ABOUT THAT LAST QUIZ
Calculating the Electric Field
February 6, 2016
Three point charges have equal magnitudes, two being positive and one negative. These charges
are fixed to the corners of an equilateral triangle, as the drawing shows. The magnitude of each
of the charges is 5.0 µC, and the lengths of the sides of the triangle are 4.0 cm. Calculate the
magnitude of the Electric Field at point B.
It is suggested that you read the textbook on this topic as you consider the following material.
The Electric Field is a critical concept in this course as will the Magnetic Field later in the
semester. So the concept of a “field” becomes critical from a conceptual point of view.
There are two ways to consider the forces that act on charge B. The first and seemingly the
easiest is to simply recognize that there are two (vector) forces acting on charge B; one from
charge A and one from charge C. To calculate the force on B in this way we simply use
Coulomb’s law twice.
Coulomb’s Law:
Fk
QC QB
QAQB
r

k
rCBUnit
AB

Unit
2
2
,
rAB
rCB
where the r vectors are unit vectors pointing from each of the charges pushing or pulling on
charge B to the charge B itself (C and A). The units of F are Newtons. This is needed to keep the
equation indicating a vector sum. Using the Greek letter sigma () to indicate summation, this
is often written as:
F
 k QrQ
i
i
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B
2
iB
riB
,
where the i ranges over all of the charges contributing to the force on charge B.
Notice that in the material above, the charge on B is carried throughout the calculation.
The other approach is via the Electric Field concept where we remove the charge B entirely from
the picture and calculate the effect that the other charges have on the space where B was located.
After this, we return the charge to the point to calculate the force on it. The beauty of this
approach is that we can place ANY charge at point B and calculate the force on it without going
through the whole calculation again. The Electric Field at a point (B) caused by the other
charges is the same as the force that would be on a one coulomb charge placed at that point but
watch the units.
So,
Ek
QC
QA
r

k
rCB Unit
AB

Unit
2
2
rAB
rCB
We have replaced QB by a 1 (but we omit the unit of charge at this point and will return it later).
The units of E are N/C while the units of F are simply N.
We can now calculate the Electric Field at B in two ways.
FORCE
ELECTRIC FIELD
As we did in class, each force vector is along
one of the sides of the triangle and all of the
angles that we need to deal with are 60 degrees.
The force from A is therefore the same as the
force from C since both charges are the same.
Call this force simply F Since the two vertical
components cancel out, the net force to the left
The calculation for E is similar but here there
is no charge at point B. As before, the angles
of importance are 60 degrees. At point B, the
electric field E caused by charge A has
magnitude
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is given by
Fnet  2F cos(600 )  2  0.5F  F
And F is given by
F k
QAQB
2
rAB
9  109  5  106  5  10 6
F
N
(0.04) 2
F  141N
Ek
QA
2
rAB
9  109  5  106
E
N
2
(0.04)
N
E  2.81x107
C
Again, with the same geometry, there is an E
The Electric field is also to the left and is equal along both lines connecting point B to the
sources of the field at A and C. Both
to the force divided by the charge or
components are Ecos(600) and there are two
of them (as before) so the net field to the left
F
141N
7 N
E


2.8
x
10
is
6
QB
5 x10 C
C
Enet=E=2.81x107 N/C
Just to check,
F=qE
The force on ANY charge at point B is then
simply F=qEnet and this provides the same
result obtained in the other column.
Or
F=(5x10-5)x(2.8x107)=141N
As expected.
The equation for E is therefore:
E

i
k
Qi
riB
riB2
At this point you may be wondering why we need to mess with E when F seems sufficient.
There are many reasons for introducing the Electric field. It exists at a point in space no matter
what charges are producing it and in an electromagnetic wave (light, radio) travels through space
with no charges in the neighborhood but the E field travels along anyway. It is a real, physical
quantity which is far more fundamental than F.
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The Electric field can also be calculated using the same equation for points at which there is NO
charge. Just add the kQ/d2 terms (as vectors) using all of the charges in the neighborhood.
I hope that this helps.
JBB
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