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Applications of Gausss law
Electric potential
Center for Advanced Mathematics and Physics-NUST, Islamabad.
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Center for Advanced Mathematics and Physics-NUST, Islamabad.
Review of Gauss's Law
The total electric flux leaving a closed surface is equal to the
charge enclosed by the surface divided by 0. We can express this
directly in terms of the
mathematics we have learned,
Applying Gausss law to a sphere containing charge q:
- If the surface is sphere then the angle between Electric field
and vector area
would be zero
and also the electric field is constant over the surface
Thus,
This is Coulombs law
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Applying Gauss law: (Cylindrical symmetry)
Consider an infinity long cylindrical plastic rod with a uniform
positive linear
Charge density .
We want to determine the electric field
at a distance r from the axis of rod
In this case Gaussian surface should match the symmetry of
cylindrical shape
Thus, at every point on the cylindrical part of the Gaussian
surface,
must have
the same magnitude E and (for positively charged rod) must be
directed
radially outward.
(i)
Here
There is no flux through the end caps because
being radially directed.
and
Center for Advanced Mathematics and Physics-NUST, Islamabad.
+
+
+
+
+
+
+
+
+
+
r
h
Gaussian
surface
2r
-
So according to the Gauss law (or from equation (i))
which yields
This is electric field due to an infinitely long, straight line
of charge, at a point
that is radial distance r from the line.
The direction of
Is radially outward from the line of charge it charge is
positive
and radially inward if it is negative.
Sample problem: 24.5
Center for Advanced Mathematics and Physics-NUST, Islamabad.
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Electric potential
When a test charge q0 is placed in an electric field E created
by some other
charged object, the electric force acting on the test charge is
q0E. (If the field is produced by more than one charged object,
this force acting on the test charge is the vector sum of the
individual forces exerted on it by the various other charged
objects.) The force q0E is conservative because the individual
forces described by Coulombs law are conservative. When the test
charge is moved in the field by some external agent, the work done
by the field on the charge is equal to the negative of the work
done by the external agent causing the displacement.
For an infinitesimal displacement ds, the work done by the
electric field on the charge is
As this amount of work is done by the field, the potential
energy of the chargefield system is decreased by an amount
Center for Advanced Mathematics and Physics-NUST, Islamabad.
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Electric potential energy
where W is the work done by the static electric field.
For convenience we often define
Where is the work done by the field to move the charge particle
from infinity to its current position and .
If the system changes it configuration from an initial state I
to final state f,
The electrostatic force does work on the particles.
For conservative force, in general,
Center for Advanced Mathematics and Physics-NUST, Islamabad.
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In general we have
Electric potential energy
Center for Advanced Mathematics and Physics-NUST, Islamabad.
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Equipotential surfaces
Surfaces in space on which V is constant.
Center for Advanced Mathematics and Physics-NUST, Islamabad.
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Equipotential surfaces
If V is chosen to be the same for all adjacent equipotential
surfaces, then the electric filed is inversely proportional to the
separations of the equipotential surfaces.
Center for Advanced Mathematics and Physics-NUST, Islamabad.
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Calculating the potential from the field
or
Vi can be assigned to any convenient value such as 0.
i
f
F
path
Electric field
qo
ds
Center for Advanced Mathematics and Physics-NUST, Islamabad.
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Potential due to a point charge
P
qo
ds
ds
r
R
Now we set Vf = 0 (at infinity) and Vi =V (at R)
Thus use
Center for Advanced Mathematics and Physics-NUST, Islamabad.
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Potential due to a point charge
Center for Advanced Mathematics and Physics-NUST, Islamabad.
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Potential due to an electric dipole
Center for Advanced Mathematics and Physics-NUST, Islamabad.
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Potential due to a group of point charges
Example
Center for Advanced Mathematics and Physics-NUST, Islamabad.
2
2
4
1
)
4
(
r
q
E
q
r
E
q
dA
E
o
o
o
enclosed
pe
e
p
e
f
=
=
=
=
E
o
enclosed
q
A
d
E
e
f
=
=
.
E
h
q
enclosed
l
=
o
enclosed
q
A
d
E
e
=
.
)
2
(
rh
E
dA
E
p
=
fi
UUUW
D=-=-
r
E
o
pe
l
2
=
o
h
rh
E
e
l
p
=
)
2
(
ds
E
q
ds
F
o
.
.
=
1
0
1
4
nn
i
i
ii
i
q
VV
r
pe
=
==
ds
E
q
dU
o
.
-
=
f
UUW
==-
W
0
U
=
fi
UUUqEs
D=-=-
r
r
/
VUqEs
DD=-
r
r
f
i
s
s
VEds
D-
r
r
r
r
f
i
s
fi
s
VVVEds
D=-=-
r
r
r
r
f
i
s
fi
s
VVEds
=-
r
r
r
r
s
d
F
dW
.
=
V
o
q
dW
D
-
=
s
d
E
q
dW
o
.
=
ds
E
s
d
E
q
cos
.
=
-
=
-
R
i
f
Edr
V
V
2
4
1
r
q
E
o
pe
=
0
1
()
4
q
Vr
r
pe
=
R
q
V
r
q
dr
r
q
V
o
R
o
R
o
pe
pe
pe
4
1
1
4
1
4
0
2
=
=
-
=
-
