Physics 6B
Electric Field Examples
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17.22 Two point charges are located on the x-axis
x
as follows: charge q1 = +4 nC at
position x=0.2m and charge q2 = +5 nC at position x = -0.3m.
a)
Find the magnitude and direction of the net electric field produced by q1 and
q2 at the origin.
Find the net electric force on a charge q3=-0.6nC placed at the origin.
b)
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17.22 Two point charges are located on the x-axis
x
as follows: charge q1 = +4 nC at
position x=0.2m and charge q2 = +5 nC at position x = -0.3m.
a)
Find the magnitude and direction of the net electric field produced by q1 and
q2 at the origin.
Find the net electric force on a charge q3=-0.6nC placed at the origin.
b)
q2
x=-0.3m
q1
x=0
x
x=0.2m
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17.22 Two point charges are located on the x-axis
x
as follows: charge q1 = +4 nC at
position x=0.2m and charge q2 = +5 nC at position x = -0.3m.
a)
Find the magnitude and direction of the net electric field produced by q1 and
q2 at the origin.
Find the net electric force on a charge q3=-0.6nC placed at the origin.
b)
The electric field near a single point charge is given by the formula:
E=
kq
R2
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
q2
x=-0.3m
q1
x=0
x
x=0.2m
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x
as follows: charge q1 = +4 nC at
17.22 Two point charges are located on the x-axis
position x=0.2m and charge q2 = +5 nC at position x = -0.3m.
a)
Find the magnitude and direction of the net electric field produced by q1 and
q2 at the origin.
Find the net electric force on a charge q3=-0.6nC placed at the origin.
b)
The electric field near a single point charge is given by the formula:
E=
kq
R2
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
At the origin, q1 will produce an E-field
field vector that points
left, and q2 gives an E-field vector to the right.
E1
q2
x=-0.3m
E2
q1
x=0
x
x=0.2m
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x
as follows: charge q1 = +4 nC at
17.22 Two point charges are located on the x-axis
position x=0.2m and charge q2 = +5 nC at position x = -0.3m.
a)
Find the magnitude and direction of the net electric field produced by q1 and
q2 at the origin.
Find the net electric force on a charge q3=-0.6nC placed at the origin.
b)
The electric field near a single point charge is given by the formula:
E=
kq
R2
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
At the origin, q1 will produce an E-field
field vector that points
left, and q2 gives an E-field vector to the right.
E1
q2
x=-0.3m
E2
q1
x=0
x
x=0.2m
fields when
This is how we can put the +/- signs on the E-fields
we add them up.
Etotal = −E1 + E2
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17.22 Two point charges are located on the x-axis
x
as follows: charge q1 = +4 nC at
position x=0.2m and charge q2 = +5 nC at position x = -0.3m.
a)
Find the magnitude and direction of the net electric field produced by q1 and
q2 at the origin.
Find the net electric force on a charge q3=-0.6nC placed at the origin.
b)
The electric field near a single point charge is given by the formula:
E=
kq
R2
E1
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
At the origin, q1 will produce an E-field
field vector that points
left, and q2 gives an E-field vector to the right.
q2
x=-0.3m
E2
q1
x=0
x
x=0.2m
fields when
This is how we can put the +/- signs on the E-fields
we add them up.
Etotal = −E1 + E2
Etotal = −
(9 ⋅ 109
Nm2
C2
)(4 ⋅ 10−9 C)
2
(0.2m)
+
(9 ⋅ 109
Nm2
C2
)(5 ⋅ 10−9 C)
2
(0.3m)
= −900 NC + 500 NC
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17.22 Two point charges are located on the x-axis
x
as follows: charge q1 = +4 nC at
position x=0.2m and charge q2 = +5 nC at position x = -0.3m.
a)
Find the magnitude and direction of the net electric field produced by q1 and
q2 at the origin.
Find the net electric force on a charge q3=-0.6nC placed at the origin.
b)
The electric field near a single point charge is given by the formula:
E=
kq
R2
Etotal
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
At the origin, q1 will produce an E-field
field vector that points
left, and q2 gives an E-field vector to the right.
q2
x=-0.3m
q1
x=0
x
x=0.2m
fields when
This is how we can put the +/- signs on the E-fields
we add them up.
Etotal = −E1 + E2
Etotal = −
(9 ⋅ 109
Nm2
C2
)(4 ⋅ 10−9 C)
2
(0.2m)
Etotal = −400 NC
+
(9 ⋅ 109
Nm2
C2
)(5 ⋅ 10−9 C)
2
(0.3m)
= −900 NC + 500 NC
(This means 400 N/C in the negative x-direction)
direction)
Prepared by Vince Zaccone
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17.22 Two point charges are located on the x-axis
x
as follows: charge q1 = +4 nC at
position x=0.2m and charge q2 = +5 nC at position x = -0.3m.
a)
Find the magnitude and direction of the net electric field produced by q1 and
q2 at the origin.
Find the net electric force on a charge q3=-0.6nC placed at the origin.
b)
The electric field near a single point charge is given by the formula:
E=
kq
R2
Etotal
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
At the origin, q1 will produce an E-field
field vector that points
left, and q2 gives an E-field vector to the right.
q2
x=-0.3m
q3
x=0
q1
x
x=0.2m
fields when
This is how we can put the +/- signs on the E-fields
we add them up.
Etotal = −E1 + E2
Etotal = −
(9 ⋅ 109
Nm2
C2
)(4 ⋅ 10−9 C)
2
(0.2m)
Etotal = −400 NC
+
(9 ⋅ 109
Nm2
C2
)(5 ⋅ 10−9 C)
2
(0.3m)
= −900 NC + 500 NC
(This means 400 N/C in the negative x-direction)
direction)
For part b) all we need to do is multiply the E-field
field from part a) times the new charge q3.
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17.22 Two point charges are located on the x-axis
x
as follows: charge q1 = +4 nC at
position x=0.2m and charge q2 = +5 nC at position x = -0.3m.
a)
Find the magnitude and direction of the net electric field produced by q1 and
q2 at the origin.
Find the net electric force on a charge q3=-0.6nC placed at the origin.
b)
The electric field near a single point charge is given by the formula:
kq
E= 2
R
Etotal
Fon3
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
At the origin, q1 will produce an E-field
field vector that points
left, and q2 gives an E-field vector to the right.
q2
x=-0.3m
q3
x=0
q1
x
x=0.2m
fields when
This is how we can put the +/- signs on the E-fields
we add them up.
Etotal = −E1 + E2
Etotal = −
(9 ⋅ 109
Nm2
C2
)(4 ⋅ 10−9 C)
2
(0.2m)
Etotal = −400 NC
+
(9 ⋅ 109
Nm2
C2
)(5 ⋅ 10−9 C)
2
(0.3m)
= −900 NC + 500 NC
(This means 400 N/C in the negative x-direction)
direction)
For part b) all we need to do is multiply the E-field
field from part a) times the new charge q3.
Fonq3 = (−0.6 ⋅ 10−9 C)(−400 NC ) = +2.4 ⋅ 10−7 N
Note that this force is to the right, which is opposite the E-field
E
This is because q3 is a negative charge: E-fields are always set
up as if there are positive charges.
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17.28 Two unequal charges repel each other with a force F. If both charges are
doubled in magnitude, what will be the new force in terms of F?
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17.28 Two unequal charges repel each other with a force F. If both charges are
doubled in magnitude, what will be the new force in terms of F?
The formula for electric force between 2 charges is Felec =
kq1q2
R2
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17.28 Two unequal charges repel each other with a force F. If both charges are
doubled in magnitude, what will be the new force in terms of F?
The formula for electric force between 2 charges is Felec =
kq1q2
R2
If both charges are doubled, we will have Felec = k(2q1 )(2q2 ) = 4 ⋅ kq1q2
R2
R2
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17.28 Two unequal charges repel each other with a force F. If both charges are
doubled in magnitude, what will be the new force in terms of F?
The formula for electric force between 2 charges is Felec =
kq1q2
R2
If both charges are doubled, we will have Felec = k(2q1 )(2q2 ) = 4 ⋅ kq1q2
R2
R2
So the new force is 4 times as large.
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17.29 Two unequal charges attract each other with a force F when they are a
distance D apart. How far apart (in terms of D) must they be for the force to be 3
times as strong as F?
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Assistance Services at UCSB
17.29 Two unequal charges attract each other with a force F when they are a
distance D apart. How far apart (in terms of D) must they be for the force to be 3
times as strong as F?
The formula for electric force between 2 charges is Felec =
kq1q2
D2
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Assistance Services at UCSB
17.29 Two unequal charges attract each other with a force F when they are a
distance D apart. How far apart (in terms of D) must they be for the force to be 3
times as strong as F?
The formula for electric force between 2 charges is Felec =
kq1q2
D2
We want the force to be 3 times as strong, so we can set
kq q
kq q
3 ⋅ 12 2 = 21 2
up the force equation and solve for the new distance.
D
Dnew
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17.29 Two unequal charges attract each other with a force F when they are a
distance D apart. How far apart (in terms of D) must they be for the force to be 3
times as strong as F?
The formula for electric force between 2 charges is Felec =
kq1q2
D2
We want the force to be 3 times as strong, so we can set
kq q
kq q
3 ⋅ 12 2 = 21 2
up the force equation and solve for the new distance.
D
Dnew
Canceling and cross-multiplying, we get
2
Dnew
=
1
3
⋅ D2
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17.29 Two unequal charges attract each other with a force F when they are a
distance D apart. How far apart (in terms of D) must they be for the force to be 3
times as strong as F?
The formula for electric force between 2 charges is Felec =
kq1q2
D2
We want the force to be 3 times as strong, so we can set
kq q
kq q
3 ⋅ 12 2 = 21 2
up the force equation and solve for the new distance.
D
Dnew
Canceling and cross-multiplying, we get
2
Dnew
=
1
3
⋅ D2
Square-roots
roots of both sides gives us the answer: Dnew =
1
3
⋅D
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17.30 When two unequal point charges are
ar released a distance d from one another,
the heavier one has an acceleration a. Iff you
y
want to reduce this acceleration to 1/5
of this value, how far (in terms of d) should the charges be released?
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17.30 When two unequal point charges are
ar released a distance d from one another,
the heavier one has an acceleration a. Iff you
y
want to reduce this acceleration to 1/5
of this value, how far (in terms of d) should the charges be released?
Recall that Newton's 2nd law says that Fnet = ma.
So this is really a problem about the force on the heavier charge.
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17.30 When two unequal point charges are
ar released a distance d from one another,
the heavier one has an acceleration a. Iff you
y
want to reduce this acceleration to 1/5
of this value, how far (in terms of d) should the charges be released?
Recall that Newton's 2nd law says that Fnet = ma.
So this is really a problem about the force on the heavier charge.
The formula for electric force between 2 charges is F = kq1q2
elec
d2
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17.30 When two unequal point charges are
ar released a distance d from one another,
the heavier one has an acceleration a. Iff you
y
want to reduce this acceleration to 1/5
of this value, how far (in terms of d) should the charges be released?
Recall that Newton's 2nd law says that Fnet = ma.
So this is really a problem about the force on the heavier charge.
The formula for electric force between 2 charges is F = kq1q2
elec
d2
If we want the acceleration to be 1/5 as fast,
we need the force to be 1/5 as strong:
Fnew =
1
5
kq1q2
=
2
dnew
⋅ Fold
1
5
⋅
kq1q2
d2
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17.30 When two unequal point charges are
ar released a distance d from one another,
the heavier one has an acceleration a. Iff you
y
want to reduce this acceleration to 1/5
of this value, how far (in terms of d) should the charges be released?
Recall that Newton's 2nd law says that Fnet = ma.
So this is really a problem about the force on the heavier charge.
The formula for electric force between 2 charges is F = kq1q2
elec
d2
If we want the acceleration to be 1/5 as fast,
we need the force to be 1/5 as strong:
Fnew =
1
5
kq1q2
=
2
dnew
⋅ Fold
1
5
⋅
kq1q2
d2
2
2
We cancel common terms and cross-multiply
multiply to get dnew = 5 ⋅ d
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17.30 When two unequal point charges are
ar released a distance d from one another,
the heavier one has an acceleration a. Iff you
y
want to reduce this acceleration to 1/5
of this value, how far (in terms of d) should the charges be released?
Recall that Newton's 2nd law says that Fnet = ma.
So this is really a problem about the force on the heavier charge.
The formula for electric force between 2 charges is F = kq1q2
elec
d2
If we want the acceleration to be 1/5 as fast,
we need the force to be 1/5 as strong:
Fnew =
1
5
kq1q2
=
2
dnew
⋅ Fold
1
5
⋅
kq1q2
d2
2
2
We cancel common terms and cross-multiply
multiply to get dnew = 5 ⋅ d
Square-root of both sides: d
5 ⋅d
new =
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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC
is placed on the x-axis at x=0.8m. Find the
th magnitude and direction of the electric
field at the following points on the x-axis:
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC
is placed on the x-axis at x=0.8m. Find the
th magnitude and direction of the electric
field at the following points on the x-axis:
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
-4nC
+6nC
x
x=0
x=0.8m
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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC
is placed on the x-axis at x=0.8m. Find the
th magnitude and direction of the electric
field at the following points on the x-axis:
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point E = kQ
charge is given by the formula:
R2
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
-4nC
+6nC
x
x=0
x=0.8m
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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC
is placed on the x-axis at x=0.8m. Find the
th magnitude and direction of the electric
field at the following points on the x-axis:
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point E = kQ
charge is given by the formula:
R2
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
-4nC
+6nC
x
x=0
x=0.8m
For part a) which direction do the E-field
field vectors point?
-4nC
+6nC
a
x=0
x
x=0.8m
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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC
is placed on the x-axis at x=0.8m. Find the
th magnitude and direction of the electric
field at the following points on the x-axis:
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point E = kQ
charge is given by the formula:
R2
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
For part a) both E-field vectors point in the –x
x direction
Call the -4nC
4nC charge #1 and the +6nC charge #2
Etotal = −E1 − E2
-4nC
+6nC
x
x=0
x=0.8m
E2
Q1 =
-4nC
x=0
E1
a
Q2 =
+6nC
x
x=0.8m
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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC
is placed on the x-axis at x=0.8m. Find the
th magnitude and direction of the electric
field at the following points on the x-axis:
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point E = kQ
charge is given by the formula:
R2
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
For part a) both E-field vectors point in the –x
x direction
Call the -4nC
4nC charge #1 and the +6nC charge #2
Etotal = −E1 − E2
Etotal = −
-4nC
x
x=0
Nm2
C2
)(4 ⋅ 10−9 C)
2
(0.2m)
−
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.6m)
x=0.8m
E2
Q1 =
-4nC
x=0
(9 ⋅ 109
+6nC
E1
a
Q2 =
+6nC
x
x=0.8m
= −1050 NC
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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC
is placed on the x-axis at x=0.8m. Find the
th magnitude and direction of the electric
field at the following points on the x-axis:
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point E = kQ
charge is given by the formula:
R2
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
For part a) both E-field vectors point in the –x
x direction
Call the -4nC
4nC charge #1 and the +6nC charge #2
Etotal = −E1 − E2
Etotal = −
Nm2
C2
)(4 ⋅ 10−9 C)
2
(0.2m)
−
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.6m)
For part b) E1 points left and E2 points right
+6nC
x
x=0
x=0.8m
E2
Q1 =
-4nC
x=0
(9 ⋅ 109
Etotal = −E1 + E2
-4nC
E1
a
Q2 =
+6nC
x
x=0.8m
= −1050 NC
E2
Q1 =
-4nC
x=0
Q2 =
+6nC
E1
b
x
x=0.8m
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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC
is placed on the x-axis at x=0.8m. Find the
th magnitude and direction of the electric
field at the following points on the x-axis:
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point E = kQ
charge is given by the formula:
R2
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
For part a) both E-field vectors point in the –x
x direction
Call the -4nC
4nC charge #1 and the +6nC charge #2
Etotal = −E1 − E2
Etotal = −
(9 ⋅ 109
Nm2
C2
)(4 ⋅ 10−9 C)
2
(0.2m)
−
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.6m)
x=0
E2
Q1 =
-4nC
)(4 ⋅ 10−9 C)
2
(1.2m)
+
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.4m)
E1
a
Q2 =
+6nC
x
x=0.8m
E2
Q1 =
-4nC
x=0
Nm2
C2
x=0.8m
= −1050 NC
Etotal = −E1 + E2
(9 ⋅ 109
+6nC
x
x=0
For part b) E1 points left and E2 points right
Etotal = −
-4nC
Q2 =
+6nC
E1
b
x
x=0.8m
= +312.5 NC
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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC
is placed on the x-axis at x=0.8m. Find the
th magnitude and direction of the electric
field at the following points on the x-axis:
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point E = kQ
charge is given by the formula:
R2
-4nC
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
x
x=0
For part a) both E-field vectors point in the –x
x direction
Call the -4nC
4nC charge #1 and the +6nC charge #2
Q1 =
-4nC
Etotal = −
x=0
Nm2
C2
)(4 ⋅ 10−9 C)
2
(0.2m)
−
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.6m)
Q1 =
-4nC
x=0
Etotal = −
)(4 ⋅ 10−9 C)
(1.2m)2
+
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
(0.4m)2
= +312.5 NC
x
x=0.8m
Q2 =
+6nC
E1
b
x
x=0.8m
E2
E1
c
For part b) E1 points right and E2 points left
a
Q2 =
+6nC
E2
Etotal = −E1 + E2
Nm2
C2
E1
= −1050 NC
For part b) E1 points left and E2 points right
(9 ⋅ 109
x=0.8m
E2
Etotal = −E1 − E2
(9 ⋅ 109
+6nC
Q1 =
-4nC
x=0
Q2 =
+6nC
x
x=0.8m
Prepared by Vince Zaccone
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17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 nC
is placed on the x-axis at x=0.8m. Find the
th magnitude and direction of the electric
field at the following points on the x-axis:
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point E = kQ
charge is given by the formula:
R2
-4nC
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
x
x=0
For part a) both E-field vectors point in the –x
x direction
Call the -4nC
4nC charge #1 and the +6nC charge #2
Q1 =
-4nC
Etotal = −
x=0
Nm2
C2
)(4 ⋅ 10−9 C)
2
(0.2m)
−
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.6m)
Q1 =
-4nC
x=0
Etotal = −
)(4 ⋅ 10−9 C)
(1.2m)2
+
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
(0.4m)2
= +312.5 NC
For part b) E1 points right and E2 points left
Etotal = +E1 − E2
Etotal = +
Nm2
C2
)(4 ⋅ 10−9 C)
2
(0.2m)
−
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
E1
(1.0m)
Q1 =
-4nC
x=0
= +846 NC
x
x=0.8m
Q2 =
+6nC
E1
b
x
x=0.8m
E2
c
(9 ⋅ 109
a
Q2 =
+6nC
E2
Etotal = −E1 + E2
Nm2
C2
E1
= −1050 NC
For part b) E1 points left and E2 points right
(9 ⋅ 109
x=0.8m
E2
Etotal = −E1 − E2
(9 ⋅ 109
+6nC
Q2 =
+6nC
x
x=0.8m
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17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
y
Part a): TRY DRAWING THE E-FIELD
VECTORS ON THE DIAGRAM
2
1
x
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17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
y
Part a): both vectors point away from their charge.
Since the distances and the charges are equal, the
vectors cancel out.
E1
2
E2
1
x
Etotal = 0
Prepared by Vince Zaccone
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17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
y
Part a): both vectors point away from their charge.
Since the distances and the charges are equal, the
vectors cancel out.
E1
E2
2
1
x
Etotal = 0
y
Part b): both vectors point away from their charge.
E1
2
1
x
E2
Prepared by Vince Zaccone
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17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
y
Part a): both vectors point away from their charge.
Since the distances and the charges are equal, the
vectors cancel out.
E1
E2
2
1
x
Etotal = 0
y
Part b): both vectors point away from their charge.
E1 =
E2 =
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.15m)
9 Nm2
C2
(9 ⋅ 10
= 2400 NC
2
−9
)(6 ⋅ 10 C)
2
(0.45m)
E1
Positive x-direction
direction
= 267 NC
Positive x-direction
direction
1
x
E2
Prepared by Vince Zaccone
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17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
y
Part a): both vectors point away from their charge.
Since the distances and the charges are equal, the
vectors cancel out.
E1
E2
2
1
x
Etotal = 0
y
Part b): both vectors point away from their charge.
E1 =
E2 =
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.15m)
9 Nm2
C2
(9 ⋅ 10
= 2400 NC
2
−9
)(6 ⋅ 10 C)
2
(0.45m)
E1
Positive x-direction
direction
= 267 NC
Positive x-direction
direction
1
x
E2
Etotal = 2400 + 267 = 2667 NC
Prepared by Vince Zaccone
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17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
Part c): both vectors point away from their charge. We
will need to use vector components to add them together.
y
(- 0.15,0)
2
(0.15,0)
1
x
(0.15,- 0.4)
Prepared by Vince Zaccone
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17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
Part c): both vectors point away from their charge. We
will need to use vector components to add them together.
y
(- 0.15,0)
2
(0.15,0)
1
x
(0.15,- 0.4)
E1,y
Prepared by Vince Zaccone
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17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
Part c): both vectors point away from their charge. We
will need to use vector components to add them together.
E1 =
E1,x

E1,y
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.4m)
= 0 NC


= −337.5 NC 
y
= 337.5 NC
(- 0.15,0)
2
(0.15,0)
1
x
(0.15,- 0.4)
E1,y
Prepared by Vince Zaccone
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Assistance Services at UCSB
17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
Part c): both vectors point away from their charge. We
will need to use vector components to add them together.
E1 =
E1,x

E1,y
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.4m)
= 0 NC


= −337.5 NC 
y
= 337.5 NC
(- 0.15,0)
2
(0.15,0)
1
x
(0.15,- 0.4)
E2
E1,y
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
Part c): both vectors point away from their charge. We
will need to use vector components to add them together.
E1 =
E1,x

E1,y
E2 =
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.4m)
= 0 NC


= −337.5 NC 
9 Nm2
C2
(9 ⋅ 10
= 337.5 NC
(- 0.15,0)
2
−9
)(6 ⋅ 10 C)
2
(0.5m)
y
= 216 NC
The 0.5m in this formula for
E2 is the distance to charge 2,
using Pythagorean theorem or
from recognizing a 3-4-5
5 right
triangle when you see it.
(0.15,0)
1
x
0.4m
(0.15,- 0.4)
0.3m
E2
E1,y
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
Part c): both vectors point away from their charge. We
will need to use vector components to add them together.
E1 =
E1,x

E1,y
E2 =
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.4m)
= 0 NC


= −337.5 NC 
9 Nm2
C2
(9 ⋅ 10
= 337.5 NC
(- 0.15,0)
2
−9
)(6 ⋅ 10 C)
2
(0.5m)
y
= 216 NC
The 0.5m in this formula for
E2 is the distance to charge 2,
using Pythagorean theorem or
from recognizing a 3-4-5
5 right
triangle when you see it.
(0.15,0)
1
x
0.4m
(0.15,- 0.4)
E2,x
0.3m
E2,y
E1,y
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
Part c): both vectors point away from their charge. We
will need to use vector components to add them together.
E1 =
E1,x

E1,y
E2 =
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.4m)
= 0 NC


= −337.5 NC 
9 Nm2
C2
(9 ⋅ 10
= 337.5 NC
(- 0.15,0)
2
−9
)(6 ⋅ 10 C)
2
(0.5m)
y
= 216 NC
E2,x = ( +216 N ) ⋅ ( 3 ) = +129.6 N 

C
C
5


) ⋅ ( 4 ) = −172.8 N 
E2,y = ( +216 N
C
5
C
The 0.5m in this formula for
E2 is the distance to charge 2,
using Pythagorean theorem or
from recognizing a 3-4-5
5 right
triangle when you see it.
(0.15,0)
1
x
0.4m
(0.15,- 0.4)
E2,x
0.3m
E2,y
E1,y
Prepared by Vince Zaccone
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Assistance Services at UCSB
17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
Part c): both vectors point away from their charge. We
will need to use vector components to add them together.
E1 =
E1,x

E1,y
E2 =
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.4m)
= 0 NC


= −337.5 NC 
9 Nm2
C2
(9 ⋅ 10
= 337.5 NC
(- 0.15,0)
2
−9
)(6 ⋅ 10 C)
2
(0.5m)
y
= 216 NC
E2,x = ( +216 N ) ⋅ ( 3 ) = +129.6 N 

C
C
5


) ⋅ ( 4 ) = −172.8 N 
E2,y = ( +216 N
C
5
C
The 0.5m in this formula for
E2 is the distance to charge 2,
using Pythagorean theorem or
from recognizing a 3-4-5
5 right
triangle when you see it.
(0.15,0)
1
x
0.4m
Add together the x-components and the y-components
components separately:
N
N
N
Etotal,x = 0 C + 129.6 C = +129.6 C
(0.15,- 0.4)
E2,x
0.3m
E2,y
E1,y
Etotal,y = −337.5 NC − 172.8 NC = −510.3 NC
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
Part c): both vectors point away from their charge. We
will need to use vector components to add them together.
E1 =
E1,x

E1,y
E2 =
(9 ⋅ 109
Nm2
C2
)(6 ⋅ 10−9 C)
2
(0.4m)
= 0 NC


= −337.5 NC 
9 Nm2
C2
(9 ⋅ 10
= 337.5 NC
(- 0.15,0)
2
−9
)(6 ⋅ 10 C)
2
(0.5m)
y
= 216 NC
E2,x = ( +216 N ) ⋅ ( 3 ) = +129.6 N 

C
C
5


) ⋅ ( 4 ) = −172.8 N 
E2,y = ( +216 N
C
5
C
The 0.5m in this formula for
E2 is the distance to charge 2,
using Pythagorean theorem or
from recognizing a 3-4-5
5 right
triangle when you see it.
Add together the x-components and the y-components
components separately:
N
N
N
Etotal,x = 0 C + 129.6 C = +129.6 C
(0.15,0)
1
x
(0.15,- 0.4)
75.7º
Etotal
Etotal,y = −337.5 NC − 172.8 NC = −510.3 NC
Now find the magnitude and the angle using right triangle rules:
Etotal = (129.6)2 + (510.3)2 = 526.5 NC
tan(θ) =
510.3
⇒ θ = 75.7° below + x axis
129.6
Prepared by Vince Zaccone
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17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
Part d): TRY THIS ONE ON YOUR OWN FIRST...
y
(0,0.2)
(- 0.15,0)
2
(0.15,0)
1
x
Prepared by Vince Zaccone
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Assistance Services at UCSB
17.42 A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical
point charge is placed at (-0.15m,0m),
0.15m,0m), as shown.
Find the magnitude and direction of the net electric field at:
a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m);
(0.15m,
d) (0m,0.2m)
Part d): both vectors point away from their charge. We
will need to use vector components to add them together.
E1 =
9 Nm2
C2
(9 ⋅ 10
−9
)(6 ⋅ 10 C)
2
(0.25m)
= 864 NC
N
E1,x = −(864 NC )( 00..15

25 ) = −518.4 C 

0.20
N
N
E1,y = +(864 C )( 0.25 ) = +691.2 C 
The 0.25m in this formula is the
distance to each charge using the
Pythagorean theorem or from
recognizing a 3-4-5
5 right triangle
when you see it.
y
E1
E2
(0,0.2)
(- 0.15,0)
2
(0.15,0)
1
x
From symmetry, we can see that E2 will have
the same components, except for +/- signs.
E2,x = +(864 N )( 0.15 ) = +518.4 N 

C
C 0.25


)( 0.20 ) = +691.2 N 
E2,y = +(864 N
C 0.25
C
Now we can add the components
(the x-component should cancel out)
Etotal,x = −518.4 NC + 518.4 NC = 0 NC
Etotal,y = +691.2 NC + 691.2 NC = +1382.4 NC
The final answer should be 1382.4 N/C in the positive y-direction.
y
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