Physics 6B Electric Field Examples
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Transcript of Physics 6B Electric Field Examples
Physics 6B
Electric Field Examples
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
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
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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For Campus Learning Assistance Services at UCSB
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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For Campus Learning 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
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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For Campus Learning 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
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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For Campus Learning 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
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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For Campus Learning Assistance Services at UCSB
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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For Campus Learning Assistance Services at UCSB
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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For Campus Learning Assistance Services at UCSB
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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For Campus Learning Assistance Services at UCSB
-4nC
x=0 x=0.8m
+6nCx
-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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For Campus Learning Assistance Services at UCSB
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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For Campus Learning Assistance Services at UCSB
2RkQ
E
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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For Campus Learning Assistance Services at UCSB
2RkQ
E
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
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
2RkQ
E
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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For Campus Learning Assistance Services at UCSB
2RkQ
E
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
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
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2RkQ
E
a
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2 E1
E2
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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
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)
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x
y
Part a): TRY DRAWING THE E-FIELD VECTORS ON THE DIAGRAM 12
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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), 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)
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)
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
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
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
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)
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.
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
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
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
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
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
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
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
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)
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