physics 6b electric field examples
DESCRIPTION
Physics 6B Electric Field Examples. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB. 17.22Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. - PowerPoint PPT PresentationTRANSCRIPT
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Physics 6B
Electric Field Examples
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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17.22 Two point charges are located on the x-axis 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.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.
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q2 q1
x=0 x=0.2mx=-0.3m
x
17.22 Two point charges are located on the x-axis 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.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.
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For Campus Learning Assistance Services at UCSB
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q2 q1
x=0 x=0.2mx=-0.3m
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is away from a positive charge, and toward a negative one.
x
17.22 Two point charges are located on the x-axis 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.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.
2Rkq
E
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q2 q1
x=0 x=0.2mx=-0.3m
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 vector that points left, and q2 gives an E-field vector to the right.
E1 E2
x
17.22 Two point charges are located on the x-axis 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.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.
2Rkq
E
The electric field near a single point charge is given by the formula:
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q2 q1
x=0 x=0.2mx=-0.3m
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 vector that points left, and q2 gives an E-field vector to the right.
This is how we can put the +/- signs on the E-fields when we add them up.
E1 E2
x
17.22 Two point charges are located on the x-axis 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.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.
2Rkq
E
The electric field near a single point charge is given by the formula:
21total EEE
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q2 q1
x=0 x=0.2mx=-0.3m
E1 E2
x
17.22 Two point charges are located on the x-axis 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.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.
2Rkq
E
The electric field near a single point charge is given by the formula:
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 vector that points left, and q2 gives an E-field vector to the right.
This is how we can put the +/- signs on the E-fields when we add them up.
21total EEE
CN
CN
2
9
CNm9
2
9
CNm9
total 500900)m3.0(
)C105)(109(
)m2.0(
)C104)(109(E
2
2
2
2
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q2 q1
x=0 x=0.2mx=-0.3m
Etotal
(This means 400 N/C in the negative x-direction)
x
17.22 Two point charges are located on the x-axis 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.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.
2Rkq
E
The electric field near a single point charge is given by the formula:
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 vector that points left, and q2 gives an E-field vector to the right.
This is how we can put the +/- signs on the E-fields when we add them up.
21total EEE
CN
CN
2
9
CNm9
2
9
CNm9
total 500900)m3.0(
)C105)(109(
)m2.0(
)C104)(109(E
2
2
2
2
CN
total 400E
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For Campus Learning Assistance Services at UCSB
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q2 q3 q1
x=0 x=0.2mx=-0.3m
For part b) all we need to do is multiply the E-field from part a) times the new charge q3.
x
17.22 Two point charges are located on the x-axis 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.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.
2Rkq
E
The electric field near a single point charge is given by the formula:
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 vector that points left, and q2 gives an E-field vector to the right.
This is how we can put the +/- signs on the E-fields when we add them up.
21total EEE
CN
CN
2
9
CNm9
2
9
CNm9
total 500900)m3.0(
)C105)(109(
)m2.0(
)C104)(109(E
2
2
2
2
CN
total 400E (This means 400 N/C in the negative x-direction)
Etotal
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For Campus Learning Assistance Services at UCSB
![Page 10: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/10.jpg)
q2 q3 q1
x=0 x=0.2mx=-0.3m
Note that this force is to the right, which is opposite the E-fieldThis is because q3 is a negative charge: E-fields are always set up as if there are positive charges.
x
17.22 Two point charges are located on the x-axis 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.b) Find the net electric force on a charge q3=-0.6nC placed at the origin.
2Rkq
E
The electric field near a single point charge is given by the formula:
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 vector that points left, and q2 gives an E-field vector to the right.
This is how we can put the +/- signs on the E-fields when we add them up.
21total EEE
CN
CN
2
9
CNm9
2
9
CNm9
total 500900)m3.0(
)C105)(109(
)m2.0(
)C104)(109(E
2
2
2
2
CN
total 400E (This means 400 N/C in the negative x-direction)
For part b) all we need to do is multiply the E-field from part a) times the new charge q3.
N104.2)400)(C106.0(F 7CN9
onq3
Etotal
Fon3
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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 221
elec Rqkq
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
If both charges are doubled, we will have
221
elec Rqkq
F
221
221
elec Rqkq
4R
)q2)(q2(kF
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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
If both charges are doubled, we will have
So the new force is 4 times as large.
221
elec Rqkq
F
221
221
elec Rqkq
4R
)q2)(q2(kF
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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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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
221
elec Dqkq
F
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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
We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance.
221
elec Dqkq
F
2new
212
21
Dqkq
Dqkq
3
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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
Canceling and cross-multiplying, we get
We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance.
221
elec Dqkq
F
2new
212
21
Dqkq
Dqkq
3
2312
new DD
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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
Canceling and cross-multiplying, we get
We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance.
Square-roots of both sides gives us the answer:
221
elec Dqkq
F
2new
212
21
Dqkq
Dqkq
3
2312
new DD
DD3
1new
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17.30 When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you 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 released a distance d from one another, the heavier one has an acceleration a. If you 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 released a distance d from one another, the heavier one has an acceleration a. If you 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 is2
21elec d
qkqF
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17.30 When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you 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
If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong:
221
elec dqkq
F
221
51
2new
21
old51
new
dqkq
dqkq
FF
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17.30 When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you 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
If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong:
We cancel common terms and cross-multiply to get
221
elec dqkq
F
221
51
2new
21
old51
new
dqkq
dqkq
FF
22new d5d
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17.30 When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you 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
If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong:
We cancel common terms and cross-multiply to get
Square-root of both sides:
221
elec dqkq
F
221
51
2new
21
old51
new
dqkq
dqkq
FF
22new d5d
d5dnew
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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 magnitude and direction of the electric field at the following points on the x-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 magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
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-4nC
x=0 x=0.8m
+6nCx
![Page 28: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/28.jpg)
-4nC
x=0 x=0.8m
+6nCx
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is away from a positive charge, and toward a negative one.
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 magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
2RkQ
E
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a
For part a) which direction do the E-field vectors point?
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 magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
-4nC
x=0 x=0.8m
+6nCx
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is away from a positive charge, and toward a negative one.
-4nC
x=0 x=0.8m
+6nCx
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2RkQ
E
![Page 30: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/30.jpg)
a
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1
E2
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 magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is away from a positive charge, and toward a negative one.
-4nC
x=0 x=0.8m
+6nCx
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
21total EEE
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2RkQ
E
![Page 31: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/31.jpg)
a
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1
E2
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 magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is away from a positive charge, and toward a negative one.
-4nC
x=0 x=0.8m
+6nCx
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
21total EEE
CN
2
9
CNm9
2
9
CNm9
total 1050)m6.0(
)C106)(109(
)m2.0(
)C104)(109(E
2
2
2
2
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2RkQ
E
![Page 32: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/32.jpg)
a
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1
E2
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 magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is away from a positive charge, and toward a negative one.
-4nC
x=0 x=0.8m
+6nCx
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
21total EEE
CN
2
9
CNm9
2
9
CNm9
total 1050)m6.0(
)C106)(109(
)m2.0(
)C104)(109(E
2
2
2
2
For part b) E1 points left and E2 points right b
E1
E2
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
21total EEE
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2RkQ
E
![Page 33: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/33.jpg)
a
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1
E2
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 magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is away from a positive charge, and toward a negative one.
-4nC
x=0 x=0.8m
+6nCx
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
21total EEE
CN
2
9
CNm9
2
9
CNm9
total 1050)m6.0(
)C106)(109(
)m2.0(
)C104)(109(E
2
2
2
2
For part b) E1 points left and E2 points right b
E1
E2
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
21total EEE
CN
2
9
CNm9
2
9
CNm9
total 5.312)m4.0(
)C106)(109(
)m2.1(
)C104)(109(E
2
2
2
2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
2RkQ
E
![Page 34: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/34.jpg)
a
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1
E2
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 magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is away from a positive charge, and toward a negative one.
-4nC
x=0 x=0.8m
+6nCx
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
21total EEE
CN
2
9
CNm9
2
9
CNm9
total 1050)m6.0(
)C106)(109(
)m2.0(
)C104)(109(E
2
2
2
2
For part b) E1 points left and E2 points right b
E1
E2
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
21total EEE
CN
2
9
CNm9
2
9
CNm9
total 5.312)m4.0(
)C106)(109(
)m2.1(
)C104)(109(E
2
2
2
2
For part b) E1 points right and E2 points leftc
E1
E2
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
2RkQ
E
![Page 35: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/35.jpg)
a
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1
E2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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 magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is away from a positive charge, and toward a negative one.
-4nC
x=0 x=0.8m
+6nCx
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
21total EEE
CN
2
9
CNm9
2
9
CNm9
total 1050)m6.0(
)C106)(109(
)m2.0(
)C104)(109(E
2
2
2
2
For part b) E1 points left and E2 points right b
E1
E2
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
21total EEE
CN
2
9
CNm9
2
9
CNm9
total 5.312)m4.0(
)C106)(109(
)m2.1(
)C104)(109(E
2
2
2
2
For part b) E1 points right and E2 points leftc
E1
E2
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
21total EEE
CN
2
9
CNm9
2
9
CNm9
total 846)m0.1(
)C106)(109(
)m2.0(
)C104)(109(E
2
2
2
2
2RkQ
E
![Page 36: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/36.jpg)
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), 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); d) (0m,0.2m)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
![Page 37: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/37.jpg)
x
y
Part a): TRY DRAWING THE E-FIELD VECTORS ON THE DIAGRAM 12
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), 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); d) (0m,0.2m)
![Page 38: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/38.jpg)
x
y
Part a): both vectors point away from their charge.Since the distances and the charges are equal, the vectors cancel out.
Etotal = 0
12
E1 E2
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), 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); d) (0m,0.2m)
![Page 39: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/39.jpg)
x
y
x
y
12
2 1
Part b): both vectors point away from their charge.
E1
E2
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), 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); d) (0m,0.2m)
E1 E2Part a): both vectors point away from their charge.Since the distances and the charges are equal, the vectors cancel out.
Etotal = 0
![Page 40: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/40.jpg)
x
y
x
y
12
2 1
Positive x-direction
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), 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); d) (0m,0.2m)
E1
E2
Part b): both vectors point away from their charge.
Part a): both vectors point away from their charge.Since the distances and the charges are equal, the vectors cancel out.
Etotal = 0
Positive x-direction
CN
2
9
CNm9
1 2400)m15.0(
)C106)(109(E
2
2
CN
2
9
CNm9
2 267)m45.0(
)C106)(109(E
2
2
E1 E2
![Page 41: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/41.jpg)
x
y
x
y
12
2 1
Positive x-direction
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), 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); d) (0m,0.2m)
E1
E2
Part b): both vectors point away from their charge.
Part a): both vectors point away from their charge.Since the distances and the charges are equal, the vectors cancel out.
Etotal = 0
Positive x-direction
CN
2
9
CNm9
1 2400)m15.0(
)C106)(109(E
2
2
CN
2
9
CNm9
2 267)m45.0(
)C106)(109(E
2
2
CN
total 26672672400E
E1 E2
![Page 42: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/42.jpg)
x
yPart c): both vectors point away from their charge. We will need to use vector components to add them together.
12
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
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), 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); d) (0m,0.2m)
![Page 43: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/43.jpg)
x
y
12
E1,y
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
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), 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); d) (0m,0.2m)
Part c): both vectors point away from their charge. We will need to use vector components to add them together.
![Page 44: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/44.jpg)
x
y
12
E1,y
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
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), 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); d) (0m,0.2m)
Part c): both vectors point away from their charge. We will need to use vector components to add them together.
CN
2
9
CNm9
1 5.337)m4.0(
)C106)(109(E
2
2
CN
y,1
CN
x,1
5.337E
0E
![Page 45: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/45.jpg)
x
y
12
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
E2
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), 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); 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,y
CN
2
9
CNm9
1 5.337)m4.0(
)C106)(109(E
2
2
CN
y,1
CN
x,1
5.337E
0E
![Page 46: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/46.jpg)
x
y
12
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.
0.4m
0.3m
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), 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); d) (0m,0.2m)
E2
E1,y
Part c): both vectors point away from their charge. We will need to use vector components to add them together.
CN
2
9
CNm9
2 216)m5.0(
)C106)(109(E
2
2
CN
2
9
CNm9
1 5.337)m4.0(
)C106)(109(E
2
2
CN
y,1
CN
x,1
5.337E
0E
![Page 47: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/47.jpg)
x
y
12
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.
0.4m
0.3m
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), 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); d) (0m,0.2m)
E1,y
Part c): both vectors point away from their charge. We will need to use vector components to add them together.
CN
2
9
CNm9
2 216)m5.0(
)C106)(109(E
2
2
E2,x
E2,y
CN
2
9
CNm9
1 5.337)m4.0(
)C106)(109(E
2
2
CN
y,1
CN
x,1
5.337E
0E
![Page 48: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/48.jpg)
x
y
12
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.
0.4m
0.3m
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), 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); d) (0m,0.2m)
E1,y
Part c): both vectors point away from their charge. We will need to use vector components to add them together.
CN
2
9
CNm9
2 216)m5.0(
)C106)(109(E
2
2
E2,x
E2,y
CN
2
9
CNm9
1 5.337)m4.0(
)C106)(109(E
2
2
CN
y,1
CN
x,1
5.337E
0E
CN
54
CN
y,2
CN
53
CN
x,2
8.172)()216(E
6.129)()216(E
![Page 49: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/49.jpg)
x
y
12
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.
0.4m
0.3m
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), 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); d) (0m,0.2m)
E1,y
Part c): both vectors point away from their charge. We will need to use vector components to add them together.
CN
2
9
CNm9
2 216)m5.0(
)C106)(109(E
2
2
E2,x
E2,y
CN
2
9
CNm9
1 5.337)m4.0(
)C106)(109(E
2
2
CN
y,1
CN
x,1
5.337E
0E
Add together the x-components and the y-components separately:
CN
CN
CN
y,total
CN
CN
CN
x,total
3.5108.1725.337E
6.1296.1290E
CN
54
CN
y,2
CN
53
CN
x,2
8.172)()216(E
6.129)()216(E
![Page 50: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/50.jpg)
x
y
12
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.
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), 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); d) (0m,0.2m)
Part c): both vectors point away from their charge. We will need to use vector components to add them together.
CN
2
9
CNm9
2 216)m5.0(
)C106)(109(E
2
2
CN
2
9
CNm9
1 5.337)m4.0(
)C106)(109(E
2
2
CN
y,1
CN
x,1
5.337E
0E
Add together the x-components and the y-components separately:
CN
CN
CN
y,total
CN
CN
CN
x,total
3.5108.1725.337E
6.1296.1290E
Now find the magnitude and the angle using right triangle rules:
75.7º
Etotal
axisxbelow7.756.1293.510
)tan(
5.526)3.510()6.129(E CN22
total
CN
54
CN
y,2
CN
53
CN
x,2
8.172)()216(E
6.129)()216(E
![Page 51: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/51.jpg)
x
yPart d): TRY THIS ONE ON YOUR OWN FIRST...
12
(0.15,0)(- 0.15,0)
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), 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); d) (0m,0.2m)
(0,0.2)
![Page 52: Physics 6B Electric Field Examples](https://reader035.vdocuments.net/reader035/viewer/2022081505/56815805550346895dc57533/html5/thumbnails/52.jpg)
x
yPart d): both vectors point away from their charge. We will need to use vector components to add them together.
12
E1 E2
(0,0.2)
(0.15,0)(- 0.15,0)
The 0.25m in this formula is the distance to each charge using the Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it.
From symmetry, we can see that E2 will have the same components, except for +/- signs.
Now we can add the components (the x-component should cancel out)
The final answer should be 1382.4 N/C in the positive y-direction.
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), 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); d) (0m,0.2m)
CN
2
9
CNm9
1 864)m25.0(
)C106)(109(E
2
2
CN
25.020.0
CN
y,1
CN
25.015.0
CN
x,1
2.691))(864(E
4.518))(864(E
CN
25.020.0
CN
y,2
CN
25.015.0
CN
x,2
2.691))(864(E
4.518))(864(E
CN
CN
CN
y,total
CN
CN
CN
x,total
4.13822.6912.691E
04.5184.518E