18.6 The Electric Field
DEFINITION OF ELECRIC FIELD
The electric field that exists at a point is the electrostatic force experienced by a (small) test
charge q0 placed at that point divided by the charge q0 itself:

 F
E=
qo
SI Units of Electric Field: newton per coulomb (N/C)
NOTE: It is the surrounding charges that create the electric field at a given point.
The effect of the test charge itself is NEVER included in this definition
1
18.6 The Electric Field
The positive charge experiences a force which
is the vector sum of the forces exerted by the
charges on the rod and the two spheres.
This test charge should have a small
magnitude so it doesn’t affect the other charge.
Example: A Test Charge
The positive test charge has a magnitude
of 3.0x10-8C and experiences a force
of 6.0x10-8N.
(a) Find the force per coulomb that the test
charge experiences.
(b) Predict the force that a charge of
+12x10-8C would experience if it
replaced the test charge.
(a)
F 6.0 × 10−8 N
= 2 .0 N C
=
−8
qo 3.0 × 10 C
(b)
F = (2.0 N C ) 12.0 ×10 −8 C = 24 ×10 −8 N
(
)
The force experienced by a test charge q0 is
proportional to the charge q0, and its
direction is reversed if the sign of q0 is
reversed. We can think of the presence of
the rod and spheres as setting up a “force
field” that acts on any test charge introduced
2
18.6 The Electric Field
Example An Electric Field
Leads to a Force
The charges on the two metal
spheres and the ebonite rod
create an electric field at the spot
indicated. The field has a
magnitude of 2.0 N/C.
Determine the force on the
charges in (a) and (b)
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18.6 The Electric Field
Example An Electric Field Leads to a Force
The charges on the two metal spheres and the ebonite rod create an electric
field at the spot indicated. The field has a magnitude of 2.0 N/C. Determine
the force on the charges in (a) and (b)
(a)
F = qo E = (2.0 N C )(18.0 × 10−8 C) = 36 × 10−8 N
(b)
F = qo E = (2.0 N C ) 24.0 ×10 −8 C = 48 ×10 −8 N
(
)
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18.6 The Electric Field
Electric fields from different sources
add as vectors.
 

E = EA + EB
Electrical forces add as vectors.
 

F = F01 + F02 (+ ...)



F01 
F02
E01 =
, E02 =
,...
q0
q0



→ F = q0 E01 + q0 E02 ( +...)


= q0 E01 + E02 ( +...)
[
]
But by definition :


F = q0 E
 

⇒ E = E01 + E02 ( +...)
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18.6 The Electric Field
q qo 1
F
E=
=k 2
r
qo
qo
The electric field does not depend
on the test charge.
Point charge q:
E=k
q
r2
The arrows show the magnitude
(length) and direction of the electric
field from the point charge shown at
the black points.
Here a positive charge is assumed,
and the field point directly away from
the charge
The directions are reversed for a
negative charge
Note this charge q is the source of
the electric field, and NOT the test
charge q0.
This is also referred to as “Coulomb’s Law”
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18.6 The Electric Field
Example: The Electric Fields from Separate
Charges May Cancel
Two positive point charges, q1=+16μC and q2=+4.0μC
are separated in a vacuum by a 3.0m. Find the spot
on the line between the charges where the net
electric field is zero.
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18.6 The Electric Field
Example: The Electric Fields from Separate Charges May Cancel
Two positive point charges, q1=+16μC and q2=+4.0μC are separated in a vacuum by a 3.0m.
Find the spot on the line between the charges where the net electric field is zero.
Ei = k
qi
ri
2
E1 = E 2
(16 × 10 C) = k (4.0 × 10 C)
k
−6
−6
(3.0m − d )2
(16 × 10−6 C)(3.0m − d )2 = (4.0 × 10−6 C)d 2
d2
Dropping units (assume d is in meters)
4( 3 − d ) 2 = d 2 → 2( 3 − d ) = d → 6 − 2 d = d
⇒ 6 = 3d
d = +2.0 m
You can also choose the " negative"
4( 3 − d ) 2 = d 2 → 2( 3 − d ) = − d
→ 6 − 2d = − d
⇒6=d
But d = 6m is to the right of q2
so it is an " extraneous solution"
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