Physics 212 - Spring 2000
Recitation Activity 3: Electric Fields from Charge Distributions
NAME: Solution
ACTIVITY PARTNERS:
INSTRUCTOR: Redwing
REC. SECTION: _______
____________________________________
____________________________________
DATE: January 19, 2000
This activity is based on the following concepts:
 Electric field is defined as the electrostatic force on +1 C of charge; note that it is a
vector and that it is measured in units of N/C.
 To calculate the electric field from many charges, we use SUPERPOSITION:
 If we have a discrete collection of point charges, figure out the electric field
vector from each charge using Coulomb's Law and then add all the vectors.
 If we have a continuous distribution of charge, we divide up the distribution into
"differential" elements of charge, figure out the electric field from a typical
element and then use an integral to sum up all such vectors.
Exercise 1: Electric field from point charges.
The figure below shows 4 point charges located on a circle of radius R. A small test
charge - q and mass m is released from rest at the center of the circle. You are asked to
determine the acceleration of this particle as soon as it is released. Do the problem in the
following steps.
y
+Q
(a) In the figure, sketch four vectors that
represent the contributions of each of the 4
E-2Q
charges to the electric field at the center of
the circle. See figure
(b) What is the sum of the x-components of
-q
+Q
these 4 vectors?
2
2
2
kQ/R + k(2Q)/R = 3kQ/R
(c) What is the sum of the y-components of
E+Q
-2Q
these 4 vectors?
E+Q
-kQ/R2 - k(2Q)/R2 = -3kQ/R2
(d) Next, use the definition of electric field to
E-2Q
determine the magnitude of the total force F
-2Q
on the test charge - q and the acceleration a
of the test charge when it is released.
3kQ
3kqQ
3kQ
ˆ)
F  q E  (q ) 2 ( xˆ  yˆ ) 
( xˆ  yˆ )
E 
( xˆ  y
2
R
R
R2
F 3kqQ
3kqQ
(
2
)
a


( 2)
R2
m
mR2
e) Finally, in the figure sketch the path followed by the test charge after it is released.
See dotted arrow in figure
F 
x
Exercise 2: Electric Field from a continuous charge distribution
Two curved plastic rods, one of charge +q and the other charge -2q, form a circle of
radius R in an x-y plane as shown below. The charge is distributed uniformly on both
rods. Determine the magnitude and direction of the electric field E at the center of the
circle, proceeding in the steps given below.
y
+q
x
-2q
R
(a) Use a superposition argument to transform this problem into a simpler one that
involves only one semi-circle of charge. Circle the case shown below that gives the
same physical situation as the original problem.
-q
-q
-q
+q
-3q
Briefly justify your choice:
The +q semi-circle in quadrant I and II, has same effect as a –q semi-circle in
quadrant III and IV. Thus, -q added to –2q semi-circle results in net –3q.
(b) Hopefully, you figured out that the "-3 q" semi-circle was the correct one! Choose an
appropriate differential element of charge (call it "dq") on the arc and write down an
expression for the magnitude of the electric field |dE| from this element.
 = -3q/R
dq
ds
dE  k 2  k 2
R
R
(c) Use a SYMMETRY argument to argue the DIRECTION of the electric field from the
entire semi-circle.
Show your argument using a sketch and some brief
sentences.
dEx
dEx
Look at symmetric elements ds and ds’ and draw the
resulting differential vectors dE for those elements.
Breaking the differential vectors dE into x- and y- axis
components, we see that the dEx of the two elements
 
cancel and the dEy components add and are directed
downward.
ds
dEy
(d) Finally, integrate the expression that you obtained above in (b) and determine the
MAGNITUDE of the total electric field. Caution: Remember that your integral
should add appropriate COMPONENTS not the total magnitudes of the field from
each differential element!
Ignore the signs and put in proper sign of expression at the end.
dE y  dE sin   k
dE y 
k
sin d
R
ds
R2
sin 
k
k
E y   dE y   sin d 
R
R
0
 6kq
 3q
E
yˆ 
yˆ
2
R
2 2  0 R 2
180
ds = R d
k
2k
180
0 sin d  R cos 0  R 
180
3q
R  6kq
R
R 2
2k
Symmetric
element ds’