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Physics 54
Electrostatics 2
The sooner you fall behind the more time you have to catch up.
— Anonymous
Calculating E from Gauss’s law
There are a few charge distributions having sufFcient symmetry that one can use
Gauss's law to calculate the EFeld. The trick is to choose the closed “Gaussian surface”
through which the ±ux is calculated in such a way as to exploit the symmetry, making
E
⋅
d
A
simple enough that
E
can be taken outside the integral. The total ±ux then
becomes a simple product of
E
and a surface area.
Spherical symmetry
. In these cases there is a point, called the “center of symmetry”
about which the physical situation is the same in all directions. These are very
important cases, among other reasons because isolated atoms have this symmetry.
A charge distribution is spherically symmetric if its volume charge density
ρ
depends
only on the
distance
from the center of symmetry but not on the direction. The EFeld
produced by this charge re±ects the symmetry: its direction is
radial
relative to the
center of symmetry — outward/inward for positive/negative net charge — and its
magnitude
E
depends only on the
distance
from that center.
To Fnd the Feld
at distance
r
one evaluates the ±ux through a Gaussian surface which is
a sphere of radius
r
about the center of symmetry. Then, since
d
A
is radially outward,
E
⋅
d
A
=
E
r
dA
, where
E
r
is the
outward
radial component of the Feld;
E
r
is
positive
if the
Feld lines go
outward
through the surface. By the symmetry
E
r
also has the same value
at all points on the spherical surface, so the ±ux will be
E
⋅
d
A
∫
=
E
r
dA
∫
=
E
r
⋅
4
π
r
2
.
On the right side of Gauss's law we have 4
π
k
times the net amount of charge
inside
the
sphere of radius
r
. Call this amount of charge
Q
(
r
). (This may or may not be the
total
charge of the distribution.) Then from Gauss’s law
E
r
⋅
4
r
2
=
4
k
⋅
Q
(
r
)
, or
E
r
(
r
)
=
k
Q
(
r
)
r
2
.
If
Q
(
r
) is negative, this gives a negative answer, showing that
E
is directed
inward
.
PHY 54
1
Electrostatics 2
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View Full DocumentNote that this formula for
E
r
has exactly the same form as that for a point charge of
amount
Q
(
r
)
placed at the center of the distribution. This shows an important and very
useful fact about spherically symmetric distributions:
The Efeld at distance
r
From the center oF a spherically symmetric distribution is the
same as that oF a point charge located at the center, with charge
Q
(
r
) equal to the total
charge contained in a sphere oF radius
r
about the center.
If the Feld point is
outside the entire charge distribution
, then
Q
(
r
) will be simply the total
charge of the distribution. This means that the EFeld
outside
any spherical charge
distribution is that of the total charge, treated as a point charge at the center.
These conclusions allow for easy calculation of the EFeld for any spherically symmetric
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 Summer '09
 Thomas
 Physics, Charge, Electrostatics

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