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Norah Ali Al-moneef
king saud university
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king saud university
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king saud university
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q1
d
What is the potential at point P, located at the
center of the square of point charges.
Assume that d = 1.3m and the charges are
q1 = +12 n C,
q3 = +31 n C,
q2= -24 n C
q4= +17 n C
q2
d
P
d
d
q3
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Norah Ali Al-moneef
king saud university
q4
4
Two point charges each of magnitude 2.00 μC are located on the x axis.
One is at x = 1.00 m, and the other is at x = -1.00 m.
a) Determine the electric potential on the y axis at y = 0.500 m.
b) Calculate the change in electric potential energy of the system as a
third charge of –3.00 μC is brought from infinitely far away to a position
on the y axis at y = 0.500 m.
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king saud university
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A proton is moved from the negative plate to the positive plate of a parallel-plate
arrangement. The plates are 1.5cm apart, and the electric field is uniform with a
magnitude of 1500N/C.
a) How much work would be required to move a proton from the negative to the positive
plate?
b) What is the potential difference between the plates?
c) If the proton is released from rest at the positive plate, what speed will it have just
before it hits the negative plate?
W  qEx
W  (1.6 10 19 C )(1500
UE  K
N
)(.015m)
C
W  3.6 10 18 J
qV 
1
mv2
2
2qV
m
V  Ed
v
N
V  (1500
)(. 015m)
C
J
V  22.5
C
2(1.6 10 19 )( 22.5)
v
1.67 10  27
4 m
v  6.57 10
s
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king saud university
6
A uniform electric field of magnitude 290 V/m is directed in the
positive x direction. A +13.0 µC charge moves from the origin
to the point (x, y) = (20.0 cm, 50.0 cm).(a) What is the change
in the potential energy of the charge field system?
U
q
V   Ed
V 

U  qV
U  qEd  13 10 6 C  20 10  2 m  290V / m  0.000754 J
min us sign
positive charge moves in the direction of the electric field will lose electric potential energy
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Norah Ali Al-moneef
king saud university
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Gain or loss?
A negative charge gains electric potential energy
when it moves in the direction of the electric
field. Explain this.
A proton loses potential energy when moving in
the direction of the electric field, but picks up an
equal amount of kinetic energy.
Work done by a field is positive when energy is
given to an object from the field (object is
lowered in a gravitational field or moved in the
direction of E filed lines). This occurs when Uf <Ui
so, W=-ΔU
1- The electric field has a magnitude of 3.0 N/m at a distance of 60 cm from a point
charge. What is the charge?
(a) 1.4 nC
(b) 120 pC
(c) 36 mC
(d) 12 C
(e) 3.0 nC
Q
E r2
EK 2
 Q
r
K
3  60 10  2
Q
 120 pC
9
9 10
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king saud university
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The gap between electrodes in a spark plug is 0.06cm, To
produce an electric spark in a gasoline air mixture, an E field of
3x106V/m must be achieved. When starting the car, what
minimum potential difference must be supplied by the ignition
circuit?
V  Ed
d  0.0006m
E  3 106 V /m
V  Ed
V  3 10 6 V /m0.0006m

V  1.8kV

Calculate the speed of a proton
that is accelerated from rest
U
through a potential difference
of
V 
q
120V. An electron?
V  120V
v0  0
v?
mp  1.67 1027 kg
me  9.111031kg
e  1.602 19 C

1
qV  U  KE  mv2
2

2qV
v
m
vp 1.5 10 m/s

5
v e  6.5 10 6 m/s
An ion accelerated through a potential
difference of 115V experiences an increase in
kinetic energy of 7.37x10-17 J.
Calculate the charge of the ion
qV  U  KE
V  115V
KE  7.37 1017 J
q?
KE
q
V

q  6.4 1019 C


How much work is done in moving Avogadro's
number of electrons from an initial point where the
electric potential is 9V to a point where the
potential is -5V?
q  N a e  (6.02 10 23 )(1.602 1019 C)
VA  9V
VB  5V
q  96440.4C
N a  6.02 1023 electrons
W?

V  VA  VB
V  9V  (5V)
V  14V
W  U  qV
 W  qV  (96440.4C)(14V)
W  1.35MJ
Compute the energy necessary to bring together the charges in the
configuration shown below:
Calculate the electric potential
energy between each pair of
charges and add them together.
U  K
i
Qq
ri
 4 10 6 C  4 10 6 C 4 10 6 C  4 10 6 C 4 10 6 C  4 10 6 C 

 K 


2
2
2
20 10 m
20 10 m
20 10 m


12
16

10
2
 9 109 

72

10
J
2
20 10
a)
What is the electric potential at the point
i) 10 cm and
ii) 50 cm from a point charge of 2 µC ?
b) Find the work done to move a charge of 0.05 µC from
a point 50 cm from the point charge 2 µC to a point 10 cm
from the point charge. State whether the work done by
or on the electric field.
a)
b)
V(10cm )
kq (9.0 109 )(2 106 )
5



1
.
8

10
V
2
r
(10 10 )
V( 50cm )
kq (9.0 109 )(2 106 )
4



3
.
6

10
V
2
r
(50  10 )
W( 50cm 10 cm )  q V10 cm  V50 cm 


=-0.05  10-6 1.8  105  3.6  104  7.2  103 J
Work done on the electric field.
15
Two point charges, 8x10-9 C and -2x10-9 C are separated by 4 m. The electric field
magnitude (in units of V/m) midway between them is:
A) 9x109 B) 13,500 C) 135,000 D) 36x10-9 E) 22.5
kq1 kq2
E  E1  E 2  2  2
2
2
9 109  8 109 9 109  2 109
E

2
2
22
98 9 2 9
E  2  2  2 ( 8  2)
2
2
2
90
E
 22.5 N / C
4
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Norah Ali Al-moneef
king saud university
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