Good morning. Today we are going to review conductors in electrostatic equilibrium for AP
Physics C: Electricity and Magnetism. ♪ Flipping Physics ♪
Conductors are materials where the electrons are free to move rather easily, however,
when they are in electrostatic equilibrium, this means the charges in the object are stationary.
I thought equilibrium meant the net force acting on the object or system was zero. That means the
acceleration is zero, which means it could be at rest or moving at a constant velocity,
right? So, that means the charges do not have to be stationary;
they could be moving at a constant velocity. Electrostatics is the study of electromagnetic
phenomena that occur when there are no moving charges.
So, by definition, the charges in conductors in electrostatic equilibrium need to be stationary.
Right. There are four
things you need to remember about conductors in electrostatic equilibrium. The first is that the
electric field inside a conductor in electrostatic equilibrium equals zero. Billy, what would happen
if the electric field inside a conductor in electrostatic equilibrium were not equal to zero?
If there were an electric field inside a conductor in electrostatic equilibrium,
then the charges in the conductor would experience an electrostatic force
equal to the charge times the electric field, the charges would experience an acceleration,
and the conductor would not be in electrostatic equilibrium. Therefore,
there must be zero electric field inside a conductor in electrostatic equilibrium.
Well done, Billy. Notice that this means that anything inside a conductor in electrostatic
equilibrium is shielded from all external electric fields. This is called electrostatic shielding and
can be quite useful. The second thing you need to remember about conductors in electrostatic
equilibrium is that all excess charges are located on a surface of the conductor. To prove this,
let’s do an example with a solid conducting sphere. Let’s draw a Gaussian surface in the
shape of a sphere which is concentric with the solid spherical conductor but with a radius that
is a little bit smaller than the radius of the solid spherical conductor. I’ve exaggerated the
difference in spherical radii in the illustration so you can actually see the Gaussian sphere,
however, remember the Gaussian sphere is only a tiny bit smaller than the solid conducting sphere.
Bobby, please use Gauss’ law to prove the excess charges in the solid conducting sphere
are concentrated on the surface of the solid conducting sphere.
Okay. Gauss’ law states that the electric flux through a closed surface equals the
closed surface integral of the dot product of the electric field through the closed surface and the
infinitesimal area vector, dA. And it equals the charge enclosed in the Gaussian surface divided by
the permittivity of free space. Oh, but we already know the electric field inside the Gaussian sphere
equals zero, you just showed that, which means the electric flux through the Gaussian sphere
equals zero, which means the charge enclosed in the Gaussian sphere equals zero. Therefore,
all of the excess charges must be outside the Gaussian sphere and therefore all the excess
charges must be on the surface of the conductor. Thank you, Bobby. The third thing you need to
remember about conductors in electrostatic equilibrium is that the electric field which
is just outside the surface of a conductor in electrostatic equilibrium is perpendicular to
the surface and equals the local surface charge density divided by permittivity of free space.
“Local surface charge density” means the surface charge density of the conductor
in electrostatic equilibrium right near the location of the electric field.
Let’s start with the direction of the electric field at the surface of the conductor
in electrostatic equilibrium. It is always perpendicular to the surface.
If the electric field had a component parallel to the surface of the conductor, the charges
would have an electrostatic force acting on them parallel to the surface which would cause the
charges to accelerate, and the conductor would no longer be in electrostatic equilibrium. Therefore,
the electric field at the surface of a conductor in electrostatic equilibrium must be perpendicular
to the surface. Also, because equipotential surfaces are always perpendicular to the electric
field, the surface of a conductor in electrostatic equilibrium must be an equipotential surface.
You could also determine the electric potential difference between two points, A and B,
for example, on the surface of a conductor in electrostatic equilibrium. That equals the
negative of the integral from point A to point B of the dot product of the electric field and the
infinitesimal linear displacement dr. Or we could use the non-dot product version of the integral
with cosine theta instead. You just showed that the electric field is always at a right angle to
the surface, which means the angle between the electric field and the direction of dr along
the surface of the conductor in electrostatic equilibrium always equals 90 degrees. And the
cosine of 90 degrees equals zero, so the electric potential difference between any two points
on the surface of a conductor in electrostatic equilibrium equals zero which is the definition
of an equipotential surface, right mr.p? Exactly Billy. Thanks. In order to determine
the magnitude of the electric field just outside the surface of a conductor in electrostatic
equilibrium, we need to zoom in quite a bit on the surface and draw an infinitesimally small Gaussian
cylinder with its cylindrical axis normal to the surface of the conductor. Bo, please use
Gauss’ law to determine the electric field. Sure. Gauss’ law says electric flux through
a closed surface equals the closed surface integral of the dot product of the electric
field through the closed surface and the area vector, dA. We did something kind of like this
when we figured out the electric field around an infinitely large, thin plane of charges.
The total electric flux equals the electric flux through the side of the Gaussian cylinder
plus the electric flux through the left end plus the electric flux through the right end. The angle
between the area vector, dA, from the side of the Gaussian cylinder and the electric field is
always 90 degrees. The angle between the area vector, dA, from the left end of the Gaussian
cylinder and the electric field is always zero degrees; they are in the same direction.
And the angle between the area vector, dA, from the right end of the Gaussian cylinder and
actually, the electric field inside a conductor in electrostatic equilibrium always equals zero,
so there is zero electric flux through the right end of the Gaussian cylinder.
The electric field is uniform over the left end of the Gaussian cylinder,
which means we can take electric field out from the integral.
Actually, because the Gaussian cylinder is infinitesimally small, even if the electric
field is not uniform, we can still take it out from the integral because it would be uniform over
the infinitesimally small area dA, right mr.p? (That is correct Billy, thanks.
Bo, please keep going.) Sure. The integral of dA for just the left end is
just the area of one end of the Gaussian cylinder. We know surface charge density equals charge over
area. Which means, for this Gaussian cylinder, the local surface charge density equals the
charge inside the Gaussian cylinder over the area of one end of the Gaussian cylinder.
That means the charge inside the Gaussian cylinder equals local surface charge density times the
area of one end. We can substitute that in for the charge inside the Gaussian cylinder and …
Everybody brought the area of the end of the Gaussian cylinder to the party!
And the electric field just outside a conductor in electrostatic equilibrium
equals the surface charge density right at that location over permittivity of free space.
Thank you, Bo. Well done. The last of the four things you need to remember about conductors in
electrostatic equilibrium is that, for an irregular shape, the local surface charge
density is at its maximum where the radius of curvature is at its minimum. In other words,
the largest number of excess charges per area will be where the radius of curvature is the smallest.
To prove this, we have two conducting spheres connected by a long conducting wire with the whole
system in electrostatic equilibrium. This system is a conductor in electrostatic equilibrium.
In other words, when two conductors are brought into contact with one another,
the charges redistribute such that both conductors are at the same electric potential. Please realize
this happens so quickly that the time for this to occur is considered to be negligible.
Oh right. Both conductors have the same electric potential because
all the charges are on the surfaces of the conductors in electrostatic equilibrium
and the surfaces of conductors in electrostatic equilibrium are equipotential surfaces,
so when two charged conductors touch one another, the excess charges redistribute
themselves to make it so both conductors are at the same electric potential.
Exactly Billy. For our example the radius of sphere 2 is smaller than the radius of sphere 1,
and the distance, d, between the two spheres is much, much larger than either radius. Because this
is a conductor in electrostatic equilibrium, the electric potential on the surface of each
sphere has the same value of the Coulomb constant times charge all over the radius of each sphere.
Everybody brought the Coulomb constant to the party!
Sure. We can then rearrange the equation to get the charge on sphere 1
equals the ratio of the radii of the two spheres, 1 over 2, times the charge on
sphere 2. Because radius 2 is less than radius 1, radius 1 divided by radius 2 is greater than 1.
That means the excess charge on sphere 1 is greater than the excess charge on sphere 2.
The magnitude of the electric field at the surface of each sphere equals the
Coulomb constant times charge divided by the square of the radius of the sphere. We can use
those equations to get the ratio of the electric fields at the surface of each sphere, 1 over 2.
We can simplify that equation by multiplying by the multiplicative
inverse of the denominator. The Coulomb constant cancels out on the top and bottom. For charge 1,
we can substitute in radius 1 over radius 2 times charge 2. And now, a lot of variables cancel out.
Charge 2 cancels out. One of the radii for sphere 2 cancels out. And one of the radii for sphere
1 cancels out. And all we are left with is radius 2 over radius 1. That means electric field 2
equals radius 1 over radius 2 times electric field 1. And, again, because radius 2 is smaller than
radius 1, radius 1 over radius 2 is greater than 1. That means the electric field at the surface of
sphere 2 is more than the electric field at the surface of sphere 1. And, because we previously
showed that the electric field at the surface of a conductor in electrostatic equilibrium equals the
local surface charge density over permittivity of free space, that means the surface charge density
of sphere 2 is more than the surface charge density of sphere 1. We have just shown that,
where the radius of curvature of a conductor in electrostatic equilibrium is smaller, there is a
larger surface charge density. In other words, where the local surface charge density is at a
maximum, the radius of curvature of the conductor in electrostatic equilibrium is at a minimum.
That concludes my review of conductors in electrostatic equilibrium for AP Physic C:
Electricity and Magnetism. Next time we are going to review capacitors.
Thank you very much for learning with me today, I enjoy learning with you.