SERWAY 23.14
Three charged particles are at the corners of an equilateral triangle as shown.
(A) Calculate the E-field at the position of the 2.00 μC charge, due to the other two
charges.
This problem is telling you to ignore the 2.00 μC charge, put yourself at the origin, and
determine what the E-field.
In that case, the E-field comes from two contributors, the -4.00 μC charge and the 7.00
μC charge. We’ll determine their contributions separately, and then add them together
to get the net field at the origin.
The 7.00 μC charge is positive, so it generates an E-field that points directly away from
it. At the origin, this would be down-and-to-the-left. Let’s first find the magnitude of
the E-field from the 7.00 μC charge.
EO,7.00C  k e
q7.00C
r2
2
6

N
9 Nm  7.00  10 C

  8.99  10
 2.52  10 5
2 
2
C
C  0.500m

We know this magnitude points down-and-to-the-left. Let’s decompose that magnitude
into its components along the x and y axes.


E x , 7.00C   2.52  10 5



E y , 7.00C   2.52  10 5

N

5 N ˆ
 sin 30iˆ  1.26  10
i
C
C

N

5 N ˆ
 cos30 ˆj   2.18  10
j
C
C

Thus:

N 
N

EO , 7.00C  1.26  10 5 iˆ   2.18  10 5  ˆj
C 
C

Luckily, finding the field for the -4.00 μC charge is easier. Now we’re dealing with a
negative charge, so it will generate a ‘pulling’ E-field at the origin. Since the -4.00 μC
charge is located on the x axis, its field only points in the positive x direction.
Finding the magnitude of the E-field due to the -4.00 μC charge:
EO , 4.00C  k e
q  4.00C
r2

Nm 2  4.00  10 6 C
N

  8.99  10 9
 1.44  10 5
2 
2
C
C  0.500m 

We know this must point in the positive x direction, so:

N

EO , 4.00C  1.44  10 5 iˆ
C

Now, the net field at the origin will be the algebraic sum of the two contributing fields:



N 
N 
N

EO  EO , 7.00C  EO , 4.00C  1.26  10 5 iˆ   2.18  10 5  ˆj  1.44  10 5 iˆ
C 
C 
C


N 
N

EO   0.18  10 5 iˆ   2.18  10 5  ˆj
C 
C

(B) Use the answer for part (A) to determine the force the 2.00 μC charge will experience
at the origin.
This is easy, it just goes back to the definition of the electric field:

 Fe
E
q
Therefore, since we know the E-field at the origin, and we know how much charge the
2.00 μC charge has, we can rearrange the definition of the electric field to solve for
force:



N 
N 
Fe  qE  2.00  10 6 C   0.18  10 5 iˆ   2.18  10 5  ˆj 
C 
C 


Fe  0.036 N iˆ  0.436N  ˆj

