Problem 1
. A water droplet of radius 0.015 mm remains hanging stationary in the air. If
the electric field of Earth is 100 N/C, how many excess electrons must the water droplet
have? In which direction the electric field of Earth must be directed for that problem to
make any sense?
First of all, if the droplet hangs stationary in the air, then it means electrostatic force and
gravitation force cancel each other. Also, since we are told that there are excess electrons
(negatively charged) in the droplet, the only way we can have electrostatic force pointing
up is to require electric field to be pointing DOWN.
The number of excess electrons is N=q/e (where e=

1.6x10
19
C, charge of the electron)
E
q+m
g=
0, Eq=

mg=
ρ
Vg=
ρ4/3 π
r
3
g
N=
ρ4/3 π
r
3
g /E e = 1000 kg/m
3
4/3
π
(0.015
.
10
3
m)
3
9.8 m/s
2
/ 100N/C (1.6x10
19
C)
=8.65x10
6
electrons.
(Kg/m
3
)m
3
m/s
2
/ (N/C) C = (kgm/s
2
)/N = dimensionless (N=kgm/s
2
as F=ma)
Problem 2.
An electric dipole is formed from two charges, +/q, spaced 1 cm apart. The
dipole is centered at the origin, oriented along the yaxis. The electric field strength at the
point (x,y)=(0cm, 10cm) is 360 N/C. What is the charge q? What is the electric field
strength at the point (x,y)=(10cm, 0cm)?
mg
F
E
=qE
E
Ne
(0,10)
(10,0)
+

E+
E
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
First let’s consider the electric filed at the point (0,10). This field may be represented as a
sum of two unequal components, one due to positive charge of the dipole, and another
due to negative charge of the dipole. They partially cancel each other, as they point in
different directions.
)
1
1
(
)
10
,
0
(
2
,
1
2
,
1
2
,
1
2
,
1

+
+


+
+

=
+
=
r
r
q
k
r
q
k
r
q
k
E
e
e
e
where r
1,+
=9.5 cm=0.095 m and r
1,
=10.5 cm=0.105 m
C
m
m
C
Nm
C
N
r
r
k
E
q
e
9
2
2
2
2
9
2
,
1
2
,
1
10
2
)
105
.
0
(
1
)
095
.
0
(
1
/
10
99
.
8
/
360
1
1
)
10
,
0
(


+
+
⋅
=

⋅
=

=
or 2 nC
And now we can just use the textbook example (turned by 90 degrees) with a=0.5 cm and
x=10 cm. (there was y in the example, but I will use x for the sake of consistency with
my own figure).
The electric field vector at the point (0,10) will be pointing down and
C
N
m
m
C
m
C
Nm
x
a
aq
k
E
e
y
/
8
.
179
)
)
005
.
0
(
)
1
.
0
((
10
2
005
.
0
2
/
10
99
.
8
)
(
2
2
/
3
2
2
9
2
2
9
2
/
3
2
2
=
+
⋅
⋅
⋅
⋅
=
+
=

+
NB: It is also an acceptable solution to use the simplified formula for x>>a (y>>a in the
textbook example)
As long as the units in the above equations are properly crossed, the review is considered
complete.
Problem 3
:
Two 10cm charged disks face each other, 20 cm apart. The left disk is
charged to 50 nC and the right disk is charged to 50 nC. What is the electric field (both
magnitude and direction) at the midpoint between the disks? What would be the force
(again, magnitude and direction) on a 1 nC test charge placed at the midpoint?
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 Zazubovits
 Electrostatics, Electric charge, disk

Click to edit the document details