Transcript Charge and electric flux

Charge and electric flux
Charge & Electric flux
• In last chapter we asked the question "given a
charge distribution what is the electric field
produced by the distribution at a point P
• The answer can be found by representing the
distribution as an assembly of point charges
each of which produce an electric field E
which is given by the equation
Charge & Electric flux
• There is an alternative relationship between
charge distribution and electric field
• If electric field pattern is known in the in a
given region can we determine about the
charge distribution in that region?
Charge & Electric flux
• Imagine a box as shown
in fig.
• It has unknown amount
of charge
• We refer to box as a
closed surface
• How can we determine
how much electric
charge lies within the
box?
Charge & Electric flux
• Knowing that a charge
distribution produces an
electric field and that an
electric field exerts a
force on a test charge.
• We move a test charge q0
around the vicinity of the
box
• By measuring the force F
experienced by a test
charge at different point
Charge & Electric flux
• We make a three
dimensional map of the
electric field E=F/q0
outside the box
Charge & Electric flux
• To determine the
contents of a box we
actually measure E on
the surface of the box
• Fig. a shows a single +ve
point charge inside the
box
• Fig b. shows there are
two such charges
Charge & Electric flux
• The field pattern on the
surfaces of the boxes
are different in detail
but in both cases the
electric field points out
of the box
Charge & Electric flux
• Fig c show and fig. d
shows cases with one
and two –ve point
charges respectively
inside the box
• Again the details of E on
the surface of the box
are different , but in
both cases the field
points into the box
Charge & Electric flux
• The fig. in which electric field pointed outward
of the surface the electric flux is outward
• In fig. c in which the electric field points into
surface. The electric flux is inward
• Relationship : +ve charge inside the box goes
with an outward flux through the box’s
surface and –ve charge inside goes with an
inward flux
Charge & Electric flux
• What happened if there is
zero charge inside the
box?
• As shown in the fig. a
there is no E so there is
no electric flux inside or
outward
• In fig b. box containing
one +ve charge and one –
ve charge inside the box
with same magnitude
Charge & Electric flux
• The net charge inside
the box is zero. So there
is no electric flux inside
the box
Charge & Electric flux
• The box is again empty
however there is a
charge present outside
the box the box has
been placed with one
end parallel to a
uniform charged infinite
sheet which is produced
an electric field
perpendicular to the
sheet
Charge & Electric flux
• One end of the box E
points into the box on
the opposite end E
points out the box
• The inward electric flux
exactly compensates for
the outward flux on the
other part there is no
net flux in the surface
and there is no net
charge
Charge & Electric flux
• There is a relationship between magnitude of
the net charge inside the closed surface and
strength of the net flow of E over the surface
• This suggest that electric flux through the
surface of the box is directly proportional to
the magnitude of the net charge enclosed by
the box
Charge & Electric flux
• We have seen that there is relationship
between the net amount of charge inside a
closed surface and the electric flux through
the surface
• Flux through a closed surface depends on the
sign of the enclosed charge
• Charge outside the surface do not give a net
electric flux through the surface.
Charge & Electric flux
• The net electric flux is directly proportional to
the amount of the charge enclosed within the
surface but is otherwise independent of the
size of the closed surface
Calculating Electric flux
• Electric flux through a surface is a description
of whether the electric field E points into or
out of the surface
• The net electric flux is directly proportional to
the net electric charge inside the surface
Calculating Electric flux
• Fig. shows a fluid flowing
from left to right. Let’s
examine the volume flow
rate dV/dt through the wire
rectangle with area A when
the area A is perpendicular
to flow of velocity V and the
flow is the same at all
points in the fluid the
volume flow rate dV/dt is
the area A multiplied by
the flow speed v
• dV/dt = vA
Calculating Electric flux
• When the rectangle is
tilted at an angle
• So that face is not
perpendicular to V
• This area which is
outlined in red and
labeled A in fig. b is the
projection of area A
onto a surface
perpendicular to V
Calculating Electric flux
• Thus the volume flow
rate through A is
• dV/dt = vAcosφ
• if φ = 90, dv/dt =0 the
wire is edge on the flow
and no fluid pass
through the rectangle
Calculating Electric flux
• Also v cosφ is the
component of the
vector v perpendicular
to the area. Calling this
component v we can
write the volume rate
as
Calculating Electric flux
• Using the analogy between electric field and
fluid flow, we can define the electric flux in the
same way as we have just defined the
volume flow rate of a fluid. We simply replace
the fluid velocity v by the electric field E
• The symbol that we use for electric flux is ΦE
Calculating Electric flux
• Consider a flat area A
perpendicular to a
uniform electric field E
as in the fig.
• We define the electric
flux through this area to
be the product of the
field magnitude E and
the area A
• ΦE = EA
Calculating Electric flux
• Roughly speaking we can
picture the ΦE in terms
of the field lines passing
through A. increasing the
area means that more
lines of E pass through
the area, increasing the
flux
• Stronger field mean more
closely spaced lines of E
and therefore more lines
per unit area so again the
flux increases
Calculating Electric flux
• If the area A is flat but
not perpendicular to
the field E then fewer
lines pass through it
• In that case electric flux
is given by
• ΦE = EAcosΦ
Calculating Electric flux
• With a closed surface we will always choose
the direction of n to be outward we will speak
the flux out of a closed surface thus we called
outward flux
• Corresponding to a +ve value of ΦE and what
we called inward flux corresponding to –ve
value of ΦE
Calculating Electric flux
• What happened if the electric field is not
uniform but varies from point to point over
the area A
• Or what happened if A is part of a curved
surface?
• Then we divided the A into many small
elements dA each of which has a unit vector n
perpendicular to it and a vector area
• dA=ndA
Calculating Electric flux
• In this case we calculate the electric flux ΦE
Gauss’s Law
• Gauss’s law says that the total electric flux in
any closed surface is proportional to the total
electric charge inside the surface
• We’ll start with the field of a single +ve point
charge q. the field lines radiate out equally in
all direction. We place this charge at the
center of an imaginary spherical surface with
radius R. The magnitude E of the electric field
at every point on the surface is given by
Gauss’s Law
• At each point on the surface E is perpendicular
to the surface and its magnitude is the same
at every point . The total electric flux is just
the product of the field magnitude E and the
total area A = 4πR2 of the sphere
• ΦE = EA = 1/4πЄ0 .q/ R2 (4πR2) = q/Є0
• The flux is independent of the radius R of the
sphere. It depends only on the charge q
enclosed by the sphere
Gauss’s Law
• We can divide the
entire irregular surface
into element dA
compute the electric
flux EAcosΦ for each
and sum the result by
integrating
Gauss’s Law
• This equation holds for any surface of any
shape and size, provided only that it is a
closed surface enclosing the charge q. the
circle on the integral sign reminds us that
integral is always closed surface
Gauss’s Law
• Now comes the final step in obtaining the
general form of gauss’s law.
• Suppose the surface encloses not just one
point charge q but serval charges
q1,q2,q3…qn
• The total electric field E at any point is the
vector sum of the E field of the individual
charges
Gauss’s Law
• Let Qenc be the total charge enclosed by the
closed surface Qenc = q1+q2+a7q3…..
• Also let E be the total field at the position of
the surface area element dA and let E be its
component perpendicular to the plane of that
element. Then we can write general
statement of gauss’s law