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Hollow Conductor Field
Name: Paul
Status: student
Age: 16
Location: N/A
Country: N/A
Date: 10/2/2004
Question:
What I do not understand is why the intensity of the
electric field is 0 inside a charged (empty on the inside) metal sphere.
I know that the electrical charge is spread uniformly across its outer
surface ( what exactly does that have to do with it?) and that anywhere
outside E=k*q*q/(r*r), but why, for example,does E(a)=E(b)=0? (A and B
are two points from inside the sphere) Shouldn't the local electric
charge closest to A create a very high intensity field (~/(r^2); r->0 =>
E->oo) in A?; while in point B (I considered point B in the middle of the
sphere), because of the symmetry, it is clear that E=0, thus
E(a)>E(b)=0...?? I am confused; why does Coulomb's Law not apply here as
well? If we were to consider, let us say, 50 electrified particles,
placed in the form of a sphere, would it be the same as if they were on
an empty metal sphere with the same radius? I guess not, but why?
I know it has to do with the sphere being a metal conductor
(electrons moving freely inside and "stuff like that"), but my teacher
simply says: "That is how things are." So how can I understand? I know
"that's how things are", but why?
Replies:
Paul,
Try to picture a point near the top of a charged hollow sphere, on the
inside. It is true that the charges above the location are closer. Each
charge exerts a large force. What you may not realize is that there is a
many more charges below the location. Although these charges are further
away, there are many more charges trying to make the electric field at least
partially upward. Greater charge below balances greater force per charge
above.
Ken Mellendorf
Math, Science, Engineering
Illinois Central College
It is zero because
1) The field goes as 1/r^2.
2) The surface charge on the sphere's surface is uniform, thus charge is
proportional to surface area.
3) The charged regions that oppose each other (i.e., those on opposite
sides of the spherical surface from any given point inside) occupy
areas that increase as the square of the distance from the given point.
Here is a mental picture for the underlying calculus:
Draw a very small circle on the surface of the sphere, and pick any point
inside the sphere. Call the point "P". Let's say the point P is at
distance 'd' from the circle you drew. Now draw lines from every point
on the circle through the point P, and see where those lines intersect
the opposite side of the sphere's surface. The intersection will also
be a very small circle, at a distance we'll call 'D'. (If the intersection
on the other side is not a very small circle, you need to make the first
circle smaller. You're going to have to get infinitesimal.)
If the circle you drew is small enough, its area will be proportional
to d^2 (i.e., we will not have to worry about the fact that the sphere's
surface is curved), and the area of the circle on the other side will be
proportional to D^2.
Now, since charge is uniformly distributed over the spherical surface,
the total charges within the two circles are proportional to their areas.
So the field from the first area is proportional to
charge/d^2 ~ area/d^2 ~ d^d/d^2 = 1.
(I'm using '~' to mean 'proportional to'.)
Similarly, the opposing field from the other side of the sphere is
proportional to D^2/D^2 = 1, so they balance each other exactly.
Note that, as the point A approaches the surface, and 1/d^2 approaches
infinity, the charge within the circle at distance d goes to zero. So
the field contribution from the near-side circle remains finite, and it
continues to be balanced by the field contribution from the far-side
circle.
Tim Mooney
Things are the way they are for very good reasons. For example, the
electric field due to a charge falls off by an inverse square law, as you
state in your email. It is that way because we live in a three dimensional
universe and the area of a sphere in our universe increases proportionally
to the square of its radius (Area = 4*pi*R^2). So a charge at the center of
a mathematical sphere (no other charges and no conductors or other material
nearby) produces a field at the surface of the sphere over an area that
grows like the square of the distance from the charge. Therefore, the
effect per unit area has to decrease like the inverse square of the radius.
The same is true of gravity.
If you draw straight lines, say every 30 degrees, coming from a point charge
and draw circles of various radii centered on that point, you can probably
convince yourself that in a two-dimensional universe (flatland) the electric
field due to a point charge would fall off like 1/R,
As for the field inside a charged conducting sphere, your argument that the
field must be zero at the center because of the symmetry is exactly right
(and shows good understanding). When you consider a point off center,
however, the field is still zero. This is because although the charge on
the near side of the sphere is closer, less of it is effective. To see
this, consider a cone with the apex at the off center point. For each bit
of charge inside the cone on the near side, there is a larger amount of
charge inside the cone on the far side. And since the cross-sectional area
of a cone grows like the square of the distance from the apex, the relative
amounts of charge exactly balances the relative distances. If you sketch a
sphere (start with a circle) and draw some two ended cones, it will be a lot
clearer.
Fifty charges uniformly spaced over the surface of a mathematical sphere
would have almost the same effects, except that the gaps between the charges
will have some minor effects. If you had a million charges, it would act
almost identically. Remember that a charged conducting sphere probably has
more than 1.0E15 electrons uniformly spread over its surface (that is 1
followed by 15 zeros).
Best, Dick Plano, Professor of Physics emeritus, Rutgers University
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