The Electric Field
Due to a Continuous Charge Distribution
(worked examples)
finite line of charge
ring of charge
disc of charge
infinite sheet of charge
infinite line of charge
semicircle of charge
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Electric Field - Engineering Physics - Lecture Slides, Slides of Engineering Physics
This course is designed for engineers. This subject is compiled of physical applications and concepts. This lecture includes: Electric Field, Continuous Charge Distribution, Ring of Charge, Disc of Charge, Infinite Sheet of Charge, Infinite
Typology: Slides
2012/2013
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The Electric Field
Due to a Continuous Charge Distribution
(worked examples)
finite line of charge
ring of charge
disc of charge
infinite sheet of charge
infinite line of charge
semicircle of charge
Instead of talking about the next 6 slides, I am going to set up
(but not solve) one of the following two examples (you ought to
try the other for yourself). Then I will work a 2nd^ example all
the way through. The next 6 slides (which I will skip) show the
general principles I am applying.
Example: A rod of length L has a uniformly distributed total
positive charge Q. Calculate the electric field at a point P
located a distance d below the rod, along an axis through the
left end of the rod and perpendicular to the rod.
Example: A rod of length L has a uniformly distributed total
negative charge -Q. Calculate the electric field at a point P
located a distance d below the rod, along an axis through the
center of and perpendicular to the rod.
Skip to next example.
Think of a line of charge as a collection of very very tiny point
charges all lined up. The ―official‖ term for ―very very tiny‖ is
―infinitesimal.‖
So we get the electric field for the line of charge by adding the
electric fields for all the infinitesimal point charges.
What happens in calculus when you add infinitesimals? Really
really tiny infinitesimals?
- + + + + + +
Yes, I know an infinitesimal is already really really tiny, so the reallys and tinys in ―really really tiny infinitesimals‖ are redundant.
The electric field due to a small "chunk" q of charge is
2 0
1 q E = r 4 πε r
The electric field due to collection of "chunks" of charge is
i i 2 i i 0 i i
1 q E = E = r 4 πε r
unit vector from q to wherever you want to calculate E
unit vector from qi to wherever you want to calculate E
As qdq0, the sum becomes an integral.
The electric field at point P due to the charge dq is
x
dq
P
2 2 0 0
1 dq 1 dx dE = r' = r' 4 πε r' 4πε r'
r’
r'
dE
I’m assuming positively charged objects in these ―distribution of charges‖ illustrations.
Absolute value signs not needed around dq because this is a vector equation—the sign is important!
The electric field at P due to the entire line of charge is
2 0
1 λ(x) dx E = r'. 4 πε r'
The integration is carried out over the entire length of the line, which need
not be straight. Also, could be a function of position, and can be taken
outside the integral only if the charge distribution is uniform.
x
dq
P
r’
r'
E
P x
y
d L
The electric field points away from the rod. The electric field on
the axis of the rod has no y-component (why?). dE from the
charge on an infinitesimal length dx of rod is
dE
x dx^ dQ =^ ^ dx
2 2
dq dx dE = k k x x
Note: dE is in the –x direction. dE is the magnitude of dE. I’ve
used the fact that Q>0 (so dq>0) to eliminate the absolute
value signs in the starting equation.
This is a ―legal‖ starting equation (for positive dq). In fact, this is the best way to start a problem like this one.
If I don’t know the sign of dq, or if it is negative, I include the absolute value signs.
P x
y
d L
dE
x dx^ dQ =^ ^ dx
d L d+L d+L d+L
d d 2 d 2 d
dx (^) ˆ dx (^) ˆ 1 ˆ E = dE = -k i = -k i = -k i x x x
(^) (^)
(^1 1) ˆ d d L (^) ˆ L (^) ˆ kQ ˆ E = -k i = -k i= -k i= - i d L d d d L d d L d d L
^ ^ ^ ^ ^ (^) (^) ^ ^ ^ ^
This approach, where I do the whole problem all at once using unit vector notation, is not recommended for beginners. I suggest you do the x and y components separately, like I will do in lecture.
If charge is distributed over a two-dimensional surface, the
amount of charge dq on an infinitesimal piece of the surface is
dS, where is the surface density of charge ( = charge/area).
x
y
area = dS
charge dq = dS
I’m assuming positively charged objects in these ―distribution of charges‖ illustrations.
dE
The electric field at P due to the charge dq is
2 2 0 0
1 dq 1 dS dE = r' = r' 4 πε r' 4πε r'
x
y
P
r’
r'
I’m assuming positively charged objects in these ―distribution of charges‖ illustrations.
x
z P
r’
r'
2 0 V
1 (x, y, z) dV E = r'. 4 πε r'
After you have seen the previous slides, I hope you believe that
the net electric field at P due to a three-dimensional distribution
of charge is…
y
E
2
dq use dE = k r
and work with one component at a timedocsity.com
Summarizing:
2 0
1 λ dx E = r'. 4 πε r'
2 (^0) S
1 dS E = r'. 4 πε r'
2 0 V
1 dV E = r'. 4 πε r'
Charge distributed along a line:
Charge distributed over a surface:
Charge distributed inside a volume:
If the charge distribution is uniform, then , , and can be taken outside
the integrals.
Because it takes multiple repetitions for some of us (like me) to
get the message, to calculate the electric field of a charge
distribution:
2
dq use dE = k r
and work with one component at a time
This is a ―legal‖ variation of a starting equation!