A field is a way of conceptualizing and mapping the force that surrounds any object and acts on another object at a distance without apparent physical connection. For example, the gravitational field surrounding the earth (and all other masses) represents the gravitational force that would be experienced if another mass were placed at a given point within the field.
In the same way, the Coulomb force field surrounding any charge extends throughout space. Using Coulomb’s law, F=kq1q2/r2F=kq1q2/r2 size 12{F= { ital "kq" rSub { size 8{1} } q rSub { size 8{2} } } slash {r rSup { size 8{2} } } } {}, its magnitude is given by the equation
F=kqQ/r2F=kqQ/r2 size 12{F= { ital "kqQ"} slash {r rSup { size 8{2} } } } {}, for a point charge (a particle having a charge QQ size 12{Q} {}) acting on a test charge qq size 12{q} {} at a distance rr size 12{r} {} (see Figure 1). Both the magnitude and direction of the Coulomb force field depend on QQ size 12{Q} {} and the test charge qq size 12{q} {}.
To simplify things, we would prefer to have a field that depends only on QQ size 12{Q} {} and not on the test charge qq size 12{q} {}. The electric field is defined in such a manner that it represents only the charge creating it and is unique at every point in space. Specifically, the electric field EE size 12{E} {} is defined to be the ratio of the Coulomb force to the test charge:
E
=
F
q
,
E
=
F
q
,
size 12{E= { {F} over {q,} } } {}
(1)where FF size 12{F} {} is the electrostatic force (or Coulomb force) exerted on a positive test charge
qq size 12{q} {}. It is understood that
EE size 12{E} {} is in the same direction as
FF size 12{F} {}. It is also assumed that qq size 12{q} {} is so small that it does not alter the charge distribution creating the electric field. The units of electric field are newtons per coulomb (N/C). If the electric field is known, then the electrostatic force on any charge qq size 12{q} {} is simply obtained by multiplying charge times electric field, or
F
=
q
E
F
=
q
E
size 12{F=qE} {}
. Consider the electric field due to a point charge QQ size 12{Q} {}. According to Coulomb’s law, the force it exerts on a test charge
qq size 12{q} {} is
F=kqQ/r2F=kqQ/r2 size 12{F= { ital "kqQ"} slash {r rSup { size 8{2} } } } {}. Thus the magnitude of the electric field,
EE size 12{E} {}, for a point charge is
E
=
F
q

=
k

qQ
qr
2

=
k
Q
r
2
.
E
=
F
q

=
k

qQ
qr
2

=
k
Q
r
2
.
size 12{E= { {F} over {q} } =k { { ital "qQ"} over { ital "qr" rSup { size 8{2} } } } =k { {Q} over {r rSup { size 8{2} } } } } {}
(2)Since the test charge cancels, we see that
E
=
k
Q
r
2
.
E
=
k
Q
r
2
.
size 12{E=k { {Q} over {r rSup { size 8{2} } } } } {}
(3)The electric field is thus seen to depend only on the charge QQ size 12{Q} {} and the distance rr size 12{r} {}; it is completely independent of the test charge qq size 12{q} {}.
Calculate the strength and direction of the electric field EE size 12{E} {} due to a point charge of 2.00 nC (nanoCoulombs) at a distance of 5.00 mm from the charge.
Strategy
We can find the electric field created by a point charge by using the equation E=kQ/r2E=kQ/r2 size 12{E= { ital "kQ"} slash {r rSup { size 8{2} } } } {}.
Solution
Here Q=2.00×10−9Q=2.00×10−9 size 12{Q=2 "." "00" times "10" rSup { size 8{  9} } } {} C and r=5.00×10−3r=5.00×10−3 size 12{r=5 "." "00" times "10" rSup { size 8{  3} } } {} m. Entering those values into the above equation gives
E
=
k
Q
r
2
=
(
9
.
00
×
10
9
N
⋅
m
2
/C
2
)
×
(
2
.
00
×
10
−
9
C
)
(
5
.
00
×
10
−
3
m
)
2
=
7
.
20
×
10
5
N/C
.
E
=
k
Q
r
2
=
(
9
.
00
×
10
9
N
⋅
m
2
/C
2
)
×
(
2
.
00
×
10
−
9
C
)
(
5
.
00
×
10
−
3
m
)
2
=
7
.
20
×
10
5
N/C
.
alignl { stack {
size 12{E=k { {Q} over {r rSup { size 8{2} } } } } {} #
= \( 9 "." "00" times "10" rSup { size 8{9} } N cdot m rSup { size 8{2} } "/C" rSup { size 8{2} } \) times { { \( 2 "." "00" times "10" rSup { size 8{  9} } C \) } over { \( 5 "." "00" times "10" rSup { size 8{  3} } m \) rSup { size 8{2} } } } {} #
=7 "." "20" times "10" rSup { size 8{5} } "N/C" {}
} } {}
(4)
Discussion
This electric field strength is the same at any point 5.00 mm away from the charge QQ size 12{Q} {} that creates the field. It is positive, meaning that it has a direction pointing away from the charge QQ size 12{Q} {}.
What force does the electric field found in the previous example exert on a point charge of –0.250μC–0.250μC?
Strategy
Since we know the electric field strength and the charge in the field, the force on that charge can be calculated using the definition of electric field E=F/qE=F/q size 12{E= {F} slash {q} } {} rearranged to F=qEF=qE size 12{F= ital "qE"} {}.
Solution
The magnitude of the force on a charge q=−0.250μCq=−0.250μC size 12{q=  0 "." "250""μC"} {} exerted by a field of strength E=7.20×105E=7.20×105 size 12{E=7 "." "20" times "10" rSup { size 8{5} } } {} N/C is thus,
F
=
−
qE
=
(
0.250
×
10
–6
C
)
(
7.20
×
10
5
N/C
)
=
0.180 N
.
F
=
−
qE
=
(
0.250
×
10
–6
C
)
(
7.20
×
10
5
N/C
)
=
0.180 N
.
alignl { stack {
size 12{F= ital "qE"} {} #
size 12{ {}= \( "0" "." "250" times "10" rSup { size 8{"6"} } `C \) \( 7 "." "20" times "10" rSup { size 8{5} } `"N/C" \) } {} #
="0" "." "180"`N {}
} } {}
(5)Because qq is negative, the force is directed opposite to the direction of the field.
Discussion
The force is attractive, as expected for unlike charges. (The field was created by a positive charge and here acts on a negative charge.) The charges in this example are typical of common static electricity, and the modest attractive force obtained is similar to forces experienced in static cling and similar situations.