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A field is defined as a property of space in which a
material object experiences a force.
.P
Above earth, we say there is
a gravitational field at P.
Because a mass m experiences a
downward force at that point.
No force, no field; No field, no force!
m
F
The direction of the field is determined by the force.
Consider points A and B above the surface of
the earth—just points in space.
Q:What is the similarities and dissimilarities
between electric field and gravitational field?
The field at points A or B might
be found from:
F
g
m
If g is known at
every point above
the earth then the
force F on a given
mass can be found.
The magnitude and direction of
the field g is depends on the
weight, which is the force F.

A
 B
F
F
1. Now, consider point P a
distance r from +Q.
2. An electric field E exists
at P if a test charge +q
has a force F at that point.
3. The direction of the E is
the same as the direction of
a force on + (pos) charge.
E
4. The magnitude of E is
given by the formula:
N
; Units
C
F
E
q

Electric Field
++
++
+
+
++Q
.P
r
+q
F
+
E
Electric Field
++
++
+
+
++Q
.
r
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.
E
Electric Field
++
++
+
+
++Q
.
r
++q --q
F
F
Force on +q is with
field direction.
Force on -q is
against field
direction.
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.
Force on +q is with
field direction.
Force on -q is
against field
direction.
E
Electric Field
.
r
++q
F
--
-- -
-
---Q
E
Electric Field
.
r
--q
F
--
-- -
-
---Q
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
N
; Units
C
F
E
q
 
  
 
The direction of E at a point is the same as the
direction that a positive charge would move.
A +2 nC charge is placed
at a distance r from a –8 mC charge.
If the charge experiences a force of
4000 N, what is the electric field
intensity E at point P?
Electric Field
.
--
-- -
-
---Q
P
First, we note that the direction of
E is toward –Q (down).
–8 mC
E
++q
E
4000 N
-9
4000 N
2 x 10 C
F
E
q
 
+2 nC
r
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.
++
++
+
+
++Q
.
r
P
Consider a test charge +q placed
at P a distance r from Q.
The outward force on +q is:
The electric field E is therefore:
2
F kQq r
E
q q
  2
kQ
E
r

++q
F
2
kQq
F
r

+
+
++
+
+
+
+Q
.
r
P
E
2
kQ
E
r

What is the electric field intensity E at
point P, a distance of 3 m from a negative charge of –
8 nC?
.
r
P
-Q
3 m
-8 nC
E = ? First, find the magnitude:
2
2
9 -9Nm
C
2 2
(9 x 10 )(8 x 10 C)
(3 m)
kQ
E
r
 
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
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.
+
- q1
q2
q3 -
A
E1
E3
E2
ER
Vector Sum:
E = E1 + E2 + E3
Directions are based
on positive test charge.
Magnitudes are from:
2
kQ
E
r

++
++
+
+
++Q
--
-- -
-
---Q
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.
Field lines go away from positive charges and
toward negative charges.
Assailment & presentation
topics

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Electric field gauss law

  • 1. A field is defined as a property of space in which a material object experiences a force. .P Above earth, we say there is a gravitational field at P. Because a mass m experiences a downward force at that point. No force, no field; No field, no force! m F The direction of the field is determined by the force.
  • 2. Consider points A and B above the surface of the earth—just points in space. Q:What is the similarities and dissimilarities between electric field and gravitational field? The field at points A or B might be found from: F g m If g is known at every point above the earth then the force F on a given mass can be found. The magnitude and direction of the field g is depends on the weight, which is the force F.  A  B F F
  • 3. 1. Now, consider point P a distance r from +Q. 2. An electric field E exists at P if a test charge +q has a force F at that point. 3. The direction of the E is the same as the direction of a force on + (pos) charge. E 4. The magnitude of E is given by the formula: N ; Units C F E q  Electric Field ++ ++ + + ++Q .P r +q F +
  • 4. E Electric Field ++ ++ + + ++Q . r 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. E Electric Field ++ ++ + + ++Q . r ++q --q F F Force on +q is with field direction. Force on -q is against field direction.
  • 5. 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. Force on +q is with field direction. Force on -q is against field direction. E Electric Field . r ++q F -- -- - - ---Q E Electric Field . r --q F -- -- - - ---Q
  • 6. 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 N ; Units C F E q        The direction of E at a point is the same as the direction that a positive charge would move.
  • 7. A +2 nC charge is placed at a distance r from a –8 mC charge. If the charge experiences a force of 4000 N, what is the electric field intensity E at point P? Electric Field . -- -- - - ---Q P First, we note that the direction of E is toward –Q (down). –8 mC E ++q E 4000 N -9 4000 N 2 x 10 C F E q   +2 nC r 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. ++ ++ + + ++Q . r P Consider a test charge +q placed at P a distance r from Q. The outward force on +q is: The electric field E is therefore: 2 F kQq r E q q   2 kQ E r  ++q F 2 kQq F r  + + ++ + + + +Q . r P E 2 kQ E r 
  • 9. What is the electric field intensity E at point P, a distance of 3 m from a negative charge of – 8 nC? . r P -Q 3 m -8 nC E = ? First, find the magnitude: 2 2 9 -9Nm C 2 2 (9 x 10 )(8 x 10 C) (3 m) kQ E r   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
  • 10. 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. + - q1 q2 q3 - A E1 E3 E2 ER Vector Sum: E = E1 + E2 + E3 Directions are based on positive test charge. Magnitudes are from: 2 kQ E r 
  • 11. ++ ++ + + ++Q -- -- - - ---Q 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. Field lines go away from positive charges and toward negative charges.