Elektrostatika: Hukum Coulomb

Electrostatic:
Electric Field
{
Chapter 22 Halliday-Resnick 9th Ed.
Tuesday, 31 January 2012
Hint:
0. Create a sketch (diagram)
1.
Define the charge distribution, and the corresponding
spatial element of distribution ( , , , etc.)
2.
Relate the charge
with the spatial element
1.
3.
4.
5.
2.
=
=
for 1D
for 2D
Describe the expression for
on the point of interest,
substitute any possible variables with regards to point 1 &
2.
Complete the sketch by drawing the direction of , check
for symmetry and any possible cancelling components
(that needs no further attention).
Solve the integral, if the result is to be stated in total
charge Q of the distribution, replace od the with the
corresponding charge density distribution.
( )
( )
=
=
≫
( )
=
=
→
2
2
/2
+
2
=
+
≈
2
=
+
( )
/2
⁄
+
+
⁄
=
−̂
=
=
=
=
=
= / sin
=
= / cos
= / tan
=
=
tan
cos
=
=
/
cos
sin
cos
=−
=−
sin
/2
cos |
0
=
=
=
/
cos
cos
=
=
=
cos
cos
/2
sin |
− /2
Electrostatic:
Gauss’ Law
{
Chapter 23 Halliday-Resnick 9th Ed.
Tuesday, 30 January 2012
Physics: Solving seemingly complex problem →
using symmetry
Gauss’ Law:


Imaginary surface enclosing charge distribution
Relates the electric fields at points an a closed
Gaussian surface to the net charge enclosed by the
surface
Flux of Electric Field:
The electric flux through a Gaussian surface is
proportional to the net number of electric field lines
passing through that surface.
Flux of Electric Field:



For a uniform electric field
Φ= ∙
Non-uniform electric field
Φ= ∙Δ
element of area
Over a closed surface
Φ=∮
∙
Through a surface
Number of electric field
?
Flux of Electric Field:



For a uniform electric field
Φ= ∙
Non-uniform electric field
Φ= ∙Δ
element of area
Over a closed surface
Φ=∮
∙
Through a surface
Number of electric field
The figure here shows a Gaussian
cube of face area immersed in a
uniform electric field that has
the positive direction of the
axis. In terms of E and A, what is
the flux through
(a)
the front face (which is in the
plane),
(b)
The rear face,
(c)
the top face, and
(d)
the whole cube?
(a)
(b)
(c)
(d)
Φ=
Φ=
Φ=
0 (?)
∙
∙
∙
=
=
=
∙
∙
∙
cos
cos
cos
=
=
=
cos 0° =
cos 180° = −
cos 90° = 0
Cylindrical surface
Flux through a closed cylinder, uniform field ?
Flux through a closed cube,
nonuniform field
What is the total flux through the cubical
surface if the Electrical Field passing through
the surface is described as =
̂+4 ̂?
Gauss’ Law:

Relates the electric fields at points an a closed
Gaussian surface to the net charge enclosed by
the surface
Φ=
∙
=
Gauss' Law and Coulomb's Law
=
=
1
?
Coulomb's
Law Approach
=
=
=
/
cos
cos
=
=
=
cos
cos
/2
sin |
− /2
Gauss’
Law Approach
=
=
?
0. Draw the diagram
1. Select a closed surface
2. Solve ∮ ∙
3. Solve ∮
∙
=