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Charge and electrical forces (Chapter 20)
Thermal energy was said to include potential energy
associated with the forces between the atoms and
molecules in the system. That was particularly
important for liquids and solids.
However the nature of those forces was not described
in much detail.
Here we will fill that gap. The forces between atoms
and molecules are primarily electromagnetic, which
means that they involve both electrical and magnetic
phenomena.
Here we first consider the electrical forces.
Many of the principles regarding electrical
forces were understood before the structure of
atoms and molecules was known.
However I will refer to the known electrical structure
of atoms and molecules as we go along. (See the book
for a careful discussion of how much you can understand
about this without knowing anything about atoms. )
Nevertheless we begin with some experiments done
by scientists in the 18th and 19th centuries to understand
electricity. (Names include Franklin and Coulomb.)
Conclusion so far: Electrical forces can attract or
repel. Like charges repel. Unlike charges attract.
It turns out that this can be understood if there are
just two kinds of charge, positive and negative. The charges
on the plastic rubbed with wool were chosen to be <0
long ago and we stick with that.
This experiment shows that the material with which
I rubbed the plastic gets a charge opposite (that is
>0) to the charge on the charged plastic rod.
It means that either something positive rubbed off
of the plastic unto the wool or something negative
rubbed off the wool onto the plastic. It turns out that
the second option is the correct one.
But when the something negative (which turns out
to be electrons) rubbed off it must have left something
positive behind.
Conservation of charge.
Before rubbing, there was no attraction or repulsion,
so both the wool and the plastic were charge neutral
(net charge zero). After rubbing, the plastic is
negatively charged and the wool is positively charged,
but it turns out that the total charge STILL ADDS UP
to zero. No charge was created or destroyed.
This is an example of CONSERVATION OF CHARGE, a
very general law of nature: No electrical charge is
ever created or destroyed.
Instead, in the rubbing experiment,
the neutrally charged wool and plastic each
contained equal amounts of positive and negative
charge. After rubbing some of the negative charge
in the wool was transferred to the plastic but the
total amount of charge remained zero.
We understand this in terms of atomic structure:
Atoms consist of a tiny nucleus ( 10-15 m) containing
almost all the mass surrounded by a cloud of
electrons with very little mass.
The electrons have negative
charge and are relatively
easy to remove. When the
plastic is rubbed, electrons in
the wool are detached and transferred
to the plastic, Leaving more
negative than positive charge
on the plastic and more positive
than negative charge on the wool.
The positive charges in the nucleus are the protons,
mentioned earlier and a remarkable fact is that the magnitude
of the electronic charge is exactly equal to the
magnitude of the proton charge, though they
have different signs. That makes electrically neutral atoms
and molecules possible.
Insulators and Conductors.
Now I do another experiment using a metal rod.
First I check that when an uncharged glass rod
comes up to it, there is no evidence of a force,
so the metal is charge neutral.
But when I charge the glass rod positively (but don't touch
the metal one) the metal is ATTRACTED to the
glass rod. How can that be if the metal is charge
neutral?
a. some charge was transferred from the glass
to the metal.
b.the charge on the metal moved around so the
end of the metal near the glass was negative
c. some charge was transferred from the metal to
the glass
d. the charge on the glass moved around so that
the charge on the glass near the metal was positive.
Answer is b.
I was careful not to touch the rods, making a and c
unlikely.
The same thing does not happen (at least not nearly
as much) if I approach a neutral glass rod with
a negatively charged plastic rod.
We conclude that the charges in metals MOVE AROUND
EASILY, unlike the charges in glass and plastic.
The charges which move around in metals are
the electrons. I will show some evidence for that
later.
Materials in which electrons move around easily
are called conductors.
Materials in which electrons don't move around easily
are called insulators.
What if I bring a plastic rod (negatively charged) up
to a neutral metal rod?
a. It will be repelled.
b. Nothing will happen.
c. It will be attracted.
c.
Now the electrons in the metal are pushed away
from the negatively charged plastic, leaving
a net positive charge in the part of the metal
near the plastic resulting in attraction.
Summary of some properties of electrical forces.
They can be attractive or repulsive.
charges can be of two signs, + and charges of the same sign attract
charges of opposite sign repel.
The forces are long range.
All materials contain both positive charges(protons in
the nuclei) and negative charges which exactly
balance in charge neutral material.
In conductors, electrons move easily (metals)
In insulators they don't.
Summary of some properties of electrical forces.
They can be attractive or repulsive.
charges can be of two signs, + and charges of the same sign attract
charges of opposite sign repel.
The forces are long range.
All materials contain both positive charges(protons in
the nuclei) and negative charges which exactly
balance in charge neutral material.
In conductors, electrons move easily (metals)
In insulators they don't.
Summary of some properties of electrical forces.
They can be attractive or repulsive.
charges can be of two signs, + and charges of the same sign attract
charges of opposite sign repel.
The forces are long range.
All materials contain both positive charges(protons in
the nuclei) and negative charges which exactly
balance in charge neutral material.
In conductors, electrons move easily (metals)
In insulators they don't.
If atoms and molecules are charge neutral,
how could the potential energy associated
with forces between atoms be electrical?
There is more than one way that this happens.
Let's start with water. Water
molecules look
a little like this:
There is a little
extra <0 charge
on the O making
the H's slightly
net charge >0.
When a lot of water molecules exist together in
liquid water the + ends (H) are attracted to
the - ends (O) giving a loose structure looking
like this
Quantitative characterization of the electrical forces.
We measure charge in coulombs. The charge of an
electron is
-1.6 x 10-19 Coulombs and is called -e
The charge of a proton is therefore + 1.6 x 10-19 Coulombs
Consider two point charges (very small, like a nucleus).
Let them have charges q1 and q2 and suppose that
the distance between them is r12 . Then the force which
1 exerts on 2 has magnitude
K|q1||q2|/r12
2
and is along the line between the points. The direction
is determined by the signs of the charges as we have
discussed. The value of K is 9.0 x 109 Nm2/C2 .
Calculation of forces due to Coulomb's law.
It is important to remember that force is
a vector, so when more than one point charge is
near and object the resulting force is the
VECTOR sum of the from the point charges.
Those forces may not all be in the same direction.
As an example, you are working out the force
that a water molecule exerts on a sodium atom
(in a simplified model) in homework 4.
Force due to a dipole:
The charges on the dipole are +q and -q as shown.
They are aligned along the y axis a distance
a apart. We take the origin of coordinates
to be 1/2 way between the dipole charges
and place a small ('test') charge qt on the
x axis a distance d away. What is the force on
the test charge?
a. Fx =0, Fy = 2Kqqt /d2
b. Fx =0,Fy=0
c. Fx = 2Kqqt d/(d2+a2/4)3/2 , Fy=0
d. Fx =0, Fy= -Kqqt a/(d2+a2/4)3/2
e. Fx =Kqqt /d2 , Fy = -Kqqt /d2
Same dipole. Put the test charge a distance
d away along the y axis, as sketched.
What is the force on the test charge now?
a. Fx =0, Fy = 2Kqqt /d2
b. Fx =0,Fy=0
c. Fx = Kqtq(1/(d-a/2)2 -1/(d+a/2)2) , Fy=0
d. Fx =0, Fy= Kqtq(1/(d-a/2)2 -1/(d+a/2)2)
e. Fx =Kqqt /d2 , Fy = -Kqqt /d2
Force due to a model of a CO2 molecule:
Force on test charge, as before, is
a. Fx =0, Fy = 3Kqqt /d2
b. Fx =0,Fy=0
c. Fx = Kqtq(1/d2 - d/(d2+a2/4)3/2 ), Fy=0
d. Fx =0, Fy= Kqtq(1/d2 - d/(d2+a2/4)3/2 )
e. Fx =Kqqt /d2 , Fy = -Kqqt /d2
For the CO2 molecule the force was
Fx = Kqtq(1/d2 - d/(d2+a2/4)3/2 ),
Fy=0
Assuming that q>0 and qt >0,
in which direction does the force point along the xaxis?
a. to the right, toward positive x
b.to the left, toward negative x
c. to the left for small d and to the right for large d
d. to the right for small d and to the left for large d
As an exercise, you can work out the field
on a test charge placed a distance d from the
center of the model CO2 molecule along the y- axis.
(d>a/2)
If the test charge qt is positive, which way will
that force point.
a. Along y in the positive y direction.
b. Along y in the negative y direction
c. Along x in the positive x direction
d. Along x in the negative x direction
e. negative for small d and positive for large d
along the y direction.
Electric field.
By making many such calculations with a test charge
at various points around a collection of charges
(such as a dipole or a quadrupole) one can find the
forces which a test charge WOULD EXPERIENCE if
it were at each point. All those forces are going to
be proportional to the magnitude of the test charge qt .
The ELECTRIC FIELD at each point in space around
the collection of charges is DEFINED to be the
force which a test charge at that point would
experience, divided by the value qt of the test charge.
You can see that the value of the electric field will
then be independent of qt . The electric field is
regarded as a property of the space surrounding the
collection of charges.
Consider a ring of charges, very close together,
on a ring of radius r=a. There
are N small charges q placed very close together
around the ring.
A test charge qt is placed
distance d away along the axis of the ring which
we define as the x axis.
What is the electric field there?
a. Ex =0, Ey = KNq/(d2+a2)
b. Ex =KNq/(d2+a2) ,Ey=0
c. Ex = KNqd/(d2+a2/4)3/2 , Ey=0
d. Ex =0, Ey= KNqa/(d2+a2/4)3/2
e. Ex =KNq /(d2 +a2), Ey = KNq/(d2+a2)
Many rings in a plane.
Now imagine many rings filling the yz plane with
a test charge a distance d along the x axis.
We call the radius of a ring r, with r varying from
nearly zero to nearly infinite for the different rings.
Suppose that the charges on the rings are arranged
so that the number of charges per unit area
Q/A (eg per millimeter squared ) is the same everywhere.
If the distance between rings is Δr this means that the
number of charges on each ring is qN(r)= (Q/A)2πr Δr
and the contribution of the ring of radius r to the
field a distance d away from the plane is
ΔEx= (Q/A)2πKr Δr d/(d2+r2)3/2
Now to get the field arising from the entire plane of
charge density Q/A, you have to add these contributions
from the rings all up.
Electric field from a sheet of charge with charge density
Q/A: When you do the sums d DROPS OUT
and you get
Total field from a plane of charge with charge density
Q/A is
Ex= (Q/A)2πK
Exercise: Show that this has the right units to be an
electric field.
Qualitatively this result can be understood like this:
when the test charge is close to the plane, the charges in
the plane which are close to the test charge contribute
a lot, while the ones far away contribute little.
when the test charge is farther away from the plane,
the charges closes to the test charge contribute less but
the charges far from the test charge contribute more.
The result is that the electric field ends up being the same.
Numerical demonstration that the field from a
plane of charge is independent of the distance from
the plane. Sum the fields from the rings.
For the nth ring, r=nΔr and
ΔEx= (Q/A)2πKn Δr2 d/(d2+(nΔr)2)3/2=
(Q/A)2πK [n(d/Δr)/(n2+(d/Δr)2)3/2 ]
Summing up all the contributions from the
different rings:
Ex= (Q/A)2πK[
0 +(d/Δr)/(1+(d/Δr)2)3/2 + 2(d/Δr)/(4+(d/Δr)2)3/2 +
3(d/Δr)/(9+(d/Δr)2)3/2 + ... ]
I summed up 500 rings with different values of
d/Δr using excel as shown next
Summary:
The electrical force which point charge 1 exerts on
a point charge 2 a distance r12 away has
magnitude K|q1||q2|/r122 , is along the line
between the charges and is attractive if the
charges have opposite sign and repulsive if they have the
same sign. Forces on a point charge due to other
point charges can be added vectorially to get the
total force.
The electric field arising from a collection of point charges
is the force on a test charge placed at any point in
space around the charges, divided by the charge of
the test charge. The electric field due to a plane of
charge with surface charge density Q/A is
(Q/A) 2πK independent of the distance from the plane.
Field lines.
Electric fields can be visualized either by
arrows at each point in space, pointing in the direction
of the field and of length proportional to the magnitude
of the field
OR
by field lines which are continuous through parts of
the space not occupied by point charges and
are tangent to the direction of the electric field at
each point along their length. If you choose a
number for the number of field lines per coulomb
starting out from each point charge, then the density
of field lines will be proportional to the electric
field and the field lines provide a convenient way
to visualize the field. I will illustrate wtih a few cases.
C.
+Q/A
-Q/A
A.
Electric field between around two capacitor plates
In this case, which is common in electric circuits and
also, approximately, in cell wall membranes as in
the Homework 4 group problem, we have two charged
plates with charges Q/A and -Q/A separated by some
distance d. What is the electric field above and below
the capacitor (eg at points A and C)?
a. (Q/A) 4πK above and -(Q/A) 4πK below
b. (Q/A) 2πK above and -(Q/A) 2πK below
c. zero above and below
d. (Q/A) 2πK above and (Q/A) 2πK below
C.
+Q/A
-Q/A
A.
Electric field between around two capacitor plates
In this case, which is common in electric circuits and
also, approximately, in cell wall membranes as in
the Homework 4 group problem, we have two charged
plates with charges Q/A and -Q/A separated by some
distance d. What is the electric field above and below
the capacitor (eg at points A and C)?
a. (Q/A) 4πK above and -(Q/A) 4πK below
b. (Q/A) 2πK above and -(Q/A) 2πK below
c. zero above and below
d. (Q/A) 2πK above and (Q/A) 2πK below
OK now what is the field between the plates
(like at point B)?
+Q/A
B.
-Q/A
a. 0
b. (Q/A) 2πK
c. -(Q/A) 2πK
d. (Q/A) 4πK
e. -(Q/A) 4πK
Dipole between capacitor plates:
+Q/A
a
d
-q
q
-Q/A
The angle beween the bond between the charges
and the horizontal is θ. a is fixed and the dipole
is free to rotate about its center point.
What is the torque on the dipole?
a. (qQ/A) 4πK a cos θ clockwise
b. (qQ/A) 4πK a sin θ clockwise
c. (qQ/A) 4πK a cos θ counter clockwise
d. (qQ/A) 4πK a sin θ
counter clockwise
Massive charge moving between capacitor plates.
Suppose a negative charge -q on a point particle of mass m
is released at zero velocity from the bottom of the capacitor
plate. (This is what happens repeatedly in the electron
gun you have been using in the lab this week. Vacuum
in the plate, E=(Q/A)4πK.)
+Q/A
d
-q .
-Q/A
What happens to it after release?
a. sticks to the bottom plate
b. accelerates to a velocity of √2qEd/m
at the top plate.
c. moves toward the top plate
at constant speed √qEd/m
d. accelerates diagonally to the right to velocity √qEd/m
Electrical Potential Energy
The particle gained kinetic energy
(1/2)mv2 = (1/2)m(√2qEd/m)2 =qEd in passing from the bottom
to the top plate.
Energy is conserved so we infer that the particle
had an amount qEd more potential energy
when it was at the bottom than when it was at
the top. This is a new kind, electrical, of potential
energy, associated with the presence of an electric
field.
We define the voltage difference between to points
in space as the difference in the electrical potential
energy of a test charge at those two points, divided
by the test charge (including the sign). In the case
above that means that the voltage difference between
the top and the bottom of the capacitor plate is Ed .
Massive charge moving between capacitor plates.
Suppose a negative charge -q on a point particle of mass m
is released at zero velocity from the bottom of the capacitor
plate. The final speed just as it reaches the top plate
is √2qEd/m . Tne voltage between the plates is
500V and the particle is an electron with q=e=1.6 x 10-19 C
and m=9.1 x 10-31 kg. What is that speed in m/s? d=5cm.
+Q/A
d
-q .
a.2.96 x 106 m/s
b. 5.93 x 107 m/s
c.1.33 x 107 m/s
d. 7.45 x 10-5 m/s
-Q/A
B.
d
A.
+Q/A
-Q/A
In terms of the variables shown, what is the
voltage difference between points A and B?
a. VA-VB = d4πKQ/A
b.VA-VB = -d4πKQ/A
c.VA-VB = d2πKQ/A
d.VA-VB = -d2πKQ/A
e. 0
B.
d
C.
+Q/A
45o
A.
-Q/A
In terms of the variables shown, what is the
voltage difference between points A and C?
a. VA-VC = d(4πKQ/A)(1/√2 )
b.VA-VC = d(-4πKQ/A)(1/√2)
c.VA-VC = d4πKQ/A
d.VA-VC = -d4πKQ/A
e. 0
B.
d
C.
+Q/A
45o
A.
-Q/A
In terms of the variables shown, what is the
voltage difference between points B and C?
a. VB-VC = d(4πKQ/A)(1/√2 )
b.VB-VC = d(-4πKQ/A)(1/√2)
c.VB-VC = d4πKQ/A
d.VB-VC = -d4πKQ/A
e. 0
d
B.
C.
D.
E.
+Q/A
A.
-Q/A
In terms of the variables shown, what is the
voltage difference between points D and E?
a. Vd-Vf =d (2πKQ/A
b.Vd-VE = d(-2πKQ/A)
c.VD-VE = d4πKQ/A
d.VD-VE = -d4πKQ/A
e. 0
)
Massive charge moving between capacitor plates.
Suppose a POSITIVE charge +q on a point particle of mass m
is released at zero velocity from the bottom of the capacitor
plate.
+Q/A
d
+q .
-Q/A
What happens to it after release?
a. sticks to the bottom plate
b. accelerates to a velocity of √2qEd/m
at the top plate.
c. moves toward the top plate
at constant speed √qEd/m
d. accelerates diagonally to the right to velocity √qEd/m
Massive charge moving between capacitor plates.
Suppose the charge is a negatively charged
electron with charge -e = -1.6 x 10-19 C, mass m =9.1 x 10-31 kg.
The voltage V is 500 volts but the
sign of the charges on the plates is reversed from
the preceding situation. The electron has entered the capacitor
from the bottom with upward velocity v = 107 m/s. d=5cm.
-Q/A
d
-e .
v
+Q/A
What happens to it next?
a. electron starts back down through the hole it came in.
b. goes to the top gaining energy qx voltage difference.
c. goes toward the top plate losing kinetic energy
and gaining potential energy
d. goes toward the top plate gaining kinetic energy
and losing potential energy.
Massive charge moving between capacitor plates.
Suppose the charge is a negatively charged
electron with charge -e = -1.6 x 10-19 C, mass m =9.1 x 10-31 kg.
The voltage V is 500 volts but the
sign of the charges on the plates is reversed from
the preceding situation. The electron has entered the capacitor
from the bottom with upward velocity v = 106 m/s. d=5cm.
-Q/A
d
-e .
v
+Q/A
Does the electron get to the top plate?
a. yes
b. no
Massive charge moving between capacitor plates.
Suppose the charge is a negatively charged
electron with charge -e = -1.6 x 10-19 C, mass m =9.1 x 10-31 kg.
The voltage V is 500 volts but the
sign of the charges on the plates is reversed from
the preceding situation. The electron has entered the capacitor
from the bottom with upward velocity v = 107 m/s. d=5cm.
-Q/A
d
-e .
v
+Q/A
How far does it get?
a. 5cm, almost to top plate
b. hits the top plate
c. goes up 2.84 cm and starts back down.
d. goes up 2.84 cm and stays there.
y
+500V
-e,m .
v
5cm
0V
x
Same plates, but electron enters from the side.
Now what happens?
a. keeps on moving at constant velocity
the right.
b. velocity starts pointing down but speed
stays the same
c. x component of velocity stays the same and
y component starts getting more negative.
d.. x component of velocity stays the same and
y component starts getting more positve.
e. magnitude of x component of velocity
decreases and magnitude of y component
of velocity increases.
y
+20V
-e,m .
v
5cm
2cm
0V
20 cm
x
Same plates, but electron enters from the side
and the voltage is lower.
If it enters from the left with speed 107 m/s does
it emerge from between the plates at the right, or
does it hit the top plate?
a. emerges
b. hits the top plate.
I have drawn two imaginary planes between the
plates, one at height h1 and another at height h2 .
In terms of the variables given, what is the electrical
potential difference between the upper and lower
planes?
a. 500volts x (h2/h1)
b. 500volts x (h1/h2)
c. 500volts x (h2/d - h1/d)
d. 500 volts
e. depends on where the two points on the two planes are.
Electrical potential and electrical field.
In the capacitor we have been considering
,the electric field is perpendicular to the plates
everywhere.
We saw that points at the same distance from
the plates were all at the same electrical potential.
We can draw horizontal surfaces in the capacitor at
various heights. The electrical potential is the same
along each horizontal surface. In this case at least,
the potential difference ΔV between two adjacent planes
was
ΔV=
200volts x (h2/d -h1/d)= (200Volts/d)(h2 -h1) =-E (h2 -h1)
=- EΔh
so the electric field is E= - ΔV/Δh
Summary:
Inside the capacitor, at least, another way to
describe the electric field is by the relation
E= - ΔV/Δh
where the numerator is the difference between the
potentials on two equipotential surfaces and the
denominator is the distance between the surfaces,
measured along a line perpendicular to the surfaces.
The resulting electric field is perpendicular to
the equipotential surfaces.
These features turn out to be generally true of the
relation between equipotential surfaces and electric
fields as long as Δh is small .
I illustrate for one other case, a point charge
Consider a point charge with charge q.
What is the geometrical form of the equipotential
surfaces around this point charge?
a. cylinders
b. cubes
c. spheres
d. tetrahedrons
e. dodecahedrons
OK spheres. You can understand this by imaginging
moving a test charge along the a sphere around the
charge q. The electric field is perpendicular to the
surface, so moving the test charge along the
surface requires no work. Therefore the potential
is the same everywhere on the sphere's surface.
Now consider two spheres, both centered on q and
close together but separated by radial distance
Δr.
Move a test charge qt from the inner to the outer one.
If both charges are >0, the test charge is being pushed
away, so the change in potential energy is negative
and equal to -qt EΔr =qt ΔV giving the same relation
E=- ΔV/Δr . It is important that Δr is small in this case
because the field depends on r.
Potential around a point charge.
We can use the relation
E = -ΔV/Δr
to get an equation for the potential energy around
a point charge, as follows
ΔV=-EΔr =-KqΔr/r2
I claim that V(r)= Kq /r will give this relation if Δr <<r:
ΔV =Kq(1/(r+Δr)-1/r)=Kq((r-(r+Δr))/(r+Δr)r))=
-KqΔr/((r+Δr)r) ≈-KqΔr/r2
So we can take V(r)= Kq /r to be the electrical potential
around a point charge q.
Equipotential surfaces are drawn at separations corresponding
to a fixed potential difference (eg 0.1 volt) so that they
are evenly spaced between capacitor plates but
closer together near a point charge than far from it.
B.
+Q/A
A.
+Q/A
C.
One more question. In the situation shown,
what is the electric field at A, at B and at C?
at A
at B
at C
a. 4ΠKQ/A up
0
0
b. 4ΠKQ/A down
0
0
c. 2ΠKQ/A down 2ΠKQ/A up
2ΠKQ/A down
d.
0
4ΠKQ/A up
4ΠKQ/A down
e.
0
2ΠKQ/A up
2ΠKQ/A down
Electric Fields and potentials in and around
metal electrodes.
We define a metal as a material in which
charge (actually electrons) moves easily.
Later we consider moving charges in
metals. Now we consider situations
in which no charges are moving, called
equilibrium in your book.
First conclusion: Inside a metal in equilibrium,
the electric field is zero everywhere.
At any point inside the metal there are
mobile charges (electrons) If there were
an electric field there, then those charges would
be accelerating. But we are supposing that metal
is in equilibrium and no charges are moving.
Therefore the field must be zero.
Electric Fields and potentials in and around
metal electrodes.
We define a metal as a material in which
charge (actually electrons) moves easily.
Later we consider moving charges in
metals. Now we consider situations
in which no charges are moving, called
equilibrium in your book.
First conclusion: Inside a metal in equilibrium,
the electric field is zero everywhere.
At any point inside the metal there are
mobile charges (electrons) If there were
an electric field there, then those charges would
be accelerating. But we are supposing that the
metal is in equilibrium and no charges are moving.
Therefore the field must be zero.
A charged metal in equilibrium:
In a charged piece of metal in equilibrium,
the charges are all on the surface.
Why would that be? The argument that the
electric field inside the metal must be zero
is still valid. But if there were any (net)
charge anywhere inside, it would cause
a field. Therefore there cannot be any net
charge anywhere inside.
On the surface, it's different: The surface
provides a barrier to any charge there
so a charge there could experience an
electrical force pushing it out of the
metal but be held back by the surface.
Therefore any excess charge must
be at the surface.
The last question we did realised these points in
the case of a thin flat plate of metal. Just regard the
sheets of charge as the charge on the metal surface
and the region between the sheets as filled with metal.
B.
+Q/A
A.
+Q/A
C.
One more question. In the situation shown,
what is the electric field at A, at B and at C?
at A
at B
at C
d.
0
4ΠKQ/A up
4ΠKQ/A down
So we said that the charge on a metal in
equilibrium is all at its surface and that
the electric field inside is zero everywhere.
What about the electric field just at the surface
of the charged metal?
At the surface, the field does not have to be zero,
because there is another force on any charge there,
due to the presence of the surface, which is holding the
charge inside the metal and is not due to the
electric field. What can we say about the electric
field there?
The force holding the surface charges in are
perpendicular to the metal surface and
point in toward the metal. Thus the electric field
must also be perpendicular to the surface and
the resulting force must point out.
In our example of a large flat plate of charged metal,
that is what happened.
Let's look at another case, a charged metal
sphere.
If the field point is a distance r away from the
center of the sphere which is much farther
than the radius of the sphere then the field will
obviously look like the field from a point charge
of charge Q ie
Er= KQ/r2
Very close to the surface of the sphere, the
sphere will look like 2 plates of surface charge density
Q/4Πr2 and the field will be
Er= 4ΠK Q/A = KQ/r2 which is of the same form!
It turns out to be the same form in between too.
Conclusion: The field around a charged
metal sphere (but outside the sphere)
is the same as the field from a point charge with
the same charge as the charge on the sphere and
located at the center of the sphere.
Consider a solid metal sphere of radius a
charged with charge Q. What is the electric field
and electrostatic potential as a function of the distance
r from the center of the sphere?
r< a
r>a
Er
V
Er
V
a.
KQ/r2
KQ/r
KQ/r2
KQ/r
b.
0
0
KQ/r2
KQ/r
c.
0
KQ/a KQ/r2
KQ/r
d.
KQ/a2 KQ/a KQ/r2
KQ/r
e.
KQ/a2
0
KQ/r2
KQ/r