Download Electric Field of a point charge

Electric Field of a point charge
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Electric Field lines of a point charge
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Electric Dipoles
An electric dipole consists of two charges
Q, equal in magnitude and opposite in
sign, separated by a distance ℓ. The
dipole moment, 𝒑 ≡ 𝑸ℓ, points from the
negative to the positive charge.
Electric Dipoles
An electric dipole in a uniform electric field
will experience no net force, but it will, in
general, experience a torque:
Electric Dipoles
The electric field created by a dipole is the sum
of the fields created by the two charges; far from
the dipole, the field shows a 1/r3 dependence:
Extra Credit Problem
If the earth's electric field is 100 N/C downward,
what must be the charge on a 1-kg object so that
it would be "weightless"?
A) 1.0 C
B) 0.22 C
C) -9.8 ×10-2 C
D) 9.8 C
E) 9.8 × 10-2 C
Gauss’s Law
Electric Flux
Electric flux:
ΦE = 𝐸⊥ 𝐴 = 𝐸𝐴⊥ = 𝐸𝐴𝑐𝑜𝑠𝜃
ΦE = 𝐸 ∙ 𝐴
Uniform E-field
Electric flux through an area
is proportional to the total
number of field lines crossing
the area.
Calculate the electric flux through the
rectangle shown. The rectangle is 10 cm
by 20 cm, the electric field is uniform at
200 N/C, and the angle θ is 30°.
a)
b)
c)
d)
e)
𝒎𝟐
𝟐𝑵∙
𝑪
𝟐
𝒎
𝟐𝟎 𝑵 ∙
𝑪
𝟐
𝒎
𝟑. 𝟓 𝑵 ∙
𝑪
𝟐
𝒎
𝟏. 𝟕 𝑵 ∙
𝑪
𝑵𝒐𝒏𝒆 𝒂𝒓𝒆
𝒄𝒐𝒓𝒓𝒆𝒄𝒕
Electric Flux
Uniform E-field:
ΦE = 𝐸⊥ 𝐴 = 𝐸𝐴⊥ = 𝐸𝐴𝑐𝑜𝑠𝜃
ΦE = 𝐸 ∙ 𝐴
Arbitrary E-field:
𝚽𝐄 =
𝑬 ∙ 𝒅𝑨
Flux through a closed surface:
Gauss’s Law
The net number of field lines through the
surface is proportional to the charge
enclosed, and also to the flux, giving
Gauss’s law:
This can be used to find the electric field
in situations with a high degree of
symmetry.
Example
For a point charge,
Therefore,
Solving for E gives the
result we expect from
Coulomb’s law:
Using Coulomb’s law to
evaluate the integral of the
field of a point charge over
the surface of a sphere
surrounding the charge
gives:
𝑬 ∙ 𝒅𝑨 =
𝑸
𝑸
𝟏 𝑸
𝟐
𝟒𝝅𝒓 =
𝒅𝑨 =
𝟐
𝟐
𝟒𝝅𝝐𝟎 𝒓
𝝐𝟎
𝟒𝝅𝝐𝒐 𝒓
Looking at the arbitrarily shaped surface A2,
we see that the same flux passes through it
as passes through A1. Therefore, this result
should be valid for any closed surface.
Finally, if a gaussian surface encloses
several point charges, the superposition
principle shows that:
Therefore, Gauss’s law is valid for any
charge distribution. Note, however, that it
only refers to the field due to charges
within the gaussian surface – charges
outside the surface will also create fields.
Consider the two gaussian surfaces, A1 and A2,
as shown. The only charge present is the charge
Q at the center of surface A1. What is the net flux
through A2?
a) 0
b) Not 0
c) Cannot tell with given information.
A thin spherical shell of
radius r0 possesses a
total net charge Q that
is uniformly distributed
on it. Determine the
electric field at points
(a) outside the shell,
and (b) within the shell.
(c) What if the
conductor were a solid
sphere?
Solid sphere of
charge.
An electric charge Q
is distributed
uniformly throughout
a non-conducting
sphere of radius r0.
Determine the electric
field (a) outside the
sphere
(r > r0) and (b) inside
the sphere (r < r0).