Download There are only two charges, positive and negative.

There are only
two charges,
positive and
negative.
The unit of charge is
the coulomb (C). The
charge of an electron
(or a proton) is
-19
e = 1.60 x 10 C.
The smallest amount of
charge that has been
found is e, so any charge
of magnitude q is an
integer multiple of e.
q = Ne, where N is an
integer.
Ex. 1 - How many
electrons are
there in one
coulomb of
negative charge?
Friction can
transfer electric
charge, usually
by moving
electrons.
Law of conservation of
electric charge - During
any process, the net
electric charge of an
isolated system
remains constant.
Like charges repel,
unlike charges attract.
Substances that readily
conduct electric charge
are called electrical
conductors.
Materials that
conduct electric
charge poorly are
called electrical
insulators.
When one object is
given a net electric
charge by placing
it in contact with a
charged object it is
called charging by
contact.
If a charged object is held
close to another object,
and the second object is
temporarily grounded before
the first object is removed;
the second object is left
charged opposite the initial
charge.
This is charging by induction.
Charging by contact makes
both objects the same
charge. Charging by
induction will make the two
objects opposite charges.
When point charges q1
and q2 are separated
by distance r, the force
between them is
calculated by
Coulomb’s Law.
F = k•q1•q2/
2
r
k is a proportionality
constant whose value is
9
2
2
k = 8.99 x 10 N•m /C
The force is attractive
if the charges are
opposite, and
repulsive if the
charges are the
same.
k is often expressed in
terms of ε0, by writing
k = 1/(4πe0).
ε0 is called the permittivity
of free space and has a
value of ε0 = 1/(4πk) =
-12
2
2
8.85 x 10 C /(N•m ).
Ex. 2 - Two objects, whose
charges are +1.0 and -1.0 C,
are separated by 1.0 km.
Find the magnitude of the
attractive force that either
charge exerts
on the other.
Ex. 3 - In the Bohr model
of the hydrogen atom, the
electron (-e) is in orbit
about the nuclear proton (+e) at
a radius of r = 5.29 x 10-11 m.
Determine the speed of
the electron assuming
the orbit to be circular.
When more than two point
charges are involved, the
force on any one charge
is the vector sum of the
forces due to each
individual charge.
Ex. 4 - Figure 18.14a
shows three point charges
that lie along the x axis in
a vacuum. Determine the
magnitude and direction
of the net electrostatic
force on q1 .
Ex. 5 - Figure 18.15a
shows three point charges
that lie in the x,y plane in
a vacuum. Find the
magnitude and direction
of the net electrostatic
force on q1.
A small point charge,
called a test charge,
may be used to
determine the extent to
which the surrounding
charges generate a
force.
Ex. 6 - The positive test charge in
Figure 18.16 is q0 = +3.0 x 10-8 C
and experiences a force
F = 6.0 x 10-8 N in the direction
shown in the drawing. (a) Find the
force per coulomb that the test
charge experiences. (b) Using the
result of part (a), predict the force
that a charge of +12 x 10-8 C would
experience if it replaced q0.
The electric field E that exists
at a point is the electrostatic
force F experienced by a
small test charge q0 placed
at that point divided by the
charge itself:
E = F/q0
The electric field is a
vector, and its direction
is the direction of of the
force on the charge.
The unit is the newton
per coulomb (N/C).
The surrounding
charges are what
creates an electric
field at a given
point.
Ex. 7 - In figure 18.18a the
charges on the two metal spheres
and the ebonite rod create an electric
field E at the spot indicated. The field
has a magnitude of 2.0 N/C and is
directed as in the drawing. Determine
the force on a charge placed at that
spot, if the charge has a value of
(a) q0 = +18 x 10-8 C and
(b) q0 = -24 x 10-8 C.
Multiple electric fields
add in the same way
as vectors. Vector
addition can be used
to find the net field at
a particular point.
Ex. 9 - There is an
isolated charge of
q = +15 µC in a vacuum.
Using a test charge of
q0 = +0.80 µC, determine
the electric field at point P,
which is 0.20 m away.
From Coulomb’s law, the
force exerted on a test
charge q0 by a charge q is
2
F = k•q•q0/ r . Since E = F/q0,
E is also equal to k•q•q0/ r2
divided by q0. q0 cancels out,
and we are left with
2
E = k•q/ r .
2
r
E = k•q/ shows
that the electric
field does not
depend on the
test charge.
2
r
E = k•q/
If q is positive, then E
is directed away from q.
If q is negative, E is
directed toward q, since a
negative charge attracts a
positive charge.
Ex. 9 - Two positive point
charges, q1 = +16 µC and
q2 = +4.0 µC, are separated
in a vacuum by a distance of
3.0 m. Find the spot on the
line between the charges
where the net electric field is
zero.
A parallel plate capacitor
has two plates of different
charge with a space
between them. The
charges are distributed
uniformly over each plate.
The electric field
points from the
positive plate to the
negative and is
perpendicular to
both.
Using Gauss’ law, the
electric field has a
magnitude of:
E = q/ ε0A.
E = q/ ε0A = σ/ ε0
σ denotes charge
per unit area
(σ = q/A) and is
called charge
density.
In a parallel plate capacitor, the
field has the same value at all
places between the plates. The
field does not depend on the
distance from the charges, as it
does in a field created by an
isolated point charge.