1. LO1:Electric Force & Electric
Field
Lecture Slides Prepared by Physics Team
2. Learning Outcome (01) Define and solve
problems related to electricity and the
electric field.[ Chapter 23 & Chapter 24 ]
Suboutcome 01: Define the electron, charge,
and Coulomb as a unit of electric charge.
Suboutcome 02: State Coulomb’s law and apply
it to the solution of electric force problems.
Suboutcome 03: Define the electric field and
explain what determines its magnitude and
direction.
Suboutcome 04: Define the electric field
intensity expression and compute its magnitude
at a given distance from a known charge.
Suboutcome 05 : Explain Gauss’s Law, the
concept of the permittivity of a medium, and the
3. The Electric Field
1. Now, consider point P a
distance r from +Q.
+q +
P .
2. An electric field E exists
at P if a test charge +q
has a force F at that point.
E
r
+
++ +
Q
++ ++
3. The direction of the E is
the same as the direction of
a force on + (pos) charge.
4. The magnitude of E is
given by the formula:
F
Electric Field
E
F
N
; Units
q
C
4. Field is Property of Space
F .
+q +
E
r
+ ++
+Q +
++ +
Electric Field
Force on +q is with
field direction.
Force on q is
against field
direction.
.
q F
E
r
+ ++
+Q +
++ +
Electric Field
The field E at a point exists whether there is a
charge at that point or not. The direction of the
field is away from the +Q charge.
5. Field Near a Negative Charge
.
E
+q +
F r
 Q  Electric Field
Force on +q is with
field direction.
Force on q is
against field
direction.
F
.
q 
E
r
  Q Electric Field
Note that the field E in the vicinity of a negative
charge –Q is toward the charge—the direction that a
+q test charge would move.
6. The Magnitude of EField
The magnitude of the electric field intensity at a
point in space is defined as the force per unit
charge (N/C) that would be experienced by any
test charge placed at that point.
Electric Field
Intensity E
E
F
; Units
q
N
C
The direction of E at a point is the same as the
direction that a positive charge would move IF
placed at that point.
7. Example 1. A +2 nC charge is placed at
a distance r from a –8 C charge. If the
+2 nC
charge experiences a force of 4000 N,
.
+q + P
what is the electric field intensity E at
4000 N
point P?
E E r
First, we note that the direction of
E is toward –Q (down).
 Q  –8 C
 Electric Field
E
F
q
4000 N
9
2 x 10 C
E = 2 x 1012 N/C
Downward
Note: The field E would be the same for any charge
placed at point P. It is a property of that space.
8. Example 2. A constant E field of 40,000 N/C is maintained
between the two parallel plates. What are the magnitude
and direction of the force on an electron that passes
horizontally between the plates .
The Efield is downward,
and the force on e is up.
E
F
; F
q
F
qE
+ + + + + + + + +
e 
qE
Fe.
E
e 
        19
(1.6 x 10 C)(4 x 10
F = 6.40 x 1015 N, Upward
4 N
C
)
9. The EField at a distance r from a
single charge Q
Consider a test charge +q placed
at P a distance r from Q.
E
F
.
+q +. P
P
r
The outward force on +q is:
++ ++ E
+Q +
++ +
kQq
2
r
F
The electric field E is therefore:
E
F
q
kQq r
q
2
E
kQ
2
r
kQ
2
r
10. Example 3. What is the electric field intensity E
at point P, a distance of 3 m from a negative
charge of –8 nC?
First, find the magnitude:
E=?
.
9 Nm2
(9 x 10 C2 )(8 x 109C)
kQ
P
r
E
2
2
r
(3 m)
3m
Q
8 nC
E = 8.00 N/C
The direction is the same as the force on a positive
charge if it were placed at the point P: toward –Q.
E = 8.00 N, toward Q
11. The Resultant Electric Field
The resultant field E in the vicinity of a number
of point charges is equal to the vector sum of the
fields due to each charge taken individually.
Consider E for each charge.
Vector Sum:
E = E1 + E2 + E3
Magnitudes are from:
E
kQ
r2
q1 ER
E2
A
E1
q3 
E3
+
q2
Directions are based
on positive test charge.
12. Example 4. Find the resultant field at point A due
to the –3 nC charge and the +6 nC charge
arranged as shown.
E for each q is shown
with direction given.
3 nC
q1 
3 cm E 5 cm
1
E2 A 4 cm
E1
9 Nm2
C2
(9 x 10
+6 nC
+
q2
9
)(3 x 10 C)
(3 m)
E1
2
E2
kq1
; E2
2
r1
9 Nm2
C2
(9 x 10
kq2
r22
)(6 x 109C)
(4 m)2
Signs of the charges are used only to find direction of E
13. Example 4. (Cont.) Find the resultant field at
point A. The magnitudes are:
3 nC
q1 
3 cm E 5 cm
1
E1
+6 nC
E2 A 4 cm
E1 = 3.00 N, West
+
q2
E2
9 Nm2
C2
(9 x 10
(3 m)2
9 Nm2
C2
(9 x 10
2
E2
E2 = 3.38 N, North
R12 ; tan
)(6 x 109C)
(4 m)2
Next, we find vector resultant ER
ER
)(3 x 109C)
E1
E2
ER
E1
E2
14. Example 4. (Cont.) Find the resultant field at
point A using vector mathematics.
ER
E1 = 3.00 N, West
E1
E2
E
E2 = 3.38 N, North
Find vector resultant ER
(3.00 N)
2
(3.38 N)
2
4.52 N;
= 48.40 N of W; or
tan
3.38 N
3.00 N
= 131.60
Resultant Field: ER = 4.52 N; 131.60
15. Electric Field Lines
Electric Field Lines are imaginary lines drawn in
such a way that their direction at any point is the
same as the direction of the field at that point.
+
++ +
Q
++ ++
  Q 
Field lines go away from positive charges and
toward negative charges.
16. Examples of EField Lines
Two equal but
opposite charges.
Two identical
charges (both +).
Notice that lines leave + charges and enter  charges.
Also, E is strongest where field lines are most dense.
17. The Density of Field Lines
Gauss’s Law: The field E at any point in space is
proportional to the line density at that point.
Gaussian Surface
N
Line density
r
A
Radius r
N
A
18. Line Density and Spacing
Constant
Consider the field near a positive point charge q:
Then, imagine a surface (radius r) surrounding q.
Radius r
r
E is proportional to N/ A and
is equal to kq/r2 at any point.
N
A
Gaussian Surface
N
A
Define
0 E Where
E;
kq
2
r
E
as spacing constant. Then:
0
is:
1
0
4 k
19. Permittivity of Free Space
The proportionality constant for line density is
known as the permittivity and it is defined by:
1
0
4 k
8.85 x 1012
C2
2
N m
Recalling the relationship with line density, we have:
N
A
0
E or
N
Summing over entire area
A gives the total lines as:
0
E A
N=
oEA
20. Example 5. Write an equation for finding the
total number of lines N leaving a single
positive charge q.
Draw spherical Gaussian surface:
Radius r
N
r
0
E A and
N
0
EA
Substitute for E and A from:
E
Gaussian Surface
N
0 EA
0
kq
r2
q
; A = 4 r2
4 r2
q
(4 r 2 )
4 r2
N=
oqA
=q
Total number of lines is equal to the enclosed charge q.
21. Gauss’s Law
Gauss’s Law: The net number of electric field
lines crossing any closed surface in an outward
direction is numerically equal to the net total
charge within that surface.
N
0
EA
If we represent q as net enclosed
positive charge, we can write
rewrite Gauss’s law as:
q
EA
q
0
22. Example 6. How many electric field lines pass
through the Gaussian surface drawn below.
First we find the NET
charge q enclosed
by the surface:
q = (+8 –4 – 1) = +3 C
N
0
EA
q
N = +3 C = +3 x 106 lines
Gaussian surface
4 C
q1 1 C
q3 
+8 C
q2 +
q4
+
+5 C
23. Example 7. A solid sphere (R = 6 cm) having net
charge +8 C is inside a hollow shell (R = 8 cm)
having a net charge of –6 C. What is the electric
field at a distance of 12 cm from the center of the
solid sphere?
Draw Gaussian sphere at
radius of 12 cm to find E.
N
0 EA
q
q = (+8 – 6) = +2 C
q
qnet ; E
0 AE
0A
E
q
(4 r 2 )
0
Gaussian surface
6 C
8cm


12 cm

+8 C

6 cm
 
2 x 106C
12 Nm2
2
(8.85 x 10
2 )(4 )(0.12 m)
C
24. Example 7 (Cont.) What is the electric field at a
distance of 12 cm from the center of the solid sphere?
Draw Gaussian sphere at
radius of 12 cm to find E.
N
0 EA
q
Gaussian surface
q = (+8 – 6) = +2 C
q
qnet ; E
0 AE
0A
E

8cm
2 C
1.25 x 106 N C
2
0 (4 r )
12 cm
+8 C
6 C

6 cm
 
E = 1.25 MN/C
25. Summary of Formulas
The Electric Field
Intensity E:
The Electric Field
Near several charges:
Gauss’s Law for
Charge distributions.
E
F
q
0
N
Units are
C
kQ
r2
E
kQ
r2
Vector Sum
EA
q;
q
A
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