SERWAY 23.27
A uniformly charged insulating rod of length 14.0 cm is bent into the shape of a
semicircle as shown in the figure. The rod has a total charge of -7.50 μC. Find the
magnitude and direction of the electric field at O, the center of the semicircle.
At this point, you’re probably thinking “WTF.” Don’t worry, it’s not that hard.
Let’s talk direction first. Put yourself at point O. Look at the blue semicircle. Instead of
thinking about the semicircle as a continuous spread of negative charge, think of it as a
bunch of individual point charges, evenly spaced along the semicircle.
The great thing here is that we have a symmetrical situation. From the vantage point of
O, there are just as many negative point charges above as below. This means that the y
components of all contributing E-fields will cancel out at point O. We only need to
worry about the x components.
Keeping in mind the charge distribution is negative, the E-field at point O will point to
the left, along the negative i-hat direction.
So, we’ve got the directionality of this problem solved. Now, let’s tackle the magnitude.
Let’s begin by looking at the fundamental equation we always use to calculate the Efield of a continuous charge distribution. It looks like:
dq
r2
In this equation, dq represents a little tiny amount of negative charge, at some arbitrary
location on the semicircle. Since we must perform an integral with respect to dq, we
need to first convert it to something that has spatial meaning. How do we do this?
E  ke 
First, let’s figure out the linear charge density on this semicircle. Since there is a total
amount of charge -7.50 μC spread evenly across the semicircle, and since the semicircle
has a total length of 14 cm, then the linear charge density on this rod is:

Q  7.50C
C

 5.36  10 5
L
0.14m
m
That little tiny amount of charge, dq, resides in a little tiny amount of arc length on that
semicircle. Remember the definition of arc length from last semester? (ds = rdθ) We
can relate the length of that little arc to the linear charge density to re-write our variable
dq: dq  ds  rd , where r stands for the radius of the circle, and dθ stands for the
little angle subtended by that little arc-length.
Now we’ve parameterized dq in terms of spatial variables – something over which we
can integrate!
So, now, if we consider the little amount of electric field generated by the little charge
dq (the red vector in the picture), we can write its magnitude as:
dE  k e
dq
rd
 ke 2
2
r
r
But, remember, we only care about the x component of this field, so we must use some
trigonometry to decompose the red vector into its component along the x axis (the teal
vector).
dE x  k e
rd
r
2
sin   k e
d
r
sin 
Now, to get the overall field at O, we must integrate this expression with respect to θ.
ke  
sin d
r 0

ke 
k
k
2k 
Ex  
cos    e cos   cos0   e  1  1  e
r
r
r
r
0
2k 
Ex  e
r
E x   dE x 
Okay, great! This is the magnitude of the electric field. Now, we need to know the
radius of the semicircle, which is easy to find:
0.14m    r
r  0.045m
So, the magnitude of the E-field at O is:


C

2 8.99  10 9  5.36  10 5 
N
m

Ex 
 2.14  10 7
0.045m
C
Remember, we already figured out that this field points only in the negative x direction,
so the final answer is:

N

EO   2.14  10 7 iˆ
C
