Lecture 3Electric field of continuous charge distributions
August 11, 2015
Electric field of a charged rod
Example
A positive charge Q is uniformly distributed on a rod of length a. What isthe electric field along the rod, a distance b from the right end of thesegment?
Lecture 3 August 11, 2015 2 / 20
Electric field of a charged rod
Solution:Divide the rod into many, many segments. Each segment would be very,very short. Lets call their length dx .
Since one segment is very, very short, it can treated as a point particle,with a very, very small electric field
d ~E = kdq
r2
where dq is that very, very small charge of that very, very small segment.Lecture 3 August 11, 2015 3 / 20
Electric field of a charged rod
But what is the separation r? For a segment at some location x , itsseparation from point P is (a x) + b.
d ~E = kdq
[(a x) + b]2
Now we want to sum up the contributions of all the segments. Weintegrate the above expression over all segments:
~E =
all charges
kdq
[(a x) + b]2 Lecture 3 August 11, 2015 4 / 20
Electric field of a charged rod
We want the integration to be over the length of the rod. We can do thisby using the fact that the linear charge density of the rod is a constant:
dq
dx=
Q
a dq = Q
adx
So we can write the previous integration to
~E =
all charges
kdq
[(a x) + b]2 a0k
(Qa dx
)[(a x) + b]2
Lecture 3 August 11, 2015 5 / 20
Electric field of a charged rod
Performing the integration, we get
~E =kQ
a
(1
b 1
a + b
) .
What happens when b is large?
~E =kQ
a
a
b(:b
a + b)=
kQ
b2i .
The electric field is similar to a point particle.Lecture 3 August 11, 2015 6 / 20
Electric field of a charged rod
Example
A positive charge Q isdistributed uniformlyalong the y -axis betweeny = a and y = +a Findthe electric field at pointP on the x-axis at adistance x from theorigin.
Lecture 3 August 11, 2015 7 / 20
Electric field of a charged rod
Lecture 3 August 11, 2015 8 / 20
Electric field of a charged rod
Solution:Divide the rod into many, many segments. Each segment would be very,very short. Lets call their length dy .
Since one segment is very, very short, it can treated as a point particle,with a very, very small electric field magnitude
dE = kdq
r2
where dq is that very, very small charge of that very, very small segment.
Lecture 3 August 11, 2015 9 / 20
Electric field of a charged rod
But what is the separation r? For asegment at some location y , itsseparation from point P isx2 + y2.
dE = kdq
x2 + y2
Decompose the equation into its x-and y - components:
dEx = kdq
x2 + y2cos
dEy = kdq
x2 + y2sin
Lecture 3 August 11, 2015 10 / 20
Electric field of a charged rod
We can write cos and sin in terms of x and y :
cos =x
r=
xx2 + y2
sin =y
r=
yx2 + y2
Lecture 3 August 11, 2015 11 / 20
Electric field of a charged rod
So
dEx = kdq
x2 + y2cos k x dq
(x2 + y2)3/2
dEy = kdq
x2 + y2sin k y dq
(x2 + y2)3/2
Now we want to sum up the contributions of all the segments. Weintegrate the above expression over all segments:
Ex =
all charges
kx dq
(x2 + y2)3/2
Ey =
all charges
ky dq
(x2 + y2)3/2
Lecture 3 August 11, 2015 12 / 20
Electric field of a charged rod
We want the integration to be over the length of the rod. We can do thisby using the fact that the linear charge density of the rod is a constant:
dq
dy=
Q
2a dq = Q
2ady
So we can write the previous integration to
Ex =
all charges
kx dq
(x2 + y2)3/2
+aa
kx(Q2ady
)(x2 + y2)3/2
Ey =
all charges
ky dq
(x2 + y2)3/2
+aa
ky(Q2ady
)(x2 + y2)3/2
Lecture 3 August 11, 2015 13 / 20
Electric field of a charged rod
Performing the integration, we get
Ex = kQ
xx2 + a2
; Ey = 0 .
What happens when x is large?
Ex = kQ
x
:x2x2 + a2
= kQ
x2
The electric field is similar to a point particle.
Lecture 3 August 11, 2015 14 / 20
Quiz!
A positive charge Q is uniformly distributed around a conducting ring ofradius a. Find the electric field at a point P on the ring axis at a distancex from its center.
Lecture 3 August 11, 2015 15 / 20
Quiz!
Solution:Divide the ring into many, many segments. Each segment would be very,very short. Lets call their length ds.
Since one segment is very, very short, it can treated as a point particle,with a very, very small electric field magnitude
dE = kdq
r2
where dq is that very, very small charge of that very, very small segment.
Lecture 3 August 11, 2015 16 / 20
Quiz!
But what is the separation r?For a segment at somelocation y , its separationfrom point P is
x2 + a2.
dE = kdq
x2 + a2
Lecture 3 August 11, 2015 17 / 20
Quiz!
By symmetry, the net electricfield will lie on the x-axis.Thus we only need tocompute for the x-componentof the contribution of eachsegment:
dEx = kdq
x2 + a2cos
= kdq
x2 + a2x
r
= kdq
x2 + a2x
x2 + a2
= kx dq
(x2 + a2)3/2
Lecture 3 August 11, 2015 18 / 20
Quiz!
Now we want to sum up the contributions of all the segments. Weintegrate the above expression over all segments:
Ex =
all charges
kx dq
(x2 + a2)3/2
We want the integration to be over the length of the ring. We can do thisby using the fact that the linear charge density of the ring is a constant:
dq
ds=
Q
2pia dq = Q
2piads
So we can write the previous integration to
Ex =
2pia0
kx(
Q2piads
)(x2 + a2)3/2
Lecture 3 August 11, 2015 19 / 20
Quiz!
Note that x can be taken out of the integration because it is a constantfor any segment in the ring. So
Ex = kx(
Q2pia
)(x2 + a2)3/2
2pia0
ds
= kx(
Q
2pia)
(x2 + a2)3/2(2pia)
= kxQ
(x2 + a2)3/2.
What happens when x is large?
Ex = kxQ
(:x2
x2 + a2 )3/2= k
Q
x2.
The electric field is similar to a point particle.Lecture 3 August 11, 2015 20 / 20