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Physics 6BPhysics 6B
Electric Field Examples
Physics 6BPhysics 6B
Electric Field Examples
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.22 Two point charges are located on the xposition x=0.2m and charge q2 = +5 nC at position x = a) Find the magnitude and direction of the net electric field produced by q
q2 at the origin.b) Find the net electric force on a charge q
Two point charges are located on the x-axis as follows: charge q1 = +4 nC at at position x = -0.3m.
Find the magnitude and direction of the net electric field produced by q1 and
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
17.22 Two point charges are located on the xposition x=0.2m and charge q2 = +5 nC at position x = a) Find the magnitude and direction of the net electric field produced by q
q2 at the origin.b) Find the net electric force on a charge q
q2 q1
x=0 x=0.2mx=-0.3m
x
Two point charges are located on the x-axis as follows: charge q1 = +4 nC at at position x = -0.3m.
Find the magnitude and direction of the net electric field produced by q1 and
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
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awaypositive charge, and toward a negative one.
17.22 Two point charges are located on the xposition x=0.2m and charge q2 = +5 nC at position x = a) Find the magnitude and direction of the net electric field produced by q
q2 at the origin.b) Find the net electric force on a charge q
2Rkq
E = q2 q1
x=0 x=0.2mx=-0.3m
The electric field near a single point charge is given by the formula:
away from a
x
Two point charges are located on the x-axis as follows: charge q1 = +4 nC at at position x = -0.3m.
Find the magnitude and direction of the net electric field produced by q1 and
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
This is only the magnitude. The direction is awaypositive 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.
17.22 Two point charges are located on the xposition x=0.2m and charge q2 = +5 nC at position x = a) Find the magnitude and direction of the net electric field produced by q
q2 at the origin.b) Find the net electric force on a charge q
2Rkq
E =
The electric field near a single point charge is given by the formula:
q2 q1
x=0 x=0.2mx=-0.3m
away from a
field vector that points
E1 E2
x
Two point charges are located on the x-axis as follows: charge q1 = +4 nC at at position x = -0.3m.
Find the magnitude and direction of the net electric field produced by q1 and
Find the net electric force on a charge q3=-0.6nC placed at the origin.
The electric field near a single point charge is given by the formula:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
This is only the magnitude. The direction is awaypositive 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.
17.22 Two point charges are located on the xposition x=0.2m and charge q2 = +5 nC at position x = a) Find the magnitude and direction of the net electric field produced by q
q2 at the origin.b) Find the net electric force on a charge q
2Rkq
E =
The electric field near a single point charge is given by the formula:
21total EEE +−= 21total
q2 q1
x=0 x=0.2mx=-0.3m
away from a
field vector that points
fields when
E1 E2
x
Two point charges are located on the x-axis as follows: charge q1 = +4 nC at at position x = -0.3m.
Find the magnitude and direction of the net electric field produced by q1 and
Find the net electric force on a charge q3=-0.6nC placed at the origin.
The electric field near a single point charge is given by the formula:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.22 Two point charges are located on the xposition x=0.2m and charge q2 = +5 nC at position x = a) Find the magnitude and direction of the net electric field produced by q
q2 at the origin.b) Find the net electric force on a charge q
2Rkq
E =
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awaypositive 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 +−= 21total
2
9
CNm9
2
9
CNm9
total )m3.0(
)C105)(109(
)m2.0(
)C104)(109(E
2
2
2
2
−=⋅⋅
+⋅⋅
−=−−
q2 q1
x=0 x=0.2mx=-0.3m
E1 E2
x
Two point charges are located on the x-axis as follows: charge q1 = +4 nC at at position x = -0.3m.
Find the magnitude and direction of the net electric field produced by q1 and
Find the net electric force on a charge q3=-0.6nC placed at the origin.
The electric field near a single point charge is given by the formula:
away from a
field vector that points
fields when
CN
CN 500900 +−
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.22 Two point charges are located on the xposition x=0.2m and charge q2 = +5 nC at position x = a) Find the magnitude and direction of the net electric field produced by q
q2 at the origin.b) Find the net electric force on a charge q
2Rkq
E =
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awaypositive 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 +−=
(This means 400 N/C in the negative x-direction)
21total
2
9
CNm9
2
9
CNm9
total )m3.0(
)C105)(109(
)m2.0(
)C104)(109(E
2
2
2
2
−=⋅⋅
+⋅⋅
−=−−
CN
total 400E −=
q2 q1
x=0 x=0.2mx=-0.3m
Etotal
x
Two point charges are located on the x-axis as follows: charge q1 = +4 nC at at position x = -0.3m.
Find the magnitude and direction of the net electric field produced by q1 and
Find the net electric force on a charge q3=-0.6nC placed at the origin.
The electric field near a single point charge is given by the formula:
away from a
field vector that points
fields when
direction)
CN
CN 500900 +−
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.22 Two point charges are located on the xposition x=0.2m and charge q2 = +5 nC at position x = a) Find the magnitude and direction of the net electric field produced by q
q2 at the origin.b) Find the net electric force on a charge q
2Rkq
E =
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awaypositive 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 +−=
For part b) all we need to do is multiply the E-field from part a) times the new charge q
21total
2
9
CNm9
2
9
CNm9
total )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)
q2 q3 q1
x=0 x=0.2mx=-0.3m
x
Two point charges are located on the x-axis as follows: charge q1 = +4 nC at at position x = -0.3m.
Find the magnitude and direction of the net electric field produced by q1 and
Find the net electric force on a charge q3=-0.6nC placed at the origin.
The electric field near a single point charge is given by the formula:
away from a
field vector that points
fields when
Etotal
field from part a) times the new charge q3.
CN
CN 500900 +−
direction)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.22 Two point charges are located on the xposition x=0.2m and charge q2 = +5 nC at position x = a) Find the magnitude and direction of the net electric field produced by q
q2 at the origin.b) Find the net electric force on a charge q
2Rkq
E =
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awaypositive 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 +−=
Note that this force is to the right, which is opposite the EThis is because qup as if there are positive charges.
21total
2
9
CNm9
2
9
CNm9
total )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 q
N104.2)400)(C106.0(F 7CN9
onq3−− ⋅+=−⋅−=
q2 q3 q1
x=0 x=0.2mx=-0.3m
x
Two point charges are located on the x-axis as follows: charge q1 = +4 nC at at position x = -0.3m.
Find the magnitude and direction of the net electric field produced by q1 and
Find the net electric force on a charge q3=-0.6nC placed at the origin.
The electric field near a single point charge is given by the formula:
away from a
field vector that points
fields when
Etotal
Fon3
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.
CN
CN 500900 +−
direction)
field from part a) times the new charge q3.
Prepared by Vince Zaccone
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?
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?
Prepared by Vince Zaccone
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
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 =
Prepared by Vince Zaccone
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 haveelecF =
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 =
221
221
Rqkq
4R
)q2)(q2(k⋅==
Prepared by Vince Zaccone
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.
elecF =
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 =
221
221
Rqkq
4R
)q2)(q2(k⋅==
Prepared by Vince Zaccone
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?
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
Prepared by Vince Zaccone
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
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
The formula for electric force between 2 charges is 221
elec Dqkq
F =
Prepared by Vince Zaccone
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 setup the force equation and solve for the new distance.
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
The formula for electric force between 2 charges is
We want the force to be 3 times as strong, so we can setup the force equation and solve for the new distance.
221
elec Dqkq
F =
2new
21221
Dqkq
Dqkq
3 =⋅
Prepared by Vince Zaccone
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 setup the force equation and solve for the new distance.
2newD
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
The formula for electric force between 2 charges is
We want the force to be 3 times as strong, so we can setup the force equation and solve for the new distance.
221
elec Dqkq
F =
2new
21221
Dqkq
Dqkq
3 =⋅
231
new D⋅=
Prepared by Vince Zaccone
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 setup the force equation and solve for the new distance.
Square-roots of both sides gives us the answer:
2newD
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
The formula for electric force between 2 charges is
We want the force to be 3 times as strong, so we can setup the force equation and solve for the new distance.
roots of both sides gives us the answer:
221
elec Dqkq
F =
2new
21221
Dqkq
Dqkq
3 =⋅
231
new D⋅=
DD31
new ⋅=
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.30 When two unequal point charges arthe heavier one has an acceleration a. If yof this value, how far (in terms of d) should the charges be released?
are released a distance d from one another, f you want to reduce this acceleration to 1/5
of this value, how far (in terms of d) should the charges be released?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.30 When two unequal point charges arthe heavier one has an acceleration a. If yof 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.
are released a distance d from one another, f you want to reduce this acceleration to 1/5
of this value, how far (in terms of d) should the charges be released?
So this is really a problem about the force on the heavier charge.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.30 When two unequal point charges arthe heavier one has an acceleration a. If yof 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
are released a distance d from one another, f you want to reduce this acceleration to 1/5
of this value, how far (in terms of d) should the charges be released?
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 =
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.30 When two unequal point charges arthe heavier one has an acceleration a. If yof 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
51
2new
21
old51
new
dqkq
dqkq
FF
⋅=
⋅=
are released a distance d from one another, f you want to reduce this acceleration to 1/5
of this value, how far (in terms of d) should the charges be released?
So this is really a problem about the force on the heavier charge.
The formula for electric force between 2 charges is221
elec dqkq
F =
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.30 When two unequal point charges arthe heavier one has an acceleration a. If yof 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
51
2new
21
old51
new
dqkq
dqkq
FF
⋅=
⋅=
We cancel common terms and cross-multiply to get
are released a distance d from one another, f you want to reduce this acceleration to 1/5
of this value, how far (in terms of d) should the charges be released?
So this is really a problem about the force on the heavier charge.
The formula for electric force between 2 charges is221
elec dqkq
F =
multiply to get22
new d5d ⋅=
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.30 When two unequal point charges arthe heavier one has an acceleration a. If yof 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
51
2new
21
old51
new
dqkq
dqkq
FF
⋅=
⋅=
We cancel common terms and cross-multiply to get
Square-root of both sides: d5dnew ⋅=
are released a distance d from one another, f you want to reduce this acceleration to 1/5
of this value, how far (in terms of d) should the charges be released?
So this is really a problem about the force on the heavier charge.
The formula for electric force between 2 charges is221
elec dqkq
F =
multiply to get22
new d5d ⋅=
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 is placed on the x-axis at x=0.8m. Find thfield at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x=
is at the origin, and a second point charge of +6 nC the magnitude and direction of the electric
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
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 is placed on the x-axis at x=0.8m. Find thfield at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x=
is at the origin, and a second point charge of +6 nC the magnitude and direction of the electric
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
-4nC
x=0 x=0.8m
+6nCx
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awayfrom 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 is placed on the x-axis at x=0.8m. Find thfield at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x=
2RkQ
E =
-4nC
x=0 x=0.8m
+6nCx
one.
is at the origin, and a second point charge of +6 nC the magnitude and direction of the electric
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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 is placed on the x-axis at x=0.8m. Find thfield at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x=
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awayfrom a positive charge, and toward a negative one.
2RkQ
E =
a
field vectors point?
is at the origin, and a second point charge of +6 nC the magnitude and direction of the electric
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
-4nC
x=0 x=0.8m
+6nCx
one.
-4nC
x=0 x=0.8m
+6nCx
x=0 x=0.8m
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2
17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 is placed on the x-axis at x=0.8m. Find thfield at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x=
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awayfrom a positive charge, and toward a negative one.
21total EEE −−=
2RkQ
E =
21total
a
x direction4nC charge #1 and the +6nC charge #2 E1
E2
is at the origin, and a second point charge of +6 nC the magnitude and direction of the electric
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
one.
-4nC
x=0 x=0.8m
+6nCx
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
x=0 x=0.8m
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2
17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 is placed on the x-axis at x=0.8m. Find thfield at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x=
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awayfrom a positive charge, and toward a negative one.
21total EEE −−=
2RkQ
E =
21total
2
9
CNm9
2
9
CNm9
total )m6.0(
)C106)(109(
)m2.0(
)C104)(109(E
2
2
2
2
=⋅⋅
−⋅⋅
−=−−
a
x direction4nC charge #1 and the +6nC charge #2 E1
E2
is at the origin, and a second point charge of +6 nC the magnitude and direction of the electric
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
one.
-4nC
x=0 x=0.8m
+6nCx
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
x=0 x=0.8m
CN1050−
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2
17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 is placed on the x-axis at x=0.8m. Find thfield at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x=
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awayfrom a positive charge, and toward a negative one.
21total EEE −−=
2RkQ
E =
21total
2
9
CNm9
2
9
CNm9
total )m6.0(
)C106)(109(
)m2.0(
)C104)(109(E
2
2
2
2
=⋅⋅
−⋅⋅
−=−−
For part b) E1 points left and E2 points right
21total EEE +−=
a
x direction4nC charge #1 and the +6nC charge #2 E1
E2
is at the origin, and a second point charge of +6 nC the magnitude and direction of the electric
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
one.
-4nC
x=0 x=0.8m
+6nCx
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
x=0 x=0.8m
CN1050−
b
E1
E2
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2
17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 is placed on the x-axis at x=0.8m. Find thfield at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x=
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awayfrom a positive charge, and toward a negative one.
21total EEE −−=
2RkQ
E =
21total
2
9
CNm9
2
9
CNm9
total )m6.0(
)C106)(109(
)m2.0(
)C104)(109(E
2
2
2
2
=⋅⋅
−⋅⋅
−=−−
For part b) E1 points left and E2 points right
21total EEE +−=
2
9
CNm9
2
9
CNm9
total )m4.0(
)C106)(109(
)m2.1(
)C104)(109(E
2
2
2
2
=⋅⋅
+⋅⋅
−=−−
a
x direction4nC charge #1 and the +6nC charge #2 E1
E2
is at the origin, and a second point charge of +6 nC the magnitude and direction of the electric
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
one.
-4nC
x=0 x=0.8m
+6nCx
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
x=0 x=0.8m
CN1050−
b
E1
E2
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
CN5.312+
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2
17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 is placed on the x-axis at x=0.8m. Find thfield at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x=
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awayfrom a positive charge, and toward a negative one.
21total EEE −−=
2RkQ
E =
21total
2
9
CNm9
2
9
CNm9
total )m6.0(
)C106)(109(
)m2.0(
)C104)(109(E
2
2
2
2
=⋅⋅
−⋅⋅
−=−−
For part b) E1 points left and E2 points right
21total EEE +−=
2
9
CNm9
2
9
CNm9
total )m4.0(
)C106)(109(
)m2.1(
)C104)(109(E
2
2
2
2
=⋅⋅
+⋅⋅
−=−−
For part b) E1 points right and E2 points left
a
x direction4nC charge #1 and the +6nC charge #2 E1
E2
is at the origin, and a second point charge of +6 nC the magnitude and direction of the electric
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
one.
-4nC
x=0 x=0.8m
+6nCx
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
x=0 x=0.8m
CN1050−
b
E1
E2
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
CN5.312+
c
E1
E2
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For part a) both E-field vectors point in the –x directionCall the -4nC charge #1 and the +6nC charge #2
17.41 A point charge of -4 nC is at the origin, and a second point charge of +6 is placed on the x-axis at x=0.8m. Find thfield at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x=
The electric field near a single point charge is given by the formula:
This is only the magnitude. The direction is awayfrom a positive charge, and toward a negative one.
21total EEE −−=
2RkQ
E =
21total
2
9
CNm9
2
9
CNm9
total )m6.0(
)C106)(109(
)m2.0(
)C104)(109(E
2
2
2
2
=⋅⋅
−⋅⋅
−=−−
For part b) E1 points left and E2 points right
21total EEE +−=
2
9
CNm9
2
9
CNm9
total )m4.0(
)C106)(109(
)m2.1(
)C104)(109(E
2
2
2
2
=⋅⋅
+⋅⋅
−=−−
For part b) E1 points right and E2 points left
21total EEE −+=
2
9
CNm9
2
9
CNm9
total )m0.1(
)C106)(109(
)m2.0(
)C104)(109(E
2
2
2
2
=⋅⋅
−⋅⋅
+=−−
a
x direction4nC charge #1 and the +6nC charge #2 E1
E2
is at the origin, and a second point charge of +6 nC the magnitude and direction of the electric
axis: a) x=20 cm; b) x=1.20m; c) x= -20cm
one.
-4nC
x=0 x=0.8m
+6nCx
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
x=0 x=0.8m
CN1050−
b
E1
E2
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
CN5.312+
c
E1
E2
Q1 = -4nC
x=0 x=0.8m
x
Q2 = +6nC
CN846+=
17.42 A point charge of q=+6 nC is at thepoint 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,
the point (x=0.15m,y=0m), and an identical 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
Part a): TRY DRAWING THE E-FIELD VECTORS ON THE DIAGRAM
17.42 A point charge of q=+6 nC is at thepoint 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,
x
y
12
the point (x=0.15m,y=0m), and an identical 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
Part a): both vectors point away from their charge.Since the distances and the charges are equal, the vectors cancel out.
Etotal = 0
17.42 A point charge of q=+6 nC is at thepoint 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,
x
y
12
E1 E2
the point (x=0.15m,y=0m), and an identical 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
17.42 A point charge of q=+6 nC is at thepoint 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,
Part a): both vectors point away from their charge.Since the distances and the charges are equal, the vectors cancel out.
Etotal = 0
Part b): both vectors point away from their charge.
x
y
12
the point (x=0.15m,y=0m), and an identical 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
x
y
2 1
E1
E2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.42 A point charge of q=+6 nC is at thepoint 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,
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
Part b): both vectors point away from their charge.
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
=⋅⋅
=−
x
y
12
the point (x=0.15m,y=0m), and an identical 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
x
y
2 1
direction
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
E1
E2
direction
17.42 A point charge of q=+6 nC is at thepoint 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,
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
Part b): both vectors point away from their charge.
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 =+=
x
y
12
the point (x=0.15m,y=0m), and an identical 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
x
y
2 1
direction
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
E1
E2
direction
Part c): both vectors point away from their charge. We will need to use vector components to add them together.
17.42 A point charge of q=+6 nC is at thepoint 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,
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)
the point (x=0.15m,y=0m), and an identical 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)
(0.15,- 0.4)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.42 A point charge of q=+6 nC is at thepoint 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,
Part c): both vectors point away from their charge. We will need to use vector components to add them together.
x
y
12
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
the point (x=0.15m,y=0m), and an identical 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
(0.15,- 0.4)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.42 A point charge of q=+6 nC is at thepoint 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,
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
0Ex
y
12
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
the point (x=0.15m,y=0m), and an identical 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
(0.15,- 0.4)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
17.42 A point charge of q=+6 nC is at thepoint 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,
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
0Ex
y
12
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
the point (x=0.15m,y=0m), and an identical 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.
(0.15,- 0.4)
E2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
E1,y
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.
17.42 A point charge of q=+6 nC is at thepoint 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,
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
triangle when you see it.
x
y
12
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
The 0.5m in this formula for is the distance to charge 2,
using Pythagorean theorem or 5 right
triangle when you see it.
0.4m
the point (x=0.15m,y=0m), and an identical 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.
(0.15,- 0.4)triangle when you see it.
0.3m
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
E2
E1,y
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.
17.42 A point charge of q=+6 nC is at thepoint 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,
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
triangle when you see it.
x
y
12
(0.15,- 0.4)
(0.15,0)(- 0.15,0)
The 0.5m in this formula for is the distance to charge 2,
using Pythagorean theorem or 5 right
triangle when you see it.
0.4m
the point (x=0.15m,y=0m), and an identical 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.
E(0.15,- 0.4)triangle when you see it.
0.3m
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
E1,y
E2,x
E2,y
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.
17.42 A point charge of q=+6 nC is at thepoint 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,
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
+=⋅+= N3N
x,2 6.129)()216(E triangle when you see it.
−=⋅+=
+=⋅+=
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 is the distance to charge 2,
using Pythagorean theorem or 5 right
triangle when you see it.
0.4m
the point (x=0.15m,y=0m), and an identical 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.
E(0.15,- 0.4)triangle when you see it.
0.3m
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
E1,y
E2,x
E2,y
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.
17.42 A point charge of q=+6 nC is at thepoint 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,
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
+=⋅+= N3N
x,2 6.129)()216(E triangle when you see it.
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 is the distance to charge 2,
using Pythagorean theorem or 5 right
triangle when you see it.
0.4m
the point (x=0.15m,y=0m), and an identical 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.
E(0.15,- 0.4)triangle when you see it.
0.3m
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
E1,y
E2,x
E2,y
components separately:
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.
17.42 A point charge of q=+6 nC is at thepoint 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,
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
+=⋅+= N3N
x,2 6.129)()216(E triangle when you see it.
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:
axisxbelow7.756.1293.510
)tan(
5.526)3.510()6.129(E CN22
total
+°=θ⇒=θ
=+=
−=⋅+=
+=⋅+=
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 is the distance to charge 2,
using Pythagorean theorem or 5 right
triangle when you see it.
the point (x=0.15m,y=0m), and an identical 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.
(0.15,- 0.4)triangle when you see it.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
components separately:
Now find the magnitude and the angle using right triangle rules:
75.7º
Etotal
Part d): TRY THIS ONE ON YOUR OWN FIRST...
17.42 A point charge of q=+6 nC is at thepoint 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,
x
y
12
(0.15,0)(- 0.15,0)
the point (x=0.15m,y=0m), and an identical 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)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Part d): both vectors point away from their charge. We will need to use vector components to add them together.
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.
17.42 A point charge of q=+6 nC is at thepoint 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,
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
+=+= N15.0N 4.518))(864(E
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
+=+=
+=+=
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
+=++=
=+−=
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
5 right triangle
the point (x=0.15m,y=0m), and an identical 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)
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
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