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