in?
kQ1Q2 kQ3Q2

r
r
2
2


9 N m 
6
6
9 N m 
6
6
9

10
6

10
C

3

10
C
9

10
2

10
C

3

10
C









2
2
C 
C 


4
5
 0.0513J
Wyou  U 
W field  0.0513 J
U  W field  0.0513J
U  0 J  0.036 J  0.0513J
 0.0153J
Question
You have 3 charges ( q1=+6 uC, q2=-3 uC and q3=+2uC) all at an infinite
distance away from the grid whose unit indicators are measured in metres.
d) What is the total potential energy of these three assembled particles?
U  0 J  0.036 J  0.0513J
 0.0153J
U  0 J  0.036 J  0.0513J
Question
 0.0153J
You have 3 charges ( q1=+6 uC, q2=-3 uC and q3=+2uC) all at an infinite
distance away from the grid whose unit indicators are measured in metres.
e) What is the electric field’s magnitude at the origin (0, 0) of the assembled
particles q1 and q2?
Horizontal:
Vertical:
2

9 N m 
6
 9 10
  2 10 C 
2
C 
kQ 
2  2 
 2000
r
32
2

9 N m 
6

3

10
C

 9 10

2
C 
kQ
N
3  2  


1687.5
r
42
C
2
2
N 
N
N

   2000    1687.5   2616.8
C 
C
C
