Transcript Physics 6B Electric Fields - UCSB Campus Learning Assistance

Physics 6B
Electric Field Examples
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Electric charge can be either positive or negative.
Matter is chiefly comprised of electrons (negative), protons (positive) and
neutrons (electrically neutral). A neutral object will have equal numbers of
protons and electrons.
Most of the time it is the negatively-charged electrons that can move back and
forth between objects, so a negatively charged object has excess electrons, and a
positively charged object has too few electrons.
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Electric charge can be either positive or negative.
Matter is chiefly comprised of electrons (negative), protons (positive) and
neutrons (electrically neutral). A neutral object will have equal numbers of
protons and electrons.
Most of the time it is the negatively-charged electrons that can move back and
forth between objects, so a negatively charged object has excess electrons, and a
positively charged object has too few electrons.
Elementary charge: Charge is quantized, which means that the charge on any
object is always a multiple the charge on a proton (or electron).
e = 1.6 x 10-19 C This is the smallest possible charge.
Units for charge are Coulombs. The Coulomb is a very large unit, so you can
expect to see tiny values like nano-Coulombs.
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Charges interact with each other via the Electric Force.
Rules for interaction are based on the sign of the charge as follows:
* Like charges repel
* Opposite charges attract
The force is given by Coulomb’s Law:

k q1q2
Felec 
r2
k
9 Nm2
9  10
C2
Coulomb’s Constant
Notice that this is just the magnitude of the force, and the r is the center-tocenter distance between the two charges.
My advice is to not put +- signs into this formula. Instead, find the direction of
the force based on the attract/repel rules above.
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Electric Fields
This is just another (very important) way of looking
at electric forces. We find the electric field near a
charge distribution, then we can simply multiply by
any charge to find the force on that charge.

k Qq
Felec  2
r
kQ
E-field near a pointE  2
charge Q is just most
of the force formula
r


Felec  E  q
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Electric Field Lines
The charge on the right is twice the magnitude of
the charge on the left (and opposite in sign), so
there are twice as many field lines, and they point
towards the charge rather than away from it.
Electric Field of a Dipole
+q
-q
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Electric Field of a Dipole
Notice that the field lines point away from positive and
toward negative charges. This will always be true.
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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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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.
q2
x=-0.3m
q1
x=0
x
x=0.2m
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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.
The electric field near a single point charge is given by the formula:
E
kq
R2
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
q2
x=-0.3m
q1
x=0
x
x=0.2m
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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.
The electric field near a single point charge is given by the formula:
E
kq
R2
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
At the origin, q1 will produce an E-field vector that points
left, and q2 gives an E-field vector to the right.
q2
x=-0.3m
E1
x=0
E2
q1
x
x=0.2m
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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.
The electric field near a single point charge is given by the formula:
E
kq
R2
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
At the origin, q1 will produce an E-field vector that points
left, and q2 gives an E-field vector to the right.
q2
x=-0.3m
E1
x=0
E2
q1
x
x=0.2m
This is how we can put the +/- signs on the E-fields when
we add them up.
Etotal  E1  E2
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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.
The electric field near a single point charge is given by the formula:
E
kq
R2
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
At the origin, q1 will produce an E-field vector that points
left, and q2 gives an E-field vector to the right.
q2
x=-0.3m
E1
x=0
E2
q1
x
x=0.2m
This is how we can put the +/- signs on the E-fields when
we add them up.
Etotal  E1  E2
Etotal  
(9  109
Nm2
C2
)(4  109 C)
2
(0.2m)
(9  109 Nm
)(5  109 C)
C2
2

2
(0.3m)
 900 NC  500 NC
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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.
The electric field near a single point charge is given by the formula:
E
kq
R2
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
At the origin, q1 will produce an E-field vector that points
left, and q2 gives an E-field vector to the right.
q2
x=-0.3m
Etotal
x=0
q1
x
x=0.2m
This is how we can put the +/- signs on the E-fields when
we add them up.
Etotal  E1  E2
Etotal  
(9  109
Nm2
C2
)(4  109 C)
2
(0.2m)
Etotal  400 NC
(9  109 Nm
)(5  109 C)
C2
2

2
(0.3m)
 900 NC  500 NC
(This means 400 N/C in the negative x-direction)
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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.
The electric field near a single point charge is given by the formula:
E
kq
R2
This is only the magnitude. The direction is away from a
positive charge, and toward a negative one.
At the origin, q1 will produce an E-field vector that points
left, and q2 gives an E-field vector to the right.
q2
x=-0.3m
Etotal
q3
x=0
q1
x
x=0.2m
This is how we can put the +/- signs on the E-fields when
we add them up.
Etotal  E1  E2
Etotal  
(9  109
Nm2
C2
)(4  109 C)
2
(0.2m)
Etotal  400 NC
(9  109 Nm
)(5  109 C)
C2
2

2
(0.3m)
 900 NC  500 NC
(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.
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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.
The electric field near a single point charge is given by the formula:
kq
E 2
R
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.
q2
x=-0.3m
Etotal
q3
x=0
Fon3
q1
x
x=0.2m
This is how we can put the +/- signs on the E-fields when
we add them up.
Etotal  E1  E2
Etotal  
(9  109
Nm2
C2
)(4  109 C)
2
(0.2m)
Etotal  400 NC
(9  109 Nm
)(5  109 C)
C2
2

2
(0.3m)
 900 NC  500 NC
(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.
Fonq3  (0.6  109 C)(400 NC )  2.4  107N
Note that this force is to the right, which is opposite the E-field
This is because q3 is a negative charge: E-fields are always set
up as if there are positive charges.
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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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Two unequal charges repel each other with a force F. If both charges are doubled in
magnitude, what will be the new force in terms of F?
The formula for electric force between 2
charges is
Felec
kq1q2

R2
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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
Felec
kq1q2

R2
Felec
k(2q1)(2q2 )
kq1q2

 4
2
R
R2
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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
Felec
kq1q2

R2
Felec
k(2q1)(2q2 )
kq1q2

 4
2
R
R2
So the new force is 4 times as large.
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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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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?
Formula for electric force between 2 charges is
Felec 
kq1q2
D2
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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?
kq1q2
D2
Formula for electric force between 2 charges is
Felec 
We want the force to be 3 times as strong, so
we can set up the force equation and solve for
the new distance.
kq1q2 kq1q2
3
 2
2
D
Dnew
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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?
kq1q2
D2
Formula for electric force between 2 charges is
Felec 
We want the force to be 3 times as strong, so
we can set up the force equation and solve for
the new distance.
kq1q2 kq1q2
3
 2
2
D
Dnew
Canceling and cross-multiplying, we get
2
Dnew

1
3
 D2
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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?
kq1q2
D2
Formula for electric force between 2 charges is
Felec 
We want the force to be 3 times as strong, so
we can set up the force equation and solve for
the new distance.
kq1q2 kq1q2
3
 2
2
D
Dnew
Canceling and cross-multiplying, we get
Square-roots of both sides gives us the answer:
2
Dnew

Dnew 
1
3
 D2
1
3
D
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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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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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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.
formula for electric force between 2 charges is
Felec 
kq1q2
d2
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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.
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:
Felec 
Fnew 
1
5
kq1q2

2
dnew
kq1q2
d2
 Fold
1
5
kq1q2

d2
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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.
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
Felec 
Fnew 
1
5
kq1q2

2
dnew
kq1q2
d2
 Fold
1
5
kq1q2

d2
2
dnew
 5  d2
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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.
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:
Felec 
Fnew 
1
5
kq1q2

2
dnew
kq1q2
d2
 Fold
1
5
kq1q2

d2
2
dnew
 5  d2
dnew  5  d
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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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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
+6nC
x
x=0.8m
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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
kQ
The electric field near a single point
E 2
charge is given by the formula:
R
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
-4nC
x=0
+6nC
x
x=0.8m
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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
kQ
The electric field near a single point
E 2
charge is given by the formula:
R
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
-4nC
+6nC
x=0
x
x=0.8m
For part a) which direction do the E-field vectors point?
-4nC
x=0
a
+6nC
x
x=0.8m
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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
kQ
The electric field near a single point
E 2
charge is given by the formula:
R
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
For part a) both E-field vectors point in the –x direction
Call the -4nC charge #1 and the +6nC charge #2
Etotal  E1  E2
-4nC
+6nC
x=0
Q1 = E1
-4nC E
2
x=0
x
x=0.8m
a
Q2 =
+6nC
x
x=0.8m
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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
kQ
The electric field near a single point
E 2
charge is given by the formula:
R
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
For part a) both E-field vectors point in the –x direction
Call the -4nC charge #1 and the +6nC charge #2
Etotal  E1  E2
(9  109 Nm
)(4  109 C)
C2
2
(0.2m)

(9  109
Nm2
C2
)(6  109 C)
2
(0.6m)
+6nC
x=0
Q1 = E1
-4nC E
2
x=0
2
Etotal  
-4nC
x
x=0.8m
a
Q2 =
+6nC
x
x=0.8m
 1050 NC
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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
kQ
The electric field near a single point
E 2
charge is given by the formula:
R
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
For part a) both E-field vectors point in the –x direction
Call the -4nC charge #1 and the +6nC charge #2
Etotal  E1  E2
(9  109 Nm
)(4  109 C)
C2
2
(0.2m)

(9  109
Nm2
C2
)(6  109 C)
2
(0.6m)
For part b) E1 points left and E2 points right
Etotal  E1  E2
+6nC
x=0
Q1 = E1
-4nC E
2
x=0
2
Etotal  
-4nC
x
x=0.8m
a
Q2 =
+6nC
x
x=0.8m
 1050 NC
Q1 =
-4nC
x=0
Q2 =
+6nC
E1
E2
b
x
x=0.8m
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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
kQ
The electric field near a single point
E 2
charge is given by the formula:
R
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
For part a) both E-field vectors point in the –x direction
Call the -4nC charge #1 and the +6nC charge #2
Etotal  E1  E2
(9  109 Nm
)(4  109 C)
C2
2
(0.2m)

(9  109
Nm2
C2
)(6  109 C)
2
(0.6m)
Q1 =
-4nC
x=0
Etotal  E1  E2
2
Etotal  
Q1 = E1
-4nC E
2
2
(1.2m)
(9  109 Nm
)(6  109 C)
C2
x
x=0.8m
a
Q2 =
+6nC
x
x=0.8m
 1050 NC
For part b) E1 points left and E2 points right
(9  109 Nm
)(4  109 C)
C2
+6nC
x=0
x=0
2
Etotal  
-4nC
Q2 =
+6nC
E1
E2
b
x
x=0.8m
2

2
(0.4m)
 312.5 NC
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
kQ
The electric field near a single point
E 2
charge is given by the formula:
R
-4nC
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
x=0
For part a) both E-field vectors point in the –x direction
Call the -4nC charge #1 and the +6nC charge #2
Q1 = E1
-4nC E
2
Etotal  E1  E2
x=0
(9  109 Nm
)(4  109 C)
C2
2
Etotal  
2
(0.2m)

(9  109
Nm2
C2
)(6  109 C)
2
(0.6m)
Q1 =
-4nC
x=0
Etotal  E1  E2
2
Etotal  
2
(1.2m)
(9  109 Nm
)(6  109 C)
C2
x
x=0.8m
a
Q2 =
+6nC
x
x=0.8m
 1050 NC
For part b) E1 points left and E2 points right
(9  109 Nm
)(4  109 C)
C2
+6nC
Q2 =
+6nC
E1
E2
b
x
x=0.8m
2

2
(0.4m)
For part c) E1 points right and E2 points left
 312.5 NC
E1
E2
c
Q1 =
-4nC
x=0
Q2 =
+6nC
x
x=0.8m
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
kQ
The electric field near a single point
E 2
charge is given by the formula:
R
-4nC
This is only the magnitude. The direction is away
from a positive charge, and toward a negative one.
x=0
For part a) both E-field vectors point in the –x direction
Call the -4nC charge #1 and the +6nC charge #2
Q1 = E1
-4nC E
2
Etotal  E1  E2
x=0
(9  109 Nm
)(4  109 C)
C2
2
Etotal  
2
(0.2m)

(9  109
Nm2
C2
)(6  109 C)
2
(0.6m)
Q1 =
-4nC
x=0
Etotal  E1  E2
2
Etotal  
2
(1.2m)
(9  109 Nm
)(6  109 C)
C2

(9  109 Nm
)(4  109 C)
C2
2
2
(0.2m)
x=0.8m
a
Q2 =
+6nC
x
x=0.8m
Q2 =
+6nC
E1
E2
b
x
x=0.8m
2
2
(0.4m)
 312.5 NC

(9  109
Nm2
C2
)(6  109 C)
2
(1.0m)
E1
E2
c
For part c) E1 points right and E2 points left
Etotal  E1  E2
Etotal  
x
 1050 NC
For part b) E1 points left and E2 points right
(9  109 Nm
)(4  109 C)
C2
+6nC
Q1 =
-4nC
x=0
 846 NC
Q2 =
+6nC
x
x=0.8m
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
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)
y
Part a): TRY DRAWING THE E-FIELD
VECTORS ON THE DIAGRAM
2
1
x
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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)
y
Part a): both vectors point away from their charge.
Since the distances and the charges are equal, the
vectors cancel out.
2
E1
E2
1
x
Etotal = 0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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)
y
Part a): both vectors point away from their charge.
Since the distances and the charges are equal, the
vectors cancel out.
2
E2
E1
1
x
Etotal = 0
y
Part b): both vectors point away from their charge, to the right.
E2
2
1
E1
x
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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)
y
Part a): both vectors point away from their charge.
Since the distances and the charges are equal, the
vectors cancel out.
2
E2
E1
1
x
Etotal = 0
y
Part b): both vectors point away from their charge, to the right.
(9  109 Nm
)(6  109 C)
C2
2
E1 
E2 
2
(0.15m)
(9  109
Nm2
C2
)(6  109 C)
2
(0.45m)
 2400 NC
Positive x-direction
E2
2
 267 NC
Positive x-direction
1
E1
x
Etotal  2400  267  2667 NC
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
y
(- 0.15,0)
2
(0.15,0)
1
x
(0.15,- 0.4)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
y
(- 0.15,0)
2
(0.15,0)
1
x
(0.15,- 0.4)
E1,y
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
(9  109 Nm
)(6  109 C)
C2
y
2
E1 

E1,x


E1,y
2
(0.4m)
 0 NC



 337.5 NC 

 337.5 NC
(- 0.15,0)
2
(0.15,0)
1
x
(0.15,- 0.4)
E1,y
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
(9  109 Nm
)(6  109 C)
C2
y
2
E1 

E1,x


E1,y
2
(0.4m)
 0 NC



 337.5 NC 

 337.5 NC
(- 0.15,0)
2
(0.15,0)
1
x
(0.15,- 0.4)
E2
E1,y
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
(9  109 Nm
)(6  109 C)
C2
y
2
E1 

E1,x


E1,y
E2 
2
(0.4m)
 0 NC



 337.5 NC 

9 Nm2
C2
(9  10
 337.5 NC
(- 0.15,0)
2
9
)(6  10 C)
2
(0.5m)
 216 NC
The 0.5m in this formula for
E2 is the distance to charge 2,
using Pythagorean theorem or
from recognizing a 3-4-5 right
triangle when you see it.
(0.15,0)
1
x
0.4m
(0.15,- 0.4)
0.3m
E2
E1,y
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
(9  109 Nm
)(6  109 C)
C2
y
2
E1 

E1,x


E1,y
E2 
2
(0.4m)
 0 NC



 337.5 NC 

9 Nm2
C2
(9  10
 337.5 NC
(- 0.15,0)
2
9
)(6  10 C)
2
(0.5m)
 216 NC
The 0.5m in this formula for
E2 is the distance to charge 2,
using Pythagorean theorem or
from recognizing a 3-4-5 right
triangle when you see it.
(0.15,0)
1
x
0.4m
(0.15,- 0.4)
E2,x
0.3m
E2,y
E1,y
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
(9  109 Nm
)(6  109 C)
C2
y
2
E1 

E1,x


E1,y
E2 
2
(0.4m)
 0 NC



 337.5 NC 

9 Nm2
C2
(9  10
 337.5 NC
(- 0.15,0)
2
9
)(6  10 C)
2
(0.5m)
 216 NC
E2,x  ( 216 N )  ( 3 )  129.6 N 

C
5
C


)  ( 4 )  172.8 N 
E2,y  ( 216 N
C
5
C
The 0.5m in this formula for
E2 is the distance to charge 2,
using Pythagorean theorem or
from recognizing a 3-4-5 right
triangle when you see it.
(0.15,0)
1
x
0.4m
(0.15,- 0.4)
E2,x
0.3m
E2,y
E1,y
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
(9  109 Nm
)(6  109 C)
C2
y
2
E1 

E1,x


E1,y
E2 
2
(0.4m)
 0 NC



 337.5 NC 

9 Nm2
C2
(9  10
 337.5 NC
(- 0.15,0)
2
9
)(6  10 C)
2
(0.5m)
 216 NC
E2,x  ( 216 N )  ( 3 )  129.6 N 

C
5
C


)  ( 4 )  172.8 N 
E2,y  ( 216 N
C
5
C
The 0.5m in this formula for
E2 is the distance to charge 2,
using Pythagorean theorem or
from recognizing a 3-4-5 right
triangle when you see it.
(0.15,0)
1
x
0.4m
Add together the x-components and the y-components separately:
Etotal,x  0 NC  129.6 NC  129.6 NC
(0.15,- 0.4)
E2,x
0.3m
E2,y
E1,y
Etotal,y  337.5 NC  172.8 NC  510.3 NC
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
(9  109 Nm
)(6  109 C)
C2
y
2
E1 

E1,x


E1,y
E2 
2
(0.4m)
 0 NC



 337.5 NC 

9 Nm2
C2
(9  10
 337.5 NC
(- 0.15,0)
2
9
)(6  10 C)
2
(0.5m)
 216 NC
E2,x  ( 216 N )  ( 3 )  129.6 N 

C
5
C


)  ( 4 )  172.8 N 
E2,y  ( 216 N
C
5
C
The 0.5m in this formula for
E2 is the distance to charge 2,
using Pythagorean theorem or
from recognizing a 3-4-5 right
triangle when you see it.
Add together the x-components and the y-components separately:
Etotal,x  0 NC  129.6 NC  129.6 NC
(0.15,0)
1
x
(0.15,- 0.4)
75.7º
Etotal
Etotal,y  337.5 NC  172.8 NC  510.3 NC
Now find the magnitude and the angle using right triangle rules:
Etotal  (129.6)2  (510.3)2  526.5 NC
tan() 
510.3
   75.7 below  x axis
129.6
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 d): TRY THIS ONE ON YOUR OWN FIRST...
y
(0,0.2)
(- 0.15,0)
2
(0.15,0)
1
x
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 d): both vectors point away from their charge. We
will need to use vector components to add them together.
E1 
9 Nm2
C2
(9  10
9
)(6  10 C)
2
(0.25m)
 864 NC
0.15
N
N

E1,x  (864 C )(0.25)  518.4 C 



0.20
N
N
E


(
864
)(
)


691
.
2


1
,
y
C
0
.
25
C


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.
y
E1
E2
(0,0.2)
(- 0.15,0)
2
(0.15,0)
1
x
From symmetry, we can see that E2 will have
the same components, except for +/- signs.
E2,x  (864 N )( 0.15 )  518.4 N 

C 0.25
C


N )( 0.20 )  691.2 N
E


(
864


C 0.25
C
 2,y
Now we can add the components
(the x-component should cancel out)
Etotal,x  518.4 NC  518.4 NC  0 NC
Etotal,y  691.2 NC  691.2 NC  1382.4 NC
The final answer should be 1382.4 N/C in the positive y-direction.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Electric Flux
Field lines passing through a surface are called “flux”. To find the flux through a
surface, multiply field strength times the area of the surface. Here is the formula:
elec.  E  A  cos()
Notice that if the field is not perpendicular
to the area we need to multiply by cos(Φ).
Prepared by Vince Zaccone
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Electric Flux
Try this example:
A uniform electric field points in the +x direction and has magnitude 2000 N/C.
Find the electric flux through a circular window of radius 10cm, tilted at an angle
of 30° to the x-axis, as shown below.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Electric Flux
Try this example:
A uniform electric field points in the +x direction and has magnitude 2000 N/C.
Find the electric flux through a circular window of radius 10cm, tilted at an angle
of 30° to the x-axis, as shown below.
We can apply our formula directly here:
 elec.  E  A  cos()
 elec.  (2000 N )  (  (0.1m)2 )  cos(30 )
C
2
 elec.  54 Nm
C
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Electric Flux
Try this example:
A positive 3µC charge is surrounded by a spherical surface of radius 0.2m, as
shown. Find the net electric flux through the sphere.
Notice that the electric field lines intercept the sphere
at right angles, so the angle in our flux formula is 0°.
 elec.  E  A  cos()
kq
 elec.  ( )  (4r2 )  cos(0 )
r2
 elec.  4kq
We can put in formulas for the electric field near a
point charge, and the surface area for a sphere to
arrive at a nice formula.
It turns out that the flux only depends on the enclosed
charge, and furthermore, if we write it in terms of
permittivity ε0, we find the following:
q
elec.  enclosed
0
This is called Gauss’ Law, and it works out that we
don’t even need a sphere – any closed surface will do.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB